Parallel accounts (Day 2)
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Challenge Taken with permission from my University Code Challenge Contest
After finishing her studies a couple of months ago, Marie opened a bank account to start receiving the payment of her first job in town. Since then she has been performing a few transactions with it. Her first payment was $1000 dollars. With that money she paid for a dinner in which she invited her parents (The dinner cost $150 dollars), then, she did a purchase in a well-known supermarket ($80 dollars) and a hotel reservation for her vacations ($200). At the end of the month she received her payment again (1040 dollars, a little more than the previous month) and the day after she spent another $70 dollars at the supermarket.
Today, she realized that if after paying the first $80 dollars in the supermarket a second account had been created and the first one frozen, both accounts would have exactly the same balance:
$$ underbrace{1000quad -150quad -80}_{Total=770}quad underbrace{-200quad 1040quad -70}_{Total=770} $$
The event was so rare to her that she wants to continue ascertaining if the movements of her account and those of her friends have also this feature or not.
Challenge
Given a list of transactions, output the number of instants of time in which the owner of the bank account could have created a second account so that both had the same final balance.
Example: [1000, -150, -80, -200, 1040, -70]
$$ color{red}{1)quadunderbrace{}_{Total=0}quad underbrace{1000quad -150quad -80quad -200quad 1040quad -70}_{Total=1540}} $$
$$ color{red}{2)quadunderbrace{1000}_{Total=1000}quad underbrace{-150quad -80quad -200quad 1040quad -70}_{Total=540}} $$
$$ color{red}{3)quadunderbrace{1000quad -150}_{Total=850}quad underbrace{-80quad -200quad 1040quad -70}_{Total=690}} $$
$$ color{green}{4)quadunderbrace{1000quad -150quad -80}_{Total=770}quad underbrace{-200quad 1040quad -70}_{Total=770}} $$
$$ color{red}{5)quadunderbrace{1000quad -150quad -80quad-200}_{Total=570}quad underbrace{ 1040quad -70}_{Total=970}} $$
$$ color{red}{6)quadunderbrace{1000quad -150quad -80quad -200quad 1040}_{Total=1610}quad underbrace{-70}_{Total=-70}} $$
$$ color{red}{7)quadunderbrace{1000quad -150quad -80quad-200quad 1040quad -70}_{Total=1540}quad underbrace{}_{Total=0}} $$
Test Case
- Input:
1000 -150 -80 -200 1040 -70Output:1
- Input:
100 -100Output:2
- Input:
1 2 3Output:1
- Input:
10 -20 15Output:0
- Input:
15 -15 15 -15Output:3
- Input:
1Output:0
Notes
- You can assume there wont be any transaction of $0 dollars
- You can take input in any reasonable way
code-golf
asked Feb 7 at 14:45
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,69721464
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add a comment |
$begingroup$
Challenge Taken with permission from my University Code Challenge Contest
After finishing her studies a couple of months ago, Marie opened a bank account to start receiving the payment of her first job in town. Since then she has been performing a few transactions with it. Her first payment was $1000 dollars. With that money she paid for a dinner in which she invited her parents (The dinner cost $150 dollars), then, she did a purchase in a well-known supermarket ($80 dollars) and a hotel reservation for her vacations ($200). At the end of the month she received her payment again (1040 dollars, a little more than the previous month) and the day after she spent another $70 dollars at the supermarket.
Today, she realized that if after paying the first $80 dollars in the supermarket a second account had been created and the first one frozen, both accounts would have exactly the same balance:
$$ underbrace{1000quad -150quad -80}_{Total=770}quad underbrace{-200quad 1040quad -70}_{Total=770} $$
The event was so rare to her that she wants to continue ascertaining if the movements of her account and those of her friends have also this feature or not.
Challenge
Given a list of transactions, output the number of instants of time in which the owner of the bank account could have created a second account so that both had the same final balance.
Example: [1000, -150, -80, -200, 1040, -70]
$$ color{red}{1)quadunderbrace{}_{Total=0}quad underbrace{1000quad -150quad -80quad -200quad 1040quad -70}_{Total=1540}} $$
$$ color{red}{2)quadunderbrace{1000}_{Total=1000}quad underbrace{-150quad -80quad -200quad 1040quad -70}_{Total=540}} $$
$$ color{red}{3)quadunderbrace{1000quad -150}_{Total=850}quad underbrace{-80quad -200quad 1040quad -70}_{Total=690}} $$
$$ color{green}{4)quadunderbrace{1000quad -150quad -80}_{Total=770}quad underbrace{-200quad 1040quad -70}_{Total=770}} $$
$$ color{red}{5)quadunderbrace{1000quad -150quad -80quad-200}_{Total=570}quad underbrace{ 1040quad -70}_{Total=970}} $$
$$ color{red}{6)quadunderbrace{1000quad -150quad -80quad -200quad 1040}_{Total=1610}quad underbrace{-70}_{Total=-70}} $$
$$ color{red}{7)quadunderbrace{1000quad -150quad -80quad-200quad 1040quad -70}_{Total=1540}quad underbrace{}_{Total=0}} $$
Test Case
- Input:
1000 -150 -80 -200 1040 -70Output:1
- Input:
100 -100Output:2
- Input:
1 2 3Output:1
- Input:
10 -20 15Output:0
- Input:
15 -15 15 -15Output:3
- Input:
1Output:0
Notes
- You can assume there wont be any transaction of $0 dollars
- You can take input in any reasonable way
code-golf
asked Feb 7 at 14:45
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,69721464
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11
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After 6 months of frozen and newly created accounts, it is reported that Marie's banker is now interned in a sanitarium. "We are your friends. You need some rest", they said.
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– Arnauld
Feb 7 at 17:20
2
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Suggested test case of a single transaction
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– Veskah
Feb 8 at 4:53
add a comment |
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Challenge Taken with permission from my University Code Challenge Contest
After finishing her studies a couple of months ago, Marie opened a bank account to start receiving the payment of her first job in town. Since then she has been performing a few transactions with it. Her first payment was $1000 dollars. With that money she paid for a dinner in which she invited her parents (The dinner cost $150 dollars), then, she did a purchase in a well-known supermarket ($80 dollars) and a hotel reservation for her vacations ($200). At the end of the month she received her payment again (1040 dollars, a little more than the previous month) and the day after she spent another $70 dollars at the supermarket.
Today, she realized that if after paying the first $80 dollars in the supermarket a second account had been created and the first one frozen, both accounts would have exactly the same balance:
$$ underbrace{1000quad -150quad -80}_{Total=770}quad underbrace{-200quad 1040quad -70}_{Total=770} $$
The event was so rare to her that she wants to continue ascertaining if the movements of her account and those of her friends have also this feature or not.
Challenge
Given a list of transactions, output the number of instants of time in which the owner of the bank account could have created a second account so that both had the same final balance.
Example: [1000, -150, -80, -200, 1040, -70]
$$ color{red}{1)quadunderbrace{}_{Total=0}quad underbrace{1000quad -150quad -80quad -200quad 1040quad -70}_{Total=1540}} $$
$$ color{red}{2)quadunderbrace{1000}_{Total=1000}quad underbrace{-150quad -80quad -200quad 1040quad -70}_{Total=540}} $$
$$ color{red}{3)quadunderbrace{1000quad -150}_{Total=850}quad underbrace{-80quad -200quad 1040quad -70}_{Total=690}} $$
$$ color{green}{4)quadunderbrace{1000quad -150quad -80}_{Total=770}quad underbrace{-200quad 1040quad -70}_{Total=770}} $$
$$ color{red}{5)quadunderbrace{1000quad -150quad -80quad-200}_{Total=570}quad underbrace{ 1040quad -70}_{Total=970}} $$
$$ color{red}{6)quadunderbrace{1000quad -150quad -80quad -200quad 1040}_{Total=1610}quad underbrace{-70}_{Total=-70}} $$
$$ color{red}{7)quadunderbrace{1000quad -150quad -80quad-200quad 1040quad -70}_{Total=1540}quad underbrace{}_{Total=0}} $$
Test Case
- Input:
1000 -150 -80 -200 1040 -70Output:1
- Input:
100 -100Output:2
- Input:
1 2 3Output:1
- Input:
10 -20 15Output:0
- Input:
15 -15 15 -15Output:3
- Input:
1Output:0
Notes
- You can assume there wont be any transaction of $0 dollars
- You can take input in any reasonable way
code-golf
asked Feb 7 at 14:45
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,69721464
$endgroup$
Challenge Taken with permission from my University Code Challenge Contest
After finishing her studies a couple of months ago, Marie opened a bank account to start receiving the payment of her first job in town. Since then she has been performing a few transactions with it. Her first payment was $1000 dollars. With that money she paid for a dinner in which she invited her parents (The dinner cost $150 dollars), then, she did a purchase in a well-known supermarket ($80 dollars) and a hotel reservation for her vacations ($200). At the end of the month she received her payment again (1040 dollars, a little more than the previous month) and the day after she spent another $70 dollars at the supermarket.
Today, she realized that if after paying the first $80 dollars in the supermarket a second account had been created and the first one frozen, both accounts would have exactly the same balance:
$$ underbrace{1000quad -150quad -80}_{Total=770}quad underbrace{-200quad 1040quad -70}_{Total=770} $$
The event was so rare to her that she wants to continue ascertaining if the movements of her account and those of her friends have also this feature or not.
Challenge
Given a list of transactions, output the number of instants of time in which the owner of the bank account could have created a second account so that both had the same final balance.
