How to use nPn in Burst Balloon problem with the given condition?
I was trying to understand this problem Burst Balloon . Mainly selecting the most efficient balloon to burst from the given condition. Given conditions are :
Below is the condition to calculate the score.
When Balloon $B_i$ Explodes then score will be a product of $B_{i-1}$ & $B_{i+1}$ (score ${}= B_{i-1}times B_{i+1}$).
When Balloon $B_i$ Explodes and there is only left Balloon present then score will be $B_{i-1}$.
When Balloon $B_i$ Explodes and there is only right Balloon present then score will be $B_{i+1}$.
When Balloon $B_i$ explodes and there is no left and right Balloon present then score will be $B_i$.
Example:
Input: B = {1, 2, 3, 4}
Output: 20
Explanation:
For max score:
3 explodes, score= 4*2=8 (product of adjacent balloons)
2 explodes score= 4*1 + 8 = 12 (product of adjacent balloons)
1 explodes score= 4 + 12= 16 (only 4 is left on the left side)
4 explodes score = 4 + 16 = 20 (no balloons left so add 4)
score =20
other combinations will result in lesser scores.
No sorting is allowed
But when I tried to apply this on this input n = 6
values = 7, 10, 8, 20, 50, 9.
In my opinion it should be = 1331
by using 20, 8, 10, 50, 7, 9 ballons.
I've used this nPn ways (1st balloon N ways, 2nd N-1 ways …last balloon 1 ways N*(N-1)(N-2)..2*1= N! to select balloons.
But from the geekforgeeks code this output is 1400. I'm unable to understand how can be this 1400? I know this is not pure mathematical problem because it's programming related problem.
But I'm confused how this combination is actually selected
algorithms permutations combinations programming
asked Nov 23 '18 at 8:22
Anzum AAnzum A
85
add a comment |
I was trying to understand this problem Burst Balloon . Mainly selecting the most efficient balloon to burst from the given condition. Given conditions are :
Below is the condition to calculate the score.
When Balloon $B_i$ Explodes then score will be a product of $B_{i-1}$ & $B_{i+1}$ (score ${}= B_{i-1}times B_{i+1}$).
When Balloon $B_i$ Explodes and there is only left Balloon present then score will be $B_{i-1}$.
When Balloon $B_i$ Explodes and there is only right Balloon present then score will be $B_{i+1}$.
When Balloon $B_i$ explodes and there is no left and right Balloon present then score will be $B_i$.
Example:
Input: B = {1, 2, 3, 4}
Output: 20
Explanation:
For max score:
3 explodes, score= 4*2=8 (product of adjacent balloons)
2 explodes score= 4*1 + 8 = 12 (product of adjacent balloons)
1 explodes score= 4 + 12= 16 (only 4 is left on the left side)
4 explodes score = 4 + 16 = 20 (no balloons left so add 4)
score =20
other combinations will result in lesser scores.
No sorting is allowed
But when I tried to apply this on this input n = 6
values = 7, 10, 8, 20, 50, 9.
In my opinion it should be = 1331
by using 20, 8, 10, 50, 7, 9 ballons.
I've used this nPn ways (1st balloon N ways, 2nd N-1 ways …last balloon 1 ways N*(N-1)(N-2)..2*1= N! to select balloons.
But from the geekforgeeks code this output is 1400. I'm unable to understand how can be this 1400? I know this is not pure mathematical problem because it's programming related problem.
But I'm confused how this combination is actually selected
algorithms permutations combinations programming
asked Nov 23 '18 at 8:22
Anzum AAnzum A
85
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 23 '18 at 8:37
add a comment |
I was trying to understand this problem Burst Balloon . Mainly selecting the most efficient balloon to burst from the given condition. Given conditions are :
Below is the condition to calculate the score.
When Balloon $B_i$ Explodes then score will be a product of $B_{i-1}$ & $B_{i+1}$ (score ${}= B_{i-1}times B_{i+1}$).
When Balloon $B_i$ Explodes and there is only left Balloon present then score will be $B_{i-1}$.
When Balloon $B_i$ Explodes and there is only right Balloon present then score will be $B_{i+1}$.
When Balloon $B_i$ explodes and there is no left and right Balloon present then score will be $B_i$.
Example:
Input: B = {1, 2, 3, 4}
Output: 20
Explanation:
For max score:
3 explodes, score= 4*2=8 (product of adjacent balloons)
2 explodes score