Example: [1000, -150, -80, -200, 1040, -70]
$$ color{red}{1)quadunderbrace{}_{Total=0}quad underbrace{1000quad -150quad -80quad -200quad 1040quad -70}_{Total=1540}} $$
$$ color{red}{2)quadunderbrace{1000}_{Total=1000}quad underbrace{-150quad -80quad -200quad 1040quad -70}_{Total=540}} $$
$$ color{red}{3)quadunderbrace{1000quad -150}_{Total=850}quad underbrace{-80quad -200quad 1040quad -70}_{Total=690}} $$
$$ color{green}{4)quadunderbrace{1000quad -150quad -80}_{Total=770}quad underbrace{-200quad 1040quad -70}_{Total=770}} $$
$$ color{red}{5)quadunderbrace{1000quad -150quad -80quad-200}_{Total=570}quad underbrace{ 1040quad -70}_{Total=970}} $$
$$ color{red}{6)quadunderbrace{1000quad -150quad -80quad -200quad 1040}_{Total=1610}quad underbrace{-70}_{Total=-70}} $$
$$ color{red}{7)quadunderbrace{1000quad -150quad -80quad-200quad 1040quad -70}_{Total=1540}quad underbrace{}_{Total=0}} $$
Test Case
- Input:
1000 -150 -80 -200 1040 -70Output:1
- Input:
100 -100Output:2
- Input:
1 2 3Output:1
- Input:
10 -20 15Output:0
- Input:
15 -15 15 -15Output:3
- Input:
1Output:0
Notes
- You can assume there wont be any transaction of $0 dollars
- You can take input in any reasonable way
code-golf
code-golf
asked Feb 7 at 14:45
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,69721464
asked Feb 7 at 14:45
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,69721464
edited Feb 8 at 13:14
Luis felipe De jesus Munoz
asked Feb 7 at 14:45
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,69721464
asked Feb 7 at 14:45
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,69721464
asked Feb 7 at 14:45
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,69721464
4,69721464
11
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After 6 months of frozen and newly created accounts, it is reported that Marie's banker is now interned in a sanitarium. "We are your friends. You need some rest", they said.
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– Arnauld
Feb 7 at 17:20
2
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Suggested test case of a single transaction
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– Veskah
Feb 8 at 4:53
add a comment |
11
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After 6 months of frozen and newly created accounts, it is reported that Marie's banker is now interned in a sanitarium. "We are your friends. You need some rest", they said.
$endgroup$
– Arnauld
Feb 7 at 17:20
2
$begingroup$
Suggested test case of a single transaction
$endgroup$
– Veskah
Feb 8 at 4:53
11
11
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After 6 months of frozen and newly created accounts, it is reported that Marie's banker is now interned in a sanitarium. "We are your friends. You need some rest", they said.
$endgroup$
– Arnauld
Feb 7 at 17:20
$begingroup$
After 6 months of frozen and newly created accounts, it is reported that Marie's banker is now interned in a sanitarium. "We are your friends. You need some rest", they said.
$endgroup$
– Arnauld
Feb 7 at 17:20
2
2
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Suggested test case of a single transaction
$endgroup$
– Veskah
Feb 8 at 4:53
$begingroup$
Suggested test case of a single transaction
$endgroup$
– Veskah
Feb 8 at 4:53
add a comment |
20 Answers
20
active
oldest
votes
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C# (Visual C# Interactive Compiler), 63 bytes
n=>n.Append(0).Where((a,b)=>n.Take(b).Sum()*2==n.Sum()).Count()
Saved 6 bytes thanks to dana
Try it online!
answered Feb 7 at 20:55
Embodiment of IgnoranceEmbodiment of Ignorance
980119
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add a comment |
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Perl 6, 25 bytes
{+grep .sum/2,[+] 0,|$_}
Try it online!
Explanation
We just prepend a zero to the given list (0,|$_), make a sequence of partial sums with [+] (i. e. the sequence formed by the first element, the sum of first two, the sum of first three etc.), and look (grep) for any elements that are exactly equal to the half of the final account state (sum of the given list). Finally, we count them with a +.
answered Feb 7 at 17:19
RamilliesRamillies
1,641515
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05AB1E, 11 bytes
0.ø.œ2ùO€ËO
Try it online or verify all test cases.
Explanation:
0.ø # Surround the (implicit) input list with a leading and trailing 0
# i.e. [100,-100] → [0,100,-100,0]
.œ # Get all possible partitions to divide the list
# → [[[0],[100],[-100],[0]],[[0],[100],[-100,0]],[[0],[100,-100],[0]],[[0],[100,-100,0]],[[0,100],[-100],[0]],[[0,100],[-100,0]],[[0,100,-100],[0]],[[0,100,-100,0]]]
2ù # Only leave partitions consisting of 2 items
# → [[[0],[100,-100,0]],[[0,100],[-100,0]],[[0,100,-100],[0]]]
O # Take the sum of each
# → [[0,0],[100,-100],[0,0]]
€Ë # Check of each inner list if both sums are equal (1 if truthy; 0 if falsey)
# → [1,0,1]
O # Take the sum of that (and output as result)
# → 2
answered Feb 7 at 15:00
Kevin CruijssenKevin Cruijssen
38.1k557195
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Jelly, 11 6 bytes
ŻÄḤ=SS
Try it online!
answered Feb 7 at 15:32
Don't be a x-triple dotDon't be a x-triple dot
32k759199
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JavaScript (Node.js), 45 bytes
a=>!a.map(v=>o[s+=v]=-~o[s],s=0,o=[1])|o[s/2]
Try it online!
Save 4 bytes by using -~o[s]. Thanks to Shaggy.
answered Feb 8 at 10:28
tshtsh
9,01511650
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+1 for beating Arnauld :o
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– Luis felipe De jesus Munoz
Feb 8 at 12:22
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45 bytes
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– Shaggy
Feb 8 at 19:14
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@LuisfelipeDejesusMunoz, Arnauld ain't (always) unbeatable! ;)
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– Shaggy
Feb 8 at 23:45
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@Shaggy leading+is changed to!, so it could work for input[100].
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– tsh
Feb 9 at 12:00
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Ah, didn't realise we had to handle singleton arrays. Nicely fixed.
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– Shaggy
Feb 9 at 13:46
|
show 1 more comment
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Perl 5 -p, 42 41 bytes
@NahuelFouilleul saves a byte
y/ /+/;$+=eval$'==eval$`while/^|$|+/g}{
Try it online!
answered Feb 7 at 15:51
XcaliXcali
5,335520
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y/ /+/;saves 1 byte
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– Nahuel Fouilleul
Feb 7 at 16:56
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34 bytes using other approach
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– Nahuel Fouilleul
Feb 7 at 19:57
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30 bytes
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– Nahuel Fouilleul
Feb 7 at 20:03
add a comment |
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JavaScript (ES6), 52 bytes
a=>a.map(x=>n+=(s+=x)==eval(a.join`+`)-s,n=s=0)|n+!s
Try it online!
Commented
a => // a = input array
a.map(x => // for each element x in a:
n += // increment n if the following test is truthy:
(s += x) // update the left sum
== // and test whether it's equal to
eval(a.join`+`) - s, // the right sum
n = s =0 // start with n = s = 0
) // end of map()
| n // yield n; if the final sum is 0, it means that we could have
+ // created a balanced account at the beginning of the process;
!s // so, we increment n if it is
Recursive version, 54 53 bytes
f=(a,s=0)=>a+a?(s==eval(a.join`+`))+f(a,s+a.pop()):!s
Try it online!
answered Feb 7 at 14:57
ArnauldArnauld
76.1k693320
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I was just about to suggest that 52-byte version!
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– Shaggy
Feb 7 at 16:00
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@Shaggy Yeah, I discarded the non-recursive version too soon because I thought the recursive one could be shorter.
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– Arnauld
Feb 7 at 16:03
add a comment |
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APL (Dyalog Unicode), 21 bytesSBCS
Anonymous tacit prefix function
+/⊂((+/↑)=1⊥↓)¨⍨0,⍳∘≢
Try it online!
⍳ ɩndices
∘ of≢ the tally of transactions
0, prepend zero
⊂(…)¨⍨ apply the following tacit function with each of those as left argument and the entire list of transactions as right argument (⍨ swaps argument
⊂ the entire list of transactions
(…) as left argument to the below function
¨ applied to each of the indices
⍨ with swapped arguments (i.e. list on right, indices on left:
↓ drop that many from the left
1⊥ sum (lit. evaluate in base-1)
(…)= is it (0/1) equal to…
↑ take that many transactions from the left
+/ sum them
+/ sum that Boolean list to get the count of truths
answered Feb 7 at 15:08
AdámAdám
28k273199
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Batch, 84 bytes
@set s=%*
@set/as=%s: =+%,c=0
@for %%n in (0 %*)do @set/as-=%%n*2,c+=!s
@echo %c%
Takes input as command-line arguments. Explanation:
@set s=%*
Join the arguments with spaces.
@set/as=%s: =+%,c=0
Replace the spaces with +s and evaluate the result. Also clear the count.
@for %%n in (0 %*)do @set/as-=%%n*2,c+=!s
For each amount, subtract double that from the sum. If the result is zero, then this is a valid match, so increment the count. The extra zero at the beginning allows for a match before any amounts.
@echo %c%
Print the result.
answered Feb 7 at 20:58
NeilNeil
80.7k744178
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Charcoal, 15 bytes
⊞θ⁰ＩΣＥθ⁼Σθ⊗Σ✂θκ
Try it online! Link is to verbose version of code. Explanation:
θ Input list
⁰ Literal 0
⊞ Push to list
θ Augmented list
Ｅ Mapped to
θ Augmented list
✂ Sliced from
κ Current index
Σ Summed
⊗ Doubled
⁼ Equals
θ (Augmented) list
Σ Summed
Σ Sum of booleans
Ｉ Cast to string
Implicitly print
Unfortunately in Charcoal Sum() is not 0 so I have to ensure that there is always at least one element to sum.
answered Feb 7 at 21:11
NeilNeil
80.7k744178
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Python 3, 67 58 bytes
lambda l:sum(sum(l[:x])*2==sum(l)for x in range(len(l)+1))
Try it online!
-9 bytes thanks to @Don't be a x-triple dot
answered Feb 7 at 15:43
Black Owl KaiBlack Owl Kai
69011
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Summing instead of filtering will save you 7 bytes:lambda l:sum(sum(l[:x])==sum(l[x:])for x in range(len(l)+1)).
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– Don't be a x-triple dot
Feb 7 at 18:43
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sum(l[:x])*2==sum(l)saves you another 2 bytes.
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– Neil
Feb 7 at 20:49
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R, 50 37 bytes
sum(c(0,cumsum(x<-scan()))==sum(x)/2)
Try it online!
answered Feb 7 at 15:47
Kirill L.Kirill L.
4,3351422
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MATL, 9 bytes
s0GhYsE=s
Try it online!
Same approach as some other answers: prepend a zero, and check how often half the cumulative sum is equal to the total sum.
s % Total sum of (implicit) input
0Gh % Prepend 0 to another copy of the input
Ys % Cumulative sum
E= % Check element-wise equality of 2*cumulative sum with total sum
s % Sum number of `true` values
answered Feb 8 at 13:39
SanchisesSanchises
5,89212351
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Japt -x, 14 11 bytes
iT å+ ®¥nUx
Try it
iT å+ ®¥nUx :Implicit input of array U
i :Prepend
T : Zero
å+ :Cumulatively reduce by addition
® :Map each Z
¥ : Test for equality with
n : Z subtracted from
Ux : U reduced by addition
:Implicitly reduce by addition and output
answered Feb 7 at 15:14
ShaggyShaggy
19.9k21667
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PowerShell, 88 82 bytes
-6 Bytes thanks to mazzy
param($n)0..($x=$n.length)|%{$i+=+$z-eq($n[$_..$x]|measure -Su).sum;$z+=$n[$_]};$i
Try it online!
This seems like a very clumsy method but it got the job done. I'll try and revamp it in the future.
answered Feb 8 at 4:46
VeskahVeskah
93814
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you can write$i+=<predicate>insteadif(<predicate>){$i++}
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– mazzy
Feb 8 at 6:15
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bash, 52 bytes
IFS=+;for i in 0 $@;{((c+=2*(x+=i)=="$*"));};echo $c
TIO
The trick: setting IFS=+, "$*" expands to a string where arguments are delimited by +, in arithmetic expression eval to the sum
answered Feb 7 at 20:25
Nahuel FouilleulNahuel Fouilleul
2,30529
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PowerShell, 49 45 36 bytes
(($args+0|%{2*($s+=$_)})-eq$s).count
Try it online!
answered Feb 8 at 17:24
mazzymazzy
2,4651316
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Brachylog, 9 bytes
Not as good as day 1. This one loses to Jelly
{~c₂+ᵐ=}ᶜ
Explanation
{ }ᶜ # Count the ways to:
~c₂ # Split the input array in 2 ...
+ᵐ # so that their sums ...
= # are equal
Test suite:
Try it online!
answered Feb 10 at 12:47
KroppebKroppeb
1,206210
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Haskell, 46 35 bytes
f x=sum[1|a<-scanl(+)0x,a==sum x/2]
Try it online!
answered Feb 10 at 16:09
niminimi
31.8k32285
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J, 19 bytes
1#.[:(={:-])0+/@,]
Try it online!
explanation
1 #. [: (= ({: - ])) 0 +/@, ]
0 , ] NB. prepend 0 to input...
+/@ NB. and take the prefix sums...
[: ({: - ]) NB. then subtract that list
NB. from its final elm
NB. (`{:`), giving the list
NB. of suffix sums...
[: (= ( )) NB. create a 1-0 list showing
NB. where the prefix sums
NB. equal the suffix sums
1 #. NB. and take the sum.
answered 2 days ago
JonahJonah
2,121916
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20 Answers
20
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oldest
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20 Answers
20
active
oldest
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oldest
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$begingroup$
C# (Visual C# Interactive Compiler), 63 bytes
n=>n.Append(0).Where((a,b)=>n.Take(b).Sum()*2==n.Sum()).Count()
Saved 6 bytes thanks to dana
Try it online!
answered Feb 7 at 20:55
Embodiment of IgnoranceEmbodiment of Ignorance
980119
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 63 bytes
n=>n.Append(0).Where((a,b)=>n.Take(b).Sum()*2==n.Sum()).Count()
Saved 6 bytes thanks to dana
Try it online!
answered Feb 7 at 20:55
Embodiment of IgnoranceEmbodiment of Ignorance
980119
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 63 bytes
n=>n.Append(0).Where((a,b)=>n.Take(b).Sum()*2==n.Sum()).Count()
Saved 6 bytes thanks to dana
Try it online!
answered Feb 7 at 20:55
Embodiment of IgnoranceEmbodiment of Ignorance
980119
$endgroup$
C# (Visual C# Interactive Compiler), 63 bytes
n=>n.Append(0).Where((a,b)=>n.Take(b).Sum()*2==n.Sum()).Count()
Saved 6 bytes thanks to dana
Try it online!
answered Feb 7 at 20:55
Embodiment of IgnoranceEmbodiment of Ignorance
980119
edited 2 days ago
answered Feb 7 at 20:55
Embodiment of IgnoranceEmbodiment of Ignorance
980119
answered Feb 7 at 20:55
Embodiment of IgnoranceEmbodiment of Ignorance
980119
answered Feb 7 at 20:55
Embodiment of IgnoranceEmbodiment of Ignorance
980119
980119
add a comment |
add a comment |
$begingroup$
Perl 6, 25 bytes
{+grep .sum/2,[+] 0,|$_}
Try it online!
Explanation
We just prepend a zero to the given list (0,|$_), make a sequence of partial sums with [+] (i. e. the sequence formed by the first element, the sum of first two, the sum of first three etc.), and look (grep) for any elements that are exactly equal to the half of the final account state (sum of the given list). Finally, we count them with a +.
answered Feb 7 at 17:19
RamilliesRamillies
1,641515
$endgroup$
add a comment |
$begingroup$
Perl 6, 25 bytes
{+grep .sum/2,[+] 0,|$_}
Try it online!
Explanation
We just prepend a zero to the given list (0,|$_), make a sequence of partial sums with [+] (i. e. the sequence formed by the first element, the sum of first two, the sum of first three etc.), and look (grep) for any elements that are exactly equal to the half of the final account state (sum of the given list). Finally, we count them with a +.
answered Feb 7 at 17:19
RamilliesRamillies
1,641515
$endgroup$
add a comment |
$begingroup$
Perl 6, 25 bytes
{+grep .sum/2,[+] 0,|$_}
Try it online!
Explanation
We just prepend a zero to the given list (0,|$_), make a sequence of partial sums with [+] (i. e. the sequence formed by the first element, the sum of first two, the sum of first three etc.), and look (grep) for any elements that are exactly equal to the half of the final account state (sum of the given list). Finally, we count them with a +.
answered Feb 7 at 17:19
RamilliesRamillies
1,641515
$endgroup$
Perl 6, 25 bytes
{+grep .sum/2,[+] 0,|$_}
Try it online!
Explanation
We just prepend a zero to the given list (0,|$_), make a sequence of partial sums with [+] (i. e. the sequence formed by the first element, the sum of first two, the sum of first three etc.), and look (grep) for any elements that are exactly equal to the half of the final account state (sum of the given list). Finally, we count them with a +.
answered Feb 7 at 17:19
RamilliesRamillies
1,641515
answered Feb 7 at 17:19
RamilliesRamillies
1,641515
answered Feb 7 at 17:19
RamilliesRamillies
1,641515
answered Feb 7 at 17:19
RamilliesRamillies
1,641515
1,641515
add a comment |
add a comment |
$begingroup$
05AB1E, 11 bytes
0.ø.œ2ùO€ËO
Try it online or verify all test cases.
Explanation:
0.ø # Surround the (implicit) input list with a leading and trailing 0
# i.e. [100,-100] → [0,100,-100,0]
.œ # Get all possible partitions to divide the list
# → [[[0],[100],[-100],[0]],[[0],[100],[-100,0]],[[0],[100,-100],[0]],[[0],[100,-100,0]],[[0,100],[-100],[0]],[[0,100],[-100,0]],[[0,100,-100],[0]],[[0,100,-100,0]]]
2ù # Only leave partitions consisting of 2 items
# → [[[0],[100,-100,0]],[[0,100],[-100,0]],[[0,100,-100],[0]]]
O # Take the sum of each
# → [[0,0],[100,-100],[0,0]]
€Ë # Check of each inner list if both sums are equal (1 if truthy; 0 if falsey)
# → [1,0,1]
O # Take the sum of that (and output as result)
# → 2
answered Feb 7 at 15:00
Kevin CruijssenKevin Cruijssen
38.1k557195
$endgroup$
add a comment |
$begingroup$
05AB1E, 11 bytes
0.ø.œ2ùO€ËO
Try it online or verify all test cases.
Explanation:
0.ø # Surround the (implicit) input list with a leading and trailing 0
# i.e. [100,-100] → [0,100,-100,0]
.œ # Get all possible partitions to divide the list
# → [[[0],[100],[-100],[0]],[[0],[100],[-100,0]],[[0],[100,-100],[0]],[[0],[100,-100,0]],[[0,100],[-100],[0]],[[0,100],[-100,0]],[[0,100,-100],[0]],[[0,100,-100,0]]]
2ù # Only leave partitions consisting of 2 items
# → [[[0],[100,-100,0]],[[0,100],[-100,0]],[[0,100,-100],[0]]]
O # Take the sum of each
# → [[0,0],[100,-100],[0,0]]
€Ë # Check of each inner list if both sums are equal (1 if truthy; 0 if falsey)
# → [1,0,1]
O # Take the sum of that (and output as result)
# → 2
answered Feb 7 at 15:00
Kevin CruijssenKevin Cruijssen
38.1k557195
$endgroup$
add a comment |
$begingroup$
05AB1E, 11 bytes
0.ø.œ2ùO€ËO
Try it online or verify all test cases.
Explanation:
0.ø # Surround the (implicit) input list with a leading and trailing 0
# i.e. [100,-100] → [0,100,-100,0]
.œ # Get all possible partitions to divide the list
# → [[[0],[100],[-100],[0]],[[0],[100],[-100,0]],[[0],[100,-100],[0]],[[0],[100,-100,0]],[[0,100],[-100],[0]],[[0,100],[-100,0]],[[0,100,-100],[0]],[[0,100,-100,0]]]
2ù # Only leave partitions consisting of 2 items
# → [[[0],[100,-100,0]],[[0,100],[-100,0]],[[0,100,-100],[0]]]
O # Take the sum of each
# → [[0,0],[100,-100],[0,0]]
€Ë # Check of each inner list if both sums are equal (1 if truthy; 0 if falsey)
# → [1,0,1]
O # Take the sum of that (and output as result)
# → 2
answered Feb 7 at 15:00
Kevin CruijssenKevin Cruijssen
38.1k557195
$endgroup$
05AB1E, 11 bytes
0.ø.œ2ùO€ËO
Try it online or verify all test cases.
Explanation:
0.ø # Surround the (implicit) input list with a leading and trailing 0
# i.e. [100,-100] → [0,100,-100,0]
.œ # Get all possible partitions to divide the list
# → [[[0],[100],[-100],[0]],[[0],[100],[-100,0]],[[0],[100,-100],[0]],[[0],[100,-100,0]],[[0,100],[-100],[0]],[[0,100],[-100,0]],[[0,100,-100],[0]],[[0,100,-100,0]]]
2ù # Only leave partitions consisting of 2 items
# → [[[0],[100,-100,0]],[[0,100],[-100,0]],[[0,100,-100],[0]]]
O # Take the sum of each
# → [[0,0],[100,-100],[0,0]]
€Ë # Check of each inner list if both sums are equal (1 if truthy; 0 if falsey)
# → [1,0,1]
O # Take the sum of that (and output as result)
# → 2
answered Feb 7 at 15:00
Kevin CruijssenKevin Cruijssen
38.1k557195
edited Feb 7 at 15:08
answered Feb 7 at 15:00
Kevin CruijssenKevin Cruijssen
38.1k557195
answered Feb 7 at 15:00
Kevin CruijssenKevin Cruijssen
38.1k557195
answered Feb 7 at 15:00
Kevin CruijssenKevin Cruijssen
38.1k557195
38.1k557195
add a comment |
add a comment |
$begingroup$
Jelly, 11 6 bytes
ŻÄḤ=SS
Try it online!
answered Feb 7 at 15:32
Don't be a x-triple dotDon't be a x-triple dot
32k759199
$endgroup$
add a comment |
$begingroup$
Jelly, 11 6 bytes
ŻÄḤ=SS
Try it online!
answered Feb 7 at 15:32
Don't be a x-triple dotDon't be a x-triple dot
32k759199
$endgroup$
add a comment |
$begingroup$
Jelly, 11 6 bytes
ŻÄḤ=SS
Try it online!
answered Feb 7 at 15:32
Don't be a x-triple dotDon't be a x-triple dot
32k759199
$endgroup$
Jelly, 11 6 bytes
ŻÄḤ=SS
Try it online!
answered Feb 7 at 15:32
Don't be a x-triple dotDon't be a x-triple dot
32k759199
edited Feb 8 at 21:05
answered Feb 7 at 15:32
Don't be a x-triple dotDon't be a x-triple dot
32k759199
answered Feb 7 at 15:32
Don't be a x-triple dotDon't be a x-triple dot
32k759199
answered Feb 7 at 15:32
Don't be a x-triple dotDon't be a x-triple dot
32k759199
32k759199
add a comment |
add a comment |
$begingroup$
JavaScript (Node.js), 45 bytes
a=>!a.map(v=>o[s+=v]=-~o[s],s=0,o=[1])|o[s/2]
Try it online!
Save 4 bytes by using -~o[s]. Thanks to Shaggy.
answered Feb 8 at 10:28
tshtsh
9,01511650
$endgroup$
$begingroup$
+1 for beating Arnauld :o
$endgroup$
– Luis felipe De jesus Munoz
Feb 8 at 12:22
$begingroup$
45 bytes
$endgroup$
– Shaggy
Feb 8 at 19:14
$begingroup$
@LuisfelipeDejesusMunoz, Arnauld ain't (always) unbeatable! ;)
$endgroup$
– Shaggy
Feb 8 at 23:45
$begingroup$
@Shaggy leading+is changed to!, so it could work for input[100].
$endgroup$
– tsh
Feb 9 at 12:00
$begingroup$
Ah, didn't realise we had to handle singleton arrays. Nicely fixed.
$endgroup$
– Shaggy
Feb 9 at 13:46
|
show 1 more comment
$begingroup$
JavaScript (Node.js), 45 bytes
a=>!a.map(v=>o[s+=v]=-~o[s],s=0,o=[1])|o[s/2]
Try it online!
Save 4 bytes by using -~o[s]. Thanks to Shaggy.
answered Feb 8 at 10:28
tshtsh
9,01511650
$endgroup$
$begingroup$
+1 for beating Arnauld :o
$endgroup$
– Luis felipe De jesus Munoz
Feb 8 at 12:22
$begingroup$
45 bytes
$endgroup$
– Shaggy
Feb 8 at 19:14
$begingroup$
@LuisfelipeDejesusMunoz, Arnauld ain't (always) unbeatable! ;)
$endgroup$
– Shaggy
Feb 8 at 23:45
$begingroup$
@Shaggy leading+is changed to!, so it could work for input[100].
$endgroup$
– tsh
Feb 9 at 12:00
$begingroup$
Ah, didn't realise we had to handle singleton arrays. Nicely fixed.
$endgroup$
– Shaggy
Feb 9 at 13:46
|
show 1 more comment
$begingroup$
JavaScript (Node.js), 45 bytes
a=>!a.map(v=>o[s+=v]=-~o[s],s=0,o=[1])|o[s/2]
Try it online!
Save 4 bytes by using -~o[s]. Thanks to Shaggy.
answered Feb 8 at 10:28
tshtsh
9,01511650
$endgroup$
JavaScript (Node.js), 45 bytes
a=>!a.map(v=>o[s+=v]=-~o[s],s=0,o=[1])|o[s/2]
Try it online!
Save 4 bytes by using -~o[s]. Thanks to Shaggy.
answered Feb 8 at 10:28
tshtsh
9,01511650
edited Feb 9 at 11:59
answered Feb 8 at 10:28
tshtsh
9,01511650
answered Feb 8 at 10:28
tshtsh
9,01511650
answered Feb 8 at 10:28
tshtsh
9,01511650
9,01511650
$begingroup$
+1 for beating Arnauld :o
$endgroup$
– Luis felipe De jesus Munoz
Feb 8 at 12:22
$begingroup$
45 bytes
$endgroup$
– Shaggy
Feb 8 at 19:14
$begingroup$
@LuisfelipeDejesusMunoz, Arnauld ain't (always) unbeatable! ;)
$endgroup$
– Shaggy
Feb 8 at 23:45
$begingroup$
@Shaggy leading+is changed to!, so it could work for input[100].
$endgroup$
– tsh
Feb 9 at 12:00
$begingroup$
Ah, didn't realise we had to handle singleton arrays. Nicely fixed.
$endgroup$
– Shaggy
Feb 9 at 13:46
|
show 1 more comment
$begingroup$
+1 for beating Arnauld :o
$endgroup$
– Luis felipe De jesus Munoz
Feb 8 at 12:22
$begingroup$
45 bytes
$endgroup$
– Shaggy
Feb 8 at 19:14
$begingroup$
@LuisfelipeDejesusMunoz, Arnauld ain't (always) unbeatable! ;)
$endgroup$
– Shaggy
Feb 8 at 23:45
$begingroup$
@Shaggy leading+is changed to!, so it could work for input[100].
$endgroup$
– tsh
Feb 9 at 12:00
$begingroup$
Ah, didn't realise we had to handle singleton arrays. Nicely fixed.
$endgroup$
– Shaggy
Feb 9 at 13:46
$begingroup$
+1 for beating Arnauld :o
$endgroup$
– Luis felipe De jesus Munoz
Feb 8 at 12:22
$begingroup$
+1 for beating Arnauld :o
$endgroup$
– Luis felipe De jesus Munoz
Feb 8 at 12:22
$begingroup$
45 bytes
$endgroup$
– Shaggy
Feb 8 at 19:14
$begingroup$
45 bytes
$endgroup$
– Shaggy
Feb 8 at 19:14
$begingroup$
@LuisfelipeDejesusMunoz, Arnauld ain't (always) unbeatable! ;)
$endgroup$
– Shaggy
Feb 8 at 23:45
$begingroup$
@LuisfelipeDejesusMunoz, Arnauld ain't (always) unbeatable! ;)
$endgroup$
– Shaggy
Feb 8 at 23:45
$begingroup$
@Shaggy leading
+ is changed to !, so it could work for input [100].$endgroup$
– tsh
Feb 9 at 12:00
$begingroup$
@Shaggy leading
+ is changed to !, so it could work for input [100].$endgroup$
– tsh
Feb 9 at 12:00
$begingroup$
Ah, didn't realise we had to handle singleton arrays. Nicely fixed.
$endgroup$
– Shaggy
Feb 9 at 13:46
$begingroup$
Ah, didn't realise we had to handle singleton arrays. Nicely fixed.
$endgroup$
– Shaggy
Feb 9 at 13:46
|
show 1 more comment
$begingroup$
Perl 5 -p, 42 41 bytes
@NahuelFouilleul saves a byte
y/ /+/;$+=eval$'==eval$`while/^|$|+/g}{
Try it online!
answered Feb 7 at 15:51
XcaliXcali
5,335520
$endgroup$
$begingroup$
y/ /+/;saves 1 byte
$endgroup$
– Nahuel Fouilleul
Feb 7 at 16:56
$begingroup$
34 bytes using other approach
$endgroup$
– Nahuel Fouilleul
Feb 7 at 19:57
$begingroup$
30 bytes
$endgroup$
– Nahuel Fouilleul
Feb 7 at 20:03
add a comment |
$begingroup$
Perl 5 -p, 42 41 bytes
@NahuelFouilleul saves a byte
y/ /+/;$+=eval$'==eval$`while/^|$|+/g}{
Try it online!
answered Feb 7 at 15:51
XcaliXcali
5,335520
$endgroup$
$begingroup$
y/ /+/;saves 1 byte
$endgroup$
– Nahuel Fouilleul
Feb 7 at 16:56
$begingroup$
34 bytes using other approach
$endgroup$
– Nahuel Fouilleul
Feb 7 at 19:57
$begingroup$
30 bytes
$endgroup$
– Nahuel Fouilleul
Feb 7 at 20:03
add a comment |
$begingroup$
Perl 5 -p, 42 41 bytes
@NahuelFouilleul saves a byte
y/ /+/;$+=eval$'==eval$`while/^|$|+/g}{
Try it online!
answered Feb 7 at 15:51
XcaliXcali
5,335520
$endgroup$
Perl 5 -p, 42 41 bytes
@NahuelFouilleul saves a byte
y/ /+/;$+=eval$'==eval$`while/^|$|+/g}{
Try it online!
answered Feb 7 at 15:51
XcaliXcali
5,335520
edited Feb 7 at 16:58
answered Feb 7 at 15:51
XcaliXcali
5,335520
answered Feb 7 at 15:51
XcaliXcali
5,335520
answered Feb 7 at 15:51
XcaliXcali
5,335520
5,335520
$begingroup$
y/ /+/;saves 1 byte
$endgroup$
– Nahuel Fouilleul
Feb 7 at 16:56
$begingroup$
34 bytes using other approach
$endgroup$
– Nahuel Fouilleul
Feb 7 at 19:57
$begingroup$
30 bytes
$endgroup$
– Nahuel Fouilleul
Feb 7 at 20:03
add a comment |
$begingroup$
y/ /+/;saves 1 byte
$endgroup$
– Nahuel Fouilleul
Feb 7 at 16:56
$begingroup$
34 bytes using other approach
$endgroup$
– Nahuel Fouilleul
Feb 7 at 19:57
$begingroup$
30 bytes
$endgroup$
– Nahuel Fouilleul
Feb 7 at 20:03
$begingroup$
y/ /+/; saves 1 byte$endgroup$
– Nahuel Fouilleul
Feb 7 at 16:56
$begingroup$
y/ /+/; saves 1 byte$endgroup$
– Nahuel Fouilleul
Feb 7 at 16:56
$begingroup$
34 bytes using other approach
$endgroup$
– Nahuel Fouilleul
Feb 7 at 19:57
$begingroup$
34 bytes using other approach
$endgroup$
– Nahuel Fouilleul
Feb 7 at 19:57
$begingroup$
30 bytes
$endgroup$
– Nahuel Fouilleul
Feb 7 at 20:03
$begingroup$
30 bytes
$endgroup$
– Nahuel Fouilleul
Feb 7 at 20:03
add a comment |
$begingroup$
JavaScript (ES6), 52 bytes
a=>a.map(x=>n+=(s+=x)==eval(a.join`+`)-s,n=s=0)|n+!s
Try it online!
Commented
a => // a = input array
a.map(x => // for each element x in a:
n += // increment n if the following test is truthy:
(s += x) // update the left sum
== // and test whether it's equal to
eval(a.join`+`) - s, // the right sum
n = s =0 // start with n = s = 0
) // end of map()
| n // yield n; if the final sum is 0, it means that we could have
+ // created a balanced account at the beginning of the process;
!s // so, we increment n if it is
Recursive version, 54 53 bytes
f=(a,s=0)=>a+a?(s==eval(a.join`+`))+f(a,s+a.pop()):!s
Try it online!
answered Feb 7 at 14:57
ArnauldArnauld
76.1k693320
$endgroup$
$begingroup$
I was just about to suggest that 52-byte version!
$endgroup$
– Shaggy
Feb 7 at 16:00
$begingroup$
@Shaggy Yeah, I discarded the non-recursive version too soon because I thought the recursive one could be shorter.
$endgroup$
– Arnauld
Feb 7 at 16:03
add a comment |
$begingroup$
JavaScript (ES6), 52 bytes
a=>a.map(x=>n+=(s+=x)==eval(a.join`+`)-s,n=s=0)|n+!s
Try it online!
Commented
a => // a = input array
a.map(x => // for each element x in a:
n += // increment n if the following test is truthy:
(s += x) // update the left sum
== // and test whether it's equal to
eval(a.join`+`) - s, // the right sum
n = s =0 // start with n = s = 0
) // end of map()
| n // yield n; if the final sum is 0, it means that we could have
+ // created a balanced account at the beginning of the process;
!s // so, we increment n if it is
Recursive version, 54 53 bytes
f=(a,s=0)=>a+a?(s==eval(a.join`+`))+f(a,s+a.pop()):!s
Try it online!
answered Feb 7 at 14:57
ArnauldArnauld
76.1k693320
$endgroup$
$begingroup$
I was just about to suggest that 52-byte version!
$endgroup$
– Shaggy
Feb 7 at 16:00
$begingroup$
@Shaggy Yeah, I discarded the non-recursive version too soon because I thought the recursive one could be shorter.
$endgroup$
– Arnauld
Feb 7 at 16:03
add a comment |
$begingroup$
JavaScript (ES6), 52 bytes
a=>a.map(x=>n+=(s+=x)==eval(a.join`+`)-s,n=s=0)|n+!s
Try it online!
Commented
a => // a = input array
a.map(x => // for each element x in a:
n += // increment n if the following test is truthy:
(s += x) // update the left sum
== // and test whether it's equal to
eval(a.join`+`) - s, // the right sum
n = s =0 // start with n = s = 0
) // end of map()
| n // yield n; if the final sum is 0, it means that we could have
+ // created a balanced account at the beginning of the process;
!s // so, we increment n if it is
Recursive version, 54 53 bytes
f=(a,s=0)=>a+a?(s==eval(a.join`+`))+f(a,s+a.pop()):!s
Try it online!
answered Feb 7 at 14:57
ArnauldArnauld
76.1k693320
$endgroup$
JavaScript (ES6), 52 bytes
a=>a.map(x=>n+=(s+=x)==eval(a.join`+`)-s,n=s=0)|n+!s
Try it online!
Commented
a => // a = input array
a.map(x => // for each element x in a:
n += // increment n if the following test is truthy:
(s += x) // update the left sum
== // and test whether it's equal to
eval(a.join`+`) - s, // the right sum
n = s =0 // start with n = s = 0
) // end of map()
| n // yield n; if the final sum is 0, it means that we could have
+ // created a balanced account at the beginning of the process;
!s // so, we increment n if it is
Recursive version, 54 53 bytes
f=(a,s=0)=>a+a?(s==eval(a.join`+`))+f(a,s+a.pop()):!s
Try it online!
answered Feb 7 at 14:57
ArnauldArnauld
76.1k693320
edited Feb 7 at 17:12
answered Feb 7 at 14:57
ArnauldArnauld
76.1k693320
answered Feb 7 at 14:57
ArnauldArnauld
76.1k693320
answered Feb 7 at 14:57
ArnauldArnauld
76.1k693320
76.1k693320
$begingroup$
I was just about to suggest that 52-byte version!
$endgroup$
– Shaggy
Feb 7 at 16:00
$begingroup$
@Shaggy Yeah, I discarded the non-recursive version too soon because I thought the recursive one could be shorter.
$endgroup$
– Arnauld
Feb 7 at 16:03
add a comment |
$begingroup$
I was just about to suggest that 52-byte version!
$endgroup$
– Shaggy
Feb 7 at 16:00
$begingroup$
@Shaggy Yeah, I discarded the non-recursive version too soon because I thought the recursive one could be shorter.
$endgroup$
– Arnauld
Feb 7 at 16:03
$begingroup$
I was just about to suggest that 52-byte version!
$endgroup$
– Shaggy
Feb 7 at 16:00
$begingroup$
I was just about to suggest that 52-byte version!
$endgroup$
– Shaggy
Feb 7 at 16:00
$begingroup$
@Shaggy Yeah, I discarded the non-recursive version too soon because I thought the recursive one could be shorter.
$endgroup$
– Arnauld
Feb 7 at 16:03
$begingroup$
@Shaggy Yeah, I discarded the non-recursive version too soon because I thought the recursive one could be shorter.
$endgroup$
– Arnauld
Feb 7 at 16:03
add a comment |
$begingroup$
APL (Dyalog Unicode), 21 bytesSBCS
Anonymous tacit prefix function
+/⊂((+/↑)=1⊥↓)¨⍨0,⍳∘≢
Try it online!
⍳ ɩndices
∘ of≢ the tally of transactions
0, prepend zero
⊂(…)¨⍨ apply the following tacit function with each of those as left argument and the entire list of transactions as right argument (⍨ swaps argument
⊂ the entire list of transactions
(…) as left argument to the below function
¨ applied to each of the indices
⍨ with swapped arguments (i.e. list on right, indices on left:
↓ drop that many from the left
1⊥ sum (lit. evaluate in base-1)
(…)= is it (0/1) equal to…
↑ take that many transactions from the left
+/ sum them
+/ sum that Boolean list to get the count of truths
answered Feb 7 at 15:08
AdámAdám
28k273199
$endgroup$
add a comment |
$begingroup$
APL (Dyalog Unicode), 21 bytesSBCS
Anonymous tacit prefix function
+/⊂((+/↑)=1⊥↓)¨⍨0,⍳∘≢
Try it online!
⍳ ɩndices
∘ of≢ the tally of transactions
0, prepend zero
⊂(…)¨⍨ apply the following tacit function with each of those as left argument and the entire list of transactions as right argument (⍨ swaps argument
⊂ the entire list of transactions
(…) as left argument to the below function
¨ applied to each of the indices
⍨ with swapped arguments (i.e. list on right, indices on left:
↓ drop that many from the left
1⊥ sum (lit. evaluate in base-1)
(…)= is it (0/1) equal to…
↑ take that many transactions from the left
+/ sum them
+/ sum that Boolean list to get the count of truths
answered Feb 7 at 15:08
AdámAdám
28k273199
$endgroup$
add a comment |
$begingroup$
APL (Dyalog Unicode), 21 bytesSBCS
Anonymous tacit prefix function
+/⊂((+/↑)=1⊥↓)¨⍨0,⍳∘≢
Try it online!
⍳ ɩndices
∘ of≢ the tally of transactions
0, prepend zero
⊂(…)¨⍨ apply the following tacit function with each of those as left argument and the entire list of transactions as right argument (⍨ swaps argument
⊂ the entire list of transactions
(…) as left argument to the below function
¨ applied to each of the indices
⍨ with swapped arguments (i.e. list on right, indices on left:
↓ drop that many from the left
1⊥ sum (lit. evaluate in base-1)
(…)= is it (0/1) equal to…
↑ take that many transactions from the left
+/ sum them
+/ sum that Boolean list to get the count of truths
answered Feb 7 at 15:08
AdámAdám
28k273199
$endgroup$
APL (Dyalog Unicode), 21 bytesSBCS
Anonymous tacit prefix function
+/⊂((+/↑)=1⊥↓)¨⍨0,⍳∘≢
Try it online!
⍳ ɩndices
∘ of≢ the tally of transactions
0, prepend zero
⊂(…)¨⍨ apply the following tacit function with each of those as left argument and the entire list of transactions as right argument (⍨ swaps argument
⊂ the entire list of transactions
(…) as left argument to the below function
¨ applied to each of the indices
⍨ with swapped arguments (i.e. list on right, indices on left:
↓ drop that many from the left
1⊥ sum (lit. evaluate in base-1)
(…)= is it (0/1) equal to…
↑ take that many transactions from the left
+/ sum them
+/ sum that Boolean list to get the count of truths
answered Feb 7 at 15:08
AdámAdám
28k273199
edited Feb 7 at 17:58
answered Feb 7 at 15:08
AdámAdám
28k273199
answered Feb 7 at 15:08
AdámAdám
28k273199
answered Feb 7 at 15:08
AdámAdám
28k273199
28k273199
add a comment |
add a comment |
$begingroup$
Batch, 84 bytes
@set s=%*
@set/as=%s: =+%,c=0
@for %%n in (0 %*)do @set/as-=%%n*2,c+=!s
@echo %c%
Takes input as command-line arguments. Explanation:
@set s=%*
Join the arguments with spaces.
@set/as=%s: =+%,c=0
Replace the spaces with +s and evaluate the result. Also clear the count.
@for %%n in (0 %*)do @set/as-=%%n*2,c+=!s
For each amount, subtract double that from the sum. If the result is zero, then this is a valid match, so increment the count. The extra zero at the beginning allows for a match before any amounts.
@echo %c%
Print the result.
answered Feb 7 at 20:58
NeilNeil
80.7k744178
$endgroup$
add a comment |
$begingroup$
Batch, 84 bytes
@set s=%*
@set/as=%s: =+%,c=0
@for %%n in (0 %*)do @set/as-=%%n*2,c+=!s
@echo %c%
Takes input as command-line arguments. Explanation:
@set s=%*
Join the arguments with spaces.
@set/as=%s: =+%,c=0
Replace the spaces with +s and evaluate the result. Also clear the count.
@for %%n in (0 %*)do @set/as-=%%n*2,c+=!s
For each amount, subtract double that from the sum. If the result is zero, then this is a valid match, so increment the count. The extra zero at the beginning allows for a match before any amounts.
@echo %c%
Print the result.
answered Feb 7 at 20:58
NeilNeil
80.7k744178
$endgroup$
add a comment |
$begingroup$
Batch, 84 bytes
@set s=%*
@set/as=%s: =+%,c=0
@for %%n in (0 %*)do @set/as-=%%n*2,c+=!s
@echo %c%
Takes input as command-line arguments. Explanation:
@set s=%*
Join the arguments with spaces.
@set/as=%s: =+%,c=0
Replace the spaces with +s and evaluate the result. Also clear the count.
@for %%n in (0 %*)do @set/as-=%%n*2,c+=!s
For each amount, subtract double that from the sum. If the result is zero, then this is a valid match, so increment the count. The extra zero at the beginning allows for a match before any amounts.
@echo %c%
Print the result.
answered Feb 7 at 20:58
NeilNeil
80.7k744178
$endgroup$
Batch, 84 bytes
@set s=%*
@set/as=%s: =+%,c=0
@for %%n in (0 %*)do @set/as-=%%n*2,c+=!s
@echo %c%
Takes input as command-line arguments. Explanation:
@set s=%*
Join the arguments with spaces.
@set/as=%s: =+%,c=0
Replace the spaces with +s and evaluate the result. Also clear the count.
@for %%n in (0 %*)do @set/as-=%%n*2,c+=!s
For each amount, subtract double that from the sum. If the result is zero, then this is a valid match, so increment the count. The extra zero at the beginning allows for a match before any amounts.
@echo %c%
Print the result.
answered Feb 7 at 20:58
NeilNeil
80.7k744178
answered Feb 7 at 20:58
NeilNeil
80.7k744178
answered Feb 7 at 20:58
NeilNeil
80.7k744178
answered Feb 7 at 20:58
NeilNeil
80.7k744178
80.7k744178
add a comment |
add a comment |
$begingroup$
Charcoal, 15 bytes
⊞θ⁰ＩΣＥθ⁼Σθ⊗Σ✂θκ
Try it online! Link is to verbose version of code. Explanation:
θ Input list
⁰ Literal 0
⊞ Push to list
θ Augmented list
Ｅ Mapped to
θ Augmented list
✂ Sliced from
κ Current index
Σ Summed
⊗ Doubled
⁼ Equals
θ (Augmented) list
Σ Summed
Σ Sum of booleans
Ｉ Cast to string
Implicitly print
Unfortunately in Charcoal Sum() is not 0 so I have to ensure that there is always at least one element to sum.
answered Feb 7 at 21:11
NeilNeil
80.7k744178
$endgroup$
add a comment |
$begingroup$
Charcoal, 15 bytes
⊞θ⁰ＩΣＥθ⁼Σθ⊗Σ✂θκ
Try it online! Link is to verbose version of code. Explanation:
θ Input list
⁰ Literal 0
⊞ Push to list
θ Augmented list
Ｅ Mapped to
θ Augmented list
✂ Sliced from
κ Current index
Σ Summed
⊗ Doubled
⁼ Equals
θ (Augmented) list
Σ Summed
Σ Sum of booleans
Ｉ Cast to string
Implicitly print
Unfortunately in Charcoal Sum() is not 0 so I have to ensure that there is always at least one element to sum.
answered Feb 7 at 21:11
NeilNeil
80.7k744178
$endgroup$
add a comment |
$begingroup$
Charcoal, 15 bytes
⊞θ⁰ＩΣＥθ⁼Σθ⊗Σ✂θκ
Try it online! Link is to verbose version of code. Explanation:
θ Input list
⁰ Literal 0
⊞ Push to list
θ Augmented list
Ｅ Mapped to
θ Augmented list
✂ Sliced from
κ Current index
Σ Summed
⊗ Doubled
⁼ Equals
θ (Augmented) list
Σ Summed
Σ Sum of booleans
Ｉ Cast to string
Implicitly print
Unfortunately in Charcoal Sum() is not 0 so I have to ensure that there is always at least one element to sum.
answered Feb 7 at 21:11
NeilNeil
80.7k744178
$endgroup$
Charcoal, 15 bytes
⊞θ⁰ＩΣＥθ⁼Σθ⊗Σ✂θκ
Try it online! Link is to verbose version of code. Explanation:
θ Input list
⁰ Literal 0
⊞ Push to list
θ Augmented list
Ｅ Mapped to
θ Augmented list
✂ Sliced from
κ Current index
Σ Summed
⊗ Doubled
⁼ Equals
θ (Augmented) list
Σ Summed
Σ Sum of booleans
Ｉ Cast to string
Implicitly print
Unfortunately in Charcoal Sum() is not 0 so I have to ensure that there is always at least one element to sum.
answered Feb 7 at 21:11
NeilNeil
80.7k744178
answered Feb 7 at 21:11
NeilNeil
80.7k744178
answered Feb 7 at 21:11
NeilNeil
80.7k744178
answered Feb 7 at 21:11
NeilNeil
80.7k744178
80.7k744178
add a comment |
add a comment |
$begingroup$
Python 3, 67 58 bytes
lambda l:sum(sum(l[:x])*2==sum(l)for x in range(len(l)+1))
Try it online!
-9 bytes thanks to @Don't be a x-triple dot
answered Feb 7 at 15:43
Black Owl KaiBlack Owl Kai
69011
$endgroup$
1
$begingroup$
Summing instead of filtering will save you 7 bytes:lambda l:sum(sum(l[:x])==sum(l[x:])for x in range(len(l)+1)).
$endgroup$
– Don't be a x-triple dot
Feb 7 at 18:43
$begingroup$
sum(l[:x])*2==sum(l)saves you another 2 bytes.
$endgroup$
– Neil
Feb 7 at 20:49
add a comment |
$begingroup$
Python 3, 67 58 bytes
lambda l:sum(sum(l[:x])*2==sum(l)for x in range(len(l)+1))
Try it online!
-9 bytes thanks to @Don't be a x-triple dot
answered Feb 7 at 15:43
Black Owl KaiBlack Owl Kai
69011
$endgroup$
1
$begingroup$
Summing instead of filtering will save you 7 bytes:lambda l:sum(sum(l[:x])==sum(l[x:])for x in range(len(l)+1)).
$endgroup$
– Don't be a x-triple dot
Feb 7 at 18:43
$begingroup$
sum(l[:x])*2==sum(l)saves you another 2 bytes.
$endgroup$
– Neil
Feb 7 at 20:49
add a comment |
$begingroup$
Python 3, 67 58 bytes
lambda l:sum(sum(l[:x])*2==sum(l)for x in range(len(l)+1))
Try it online!
-9 bytes thanks to @Don't be a x-triple dot
answered Feb 7 at 15:43
Black Owl KaiBlack Owl Kai
69011
$endgroup$
Python 3, 67 58 bytes
lambda l:sum(sum(l[:x])*2==sum(l)for x in range(len(l)+1))
Try it online!
-9 bytes thanks to @Don't be a x-triple dot
answered Feb 7 at 15:43
Black Owl KaiBlack Owl Kai
69011
edited Feb 7 at 21:23
answered Feb 7 at 15:43
Black Owl KaiBlack Owl Kai
69011
answered Feb 7 at 15:43
Black Owl KaiBlack Owl Kai
69011
answered Feb 7 at 15:43
Black Owl KaiBlack Owl Kai
69011
69011
1
$begingroup$
Summing instead of filtering will save you 7 bytes:lambda l:sum(sum(l[:x])==sum(l[x:])for x in range(len(l)+1)).
$endgroup$
– Don't be a x-triple dot
Feb 7 at 18:43
$begingroup$
sum(l[:x])*2==sum(l)saves you another 2 bytes.
$endgroup$
– Neil
Feb 7 at 20:49
add a comment |
1
$begingroup$
Summing instead of filtering will save you 7 bytes:lambda l:sum(sum(l[:x])==sum(l[x:])for x in range(len(l)+1)).
$endgroup$
– Don't be a x-triple dot
Feb 7 at 18:43
$begingroup$
sum(l[:x])*2==sum(l)saves you another 2 bytes.
$endgroup$
– Neil
Feb 7 at 20:49
1
1
$begingroup$
Summing instead of filtering will save you 7 bytes:
lambda l:sum(sum(l[:x])==sum(l[x:])for x in range(len(l)+1)).$endgroup$
– Don't be a x-triple dot
Feb 7 at 18:43
$begingroup$
Summing instead of filtering will save you 7 bytes:
lambda l:sum(sum(l[:x])==sum(l[x:])for x in range(len(l)+1)).$endgroup$
– Don't be a x-triple dot
Feb 7 at 18:43
$begingroup$
sum(l[:x])*2==sum(l) saves you another 2 bytes.$endgroup$
– Neil
Feb 7 at 20:49
$begingroup$
sum(l[:x])*2==sum(l) saves you another 2 bytes.$endgroup$
– Neil
Feb 7 at 20:49
add a comment |
$begingroup$
R, 50 37 bytes
sum(c(0,cumsum(x<-scan()))==sum(x)/2)
Try it online!
answered Feb 7 at 15:47
Kirill L.Kirill L.
4,3351422
$endgroup$
add a comment |
$begingroup$
R, 50 37 bytes
sum(c(0,cumsum(x<-scan()))==sum(x)/2)
Try it online!
answered Feb 7 at 15:47
Kirill L.Kirill L.
4,3351422
$endgroup$
add a comment |
$begingroup$
R, 50 37 bytes
sum(c(0,cumsum(x<-scan()))==sum(x)/2)
Try it online!
answered Feb 7 at 15:47
Kirill L.Kirill L.
4,3351422
$endgroup$
R, 50 37 bytes
sum(c(0,cumsum(x<-scan()))==sum(x)/2)
Try it online!
answered Feb 7 at 15:47
Kirill L.Kirill L.
4,3351422
edited Feb 8 at 8:36
answered Feb 7 at 15:47
Kirill L.Kirill L.
4,3351422
answered Feb 7 at 15:47
Kirill L.Kirill L.
4,3351422
answered Feb 7 at 15:47
Kirill L.Kirill L.
4,3351422
4,3351422
add a comment |
add a comment |
$begingroup$
MATL, 9 bytes
s0GhYsE=s
Try it online!
Same approach as some other answers: prepend a zero, and check how often half the cumulative sum is equal to the total sum.
s % Total sum of (implicit) input
0Gh % Prepend 0 to another copy of the input
Ys % Cumulative sum
E= % Check element-wise equality of 2*cumulative sum with total sum
s % Sum number of `true` values
answered Feb 8 at 13:39
SanchisesSanchises
5,89212351
$endgroup$
add a comment |
$begingroup$
MATL, 9 bytes
s0GhYsE=s
Try it online!
Same approach as some other answers: prepend a zero, and check how often half the cumulative sum is equal to the total sum.
s % Total sum of (implicit) input
0Gh % Prepend 0 to another copy of the input
Ys % Cumulative sum
E= % Check element-wise equality of 2*cumulative sum with total sum
s % Sum number of `true` values
answered Feb 8 at 13:39
SanchisesSanchises
5,89212351
$endgroup$
add a comment |
$begingroup$
MATL, 9 bytes
s0GhYsE=s
Try it online!
Same approach as some other answers: prepend a zero, and check how often half the cumulative sum is equal to the total sum.
s % Total sum of (implicit) input
0Gh % Prepend 0 to another copy of the input
Ys % Cumulative sum
E= % Check element-wise equality of 2*cumulative sum with total sum
s % Sum number of `true` values
answered Feb 8 at 13:39
SanchisesSanchises
5,89212351
$endgroup$
MATL, 9 bytes
s0GhYsE=s
Try it online!
Same approach as some other answers: prepend a zero, and check how often half the cumulative sum is equal to the total sum.
s % Total sum of (implicit) input
0Gh % Prepend 0 to another copy of the input
Ys % Cumulative sum
E= % Check element-wise equality of 2*cumulative sum with total sum
s % Sum number of `true` values
answered Feb 8 at 13:39
SanchisesSanchises
5,89212351
answered Feb 8 at 13:39
SanchisesSanchises
5,89212351
answered Feb 8 at 13:39
SanchisesSanchises
5,89212351
answered Feb 8 at 13:39
SanchisesSanchises
5,89212351
5,89212351
add a comment |
add a comment |
$begingroup$
Japt -x, 14 11 bytes
iT å+ ®¥nUx
Try it
iT å+ ®¥nUx :Implicit input of array U
i :Prepend
T : Zero
å+ :Cumulatively reduce by addition
® :Map each Z
¥ : Test for equality with
n : Z subtracted from
Ux : U reduced by addition
:Implicitly reduce by addition and output
answered Feb 7 at 15:14
ShaggyShaggy
19.9k21667
$endgroup$
add a comment |
$begingroup$
Japt -x, 14 11 bytes
iT å+ ®¥nUx
Try it
iT å+ ®¥nUx :Implicit input of array U
i :Prepend
T : Zero
å+ :Cumulatively reduce by addition
® :Map each Z
¥ : Test for equality with
n : Z subtracted from
Ux : U reduced by addition
:Implicitly reduce by addition and output
answered Feb 7 at 15:14
ShaggyShaggy
19.9k21667
$endgroup$
add a comment |
$begingroup$
Japt -x, 14 11 bytes
iT å+ ®¥nUx
Try it
iT å+ ®¥nUx :Implicit input of array U
i :Prepend
T : Zero
å+ :Cumulatively reduce by addition
® :Map each Z
¥ : Test for equality with
n : Z subtracted from
Ux : U reduced by addition
:Implicitly reduce by addition and output
answered Feb 7 at 15:14
ShaggyShaggy
19.9k21667
$endgroup$
Japt -x, 14 11 bytes
iT å+ ®¥nUx
Try it
iT å+ ®¥nUx :Implicit input of array U
i :Prepend
T : Zero
å+ :Cumulatively reduce by addition
® :Map each Z
¥ : Test for equality with
n : Z subtracted from
Ux : U reduced by addition
:Implicitly reduce by addition and output
answered Feb 7 at 15:14
ShaggyShaggy
19.9k21667
edited Feb 8 at 16:46
answered Feb 7 at 15:14
ShaggyShaggy
19.9k21667
answered Feb 7 at 15:14
ShaggyShaggy
19.9k21667
answered Feb 7 at 15:14
ShaggyShaggy
19.9k21667
19.9k21667
add a comment |
add a comment |
$begingroup$
PowerShell, 88 82 bytes
-6 Bytes thanks to mazzy
param($n)0..($x=$n.length)|%{$i+=+$z-eq($n[$_..$x]|measure -Su).sum;$z+=$n[$_]};$i
Try it online!
This seems like a very clumsy method but it got the job done. I'll try and revamp it in the future.
answered Feb 8 at 4:46
VeskahVeskah
93814
$endgroup$
1
$begingroup$
you can write$i+=<predicate>insteadif(<predicate>){$i++}
$endgroup$
– mazzy
Feb 8 at 6:15
add a comment |
$begingroup$
PowerShell, 88 82 bytes
-6 Bytes thanks to mazzy
param($n)0..($x=$n.length)|%{$i+=+$z-eq($n[$_..$x]|measure -Su).sum;$z+=$n[$_]};$i
Try it online!
This seems like a very clumsy method but it got the job done. I'll try and revamp it in the future.
answered Feb 8 at 4:46
VeskahVeskah
93814
$endgroup$
1
$begingroup$
you can write$i+=<predicate>insteadif(<predicate>){$i++}
$endgroup$
– mazzy
Feb 8 at 6:15
add a comment |
$begingroup$
PowerShell, 88 82 bytes
-6 Bytes thanks to mazzy
param($n)0..($x=$n.length)|%{$i+=+$z-eq($n[$_..$x]|measure -Su).sum;$z+=$n[$_]};$i
Try it online!
This seems like a very clumsy method but it got the job done. I'll try and revamp it in the future.
answered Feb 8 at 4:46
VeskahVeskah
93814
$endgroup$
PowerShell, 88 82 bytes
-6 Bytes thanks to mazzy
param($n)0..($x=$n.length)|%{$i+=+$z-eq($n[$_..$x]|measure -Su).sum;$z+=$n[$_]};$i
Try it online!
This seems like a very clumsy method but it got the job done. I'll try and revamp it in the future.
answered Feb 8 at 4:46
VeskahVeskah
93814
edited Feb 8 at 21:18
answered Feb 8 at 4:46
VeskahVeskah
93814
answered Feb 8 at 4:46
VeskahVeskah
93814
answered Feb 8 at 4:46
VeskahVeskah
93814
93814
1
$begingroup$
you can write$i+=<predicate>insteadif(<predicate>){$i++}
$endgroup$
– mazzy
Feb 8 at 6:15
add a comment |
1
$begingroup$
you can write$i+=<predicate>insteadif(<predicate>){$i++}
$endgroup$
– mazzy
Feb 8 at 6:15
1
1
$begingroup$
you can write
$i+=<predicate> instead if(<predicate>){$i++}$endgroup$
– mazzy
Feb 8 at 6:15
$begingroup$
you can write
$i+=<predicate> instead if(<predicate>){$i++}$endgroup$
– mazzy
Feb 8 at 6:15
add a comment |
$begingroup$
bash, 52 bytes
IFS=+;for i in 0 $@;{((c+=2*(x+=i)=="$*"));};echo $c
TIO
The trick: setting IFS=+, "$*" expands to a string where arguments are delimited by +, in arithmetic expression eval to the sum
answered Feb 7 at 20:25
Nahuel FouilleulNahuel Fouilleul
2,30529
$endgroup$
add a comment |
$begingroup$
bash, 52 bytes
IFS=+;for i in 0 $@;{((c+=2*(x+=i)=="$*"));};echo $c
TIO
The trick: setting IFS=+, "$*" expands to a string where arguments are delimited by +, in arithmetic expression eval to the sum
answered Feb 7 at 20:25
Nahuel FouilleulNahuel Fouilleul
2,30529
$endgroup$
add a comment |
$begingroup$
bash, 52 bytes
IFS=+;for i in 0 $@;{((c+=2*(x+=i)=="$*"));};echo $c
TIO
The trick: setting IFS=+, "$*" expands to a string where arguments are delimited by +, in arithmetic expression eval to the sum
answered Feb 7 at 20:25
Nahuel FouilleulNahuel Fouilleul
2,30529
$endgroup$
bash, 52 bytes
IFS=+;for i in 0 $@;{((c+=2*(x+=i)=="$*"));};echo $c
TIO
The trick: setting IFS=+, "$*" expands to a string where arguments are delimited by +, in arithmetic expression eval to the sum
answered Feb 7 at 20:25
Nahuel FouilleulNahuel Fouilleul
2,30529
edited Feb 8 at 10:30
answered Feb 7 at 20:25
Nahuel FouilleulNahuel Fouilleul
2,30529
answered Feb 7 at 20:25
Nahuel FouilleulNahuel Fouilleul
2,30529
answered Feb 7 at 20:25
Nahuel FouilleulNahuel Fouilleul
2,30529
2,30529
add a comment |
add a comment |
$begingroup$
PowerShell, 49 45 36 bytes
(($args+0|%{2*($s+=$_)})-eq$s).count
Try it online!
answered Feb 8 at 17:24
mazzymazzy
2,4651316
$endgroup$
add a comment |
$begingroup$
PowerShell, 49 45 36 bytes
(($args+0|%{2*($s+=$_)})-eq$s).count
Try it online!
answered Feb 8 at 17:24
mazzymazzy
2,4651316
$endgroup$
add a comment |
$begingroup$
PowerShell, 49 45 36 bytes
(($args+0|%{2*($s+=$_)})-eq$s).count
Try it online!
answered Feb 8 at 17:24
mazzymazzy
2,4651316
$endgroup$
PowerShell, 49 45 36 bytes
(($args+0|%{2*($s+=$_)})-eq$s).count
Try it online!
answered Feb 8 at 17:24
mazzymazzy
2,4651316
edited Feb 9 at 10:35
answered Feb 8 at 17:24
mazzymazzy
2,4651316
answered Feb 8 at 17:24
mazzymazzy
2,4651316
answered Feb 8 at 17:24
mazzymazzy
2,4651316
2,4651316
add a comment |
add a comment |
$begingroup$
Brachylog, 9 bytes
Not as good as day 1. This one loses to Jelly
{~c₂+ᵐ=}ᶜ
Explanation
{ }ᶜ # Count the ways to:
~c₂ # Split the input array in 2 ...
+ᵐ # so that their sums ...
= # are equal
Test suite:
Try it online!
answered Feb 10 at 12:47
KroppebKroppeb
1,206210
$endgroup$
add a comment |
$begingroup$
Brachylog, 9 bytes
Not as good as day 1. This one loses to Jelly
{~c₂+ᵐ=}ᶜ
Explanation
{ }ᶜ # Count the ways to:
~c₂ # Split the input array in 2 ...
+ᵐ # so that their sums ...
= # are equal
Test suite:
Try it online!
answered Feb 10 at 12:47
KroppebKroppeb
1,206210
$endgroup$
add a comment |
$begingroup$
Brachylog, 9 bytes
Not as good as day 1. This one loses to Jelly
{~c₂+ᵐ=}ᶜ
Explanation
{ }ᶜ # Count the ways to:
~c₂ # Split the input array in 2 ...
+ᵐ # so that their sums ...
= # are equal
Test suite:
Try it online!
answered Feb 10 at 12:47
KroppebKroppeb
1,206210
$endgroup$
Brachylog, 9 bytes
Not as good as day 1. This one loses to Jelly
{~c₂+ᵐ=}ᶜ
Explanation
{ }ᶜ # Count the ways to:
~c₂ # Split the input array in 2 ...
+ᵐ # so that their sums ...
= # are equal
Test suite:
Try it online!
answered Feb 10 at 12:47
KroppebKroppeb
1,206210
answered Feb 10 at 12:47
KroppebKroppeb
1,206210
answered Feb 10 at 12:47
KroppebKroppeb
1,206210
answered Feb 10 at 12:47
KroppebKroppeb
1,206210
1,206210
add a comment |
add a comment |
$begingroup$
Haskell, 46 35 bytes
f x=sum[1|a<-scanl(+)0x,a==sum x/2]
Try it online!
answered Feb 10 at 16:09
niminimi
31.8k32285
$endgroup$
add a comment |
$begingroup$
Haskell, 46 35 bytes
f x=sum[1|a<-scanl(+)0x,a==sum x/2]
Try it online!
answered Feb 10 at 16:09
niminimi
31.8k32285
$endgroup$
add a comment |
$begingroup$
Haskell, 46 35 bytes
f x=sum[1|a<-scanl(+)0x,a==sum x/2]
Try it online!
answered Feb 10 at 16:09
niminimi
31.8k32285
$endgroup$
Haskell, 46 35 bytes
f x=sum[1|a<-scanl(+)0x,a==sum x/2]
Try it online!
answered Feb 10 at 16:09
niminimi
31.8k32285
edited Feb 10 at 18:37
answered Feb 10 at 16:09
niminimi
31.8k32285
answered Feb 10 at 16:09
niminimi
31.8k32285
answered Feb 10 at 16:09
niminimi
31.8k32285
31.8k32285
add a comment |
add a comment |
$begingroup$
J, 19 bytes
1#.[:(={:-])0+/@,]
Try it online!
explanation
1 #. [: (= ({: - ])) 0 +/@, ]
0 , ] NB. prepend 0 to input...
+/@ NB. and take the prefix sums...
[: ({: - ]) NB. then subtract that list
NB. from its final elm
NB. (`{:`), giving the list
NB. of suffix sums...
[: (= ( )) NB. create a 1-0 list showing
NB. where the prefix sums
NB. equal the suffix sums
1 #. NB. and take the sum.
answered 2 days ago
JonahJonah
2,121916
$endgroup$
add a comment |
$begingroup$
J, 19 bytes
1#.[:(={:-])0+/@,]
Try it online!
explanation
1 #. [: (= ({: - ])) 0 +/@, ]
0 , ] NB. prepend 0 to input...
+/@ NB. and take the prefix sums...
[: ({: - ]) NB. then subtract that list
NB. from its final elm
NB. (`{:`), giving the list
NB. of suffix sums...
[: (= ( )) NB. create a 1-0 list showing
NB. where the prefix sums
NB. equal the suffix sums
1 #. NB. and take the sum.
answered 2 days ago
JonahJonah
2,121916
$endgroup$
add a comment |
$begingroup$
J, 19 bytes
1#.[:(={:-])0+/@,]
Try it online!
explanation
1 #. [: (= ({: - ])) 0 +/@, ]
0 , ] NB. prepend 0 to input...
+/@ NB. and take the prefix sums...
[: ({: - ]) NB. then subtract that list
NB. from its final elm
NB. (`{:`), giving the list
NB. of suffix sums...
[: (= ( )) NB. create a 1-0 list showing
NB. where the prefix sums
NB. equal the suffix sums
1 #. NB. and take the sum.
answered 2 days ago
JonahJonah
2,121916
$endgroup$
J, 19 bytes
1#.[:(={:-])0+/@,]
Try it online!
explanation
1 #. [: (= ({: - ])) 0 +/@, ]
0 , ] NB. prepend 0 to input...
+/@ NB. and take the prefix sums...
[: ({: - ]) NB. then subtract that list
NB. from its final elm
NB. (`{:`), giving the list
NB. of suffix sums...
[: (= ( )) NB. create a 1-0 list showing
NB. where the prefix sums
NB. equal the suffix sums
1 #. NB. and take the sum.
answered 2 days ago
JonahJonah
2,121916
edited yesterday
answered 2 days ago
JonahJonah
2,121916
answered 2 days ago
JonahJonah
2,121916
answered 2 days ago
JonahJonah
2,121916
2,121916
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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11
$begingroup$
After 6 months of frozen and newly created accounts, it is reported that Marie's banker is now interned in a sanitarium. "We are your friends. You need some rest", they said.
$endgroup$
– Arnauld
Feb 7 at 17:20
2
$begingroup$
Suggested test case of a single transaction
$endgroup$
– Veskah
Feb 8 at 4:53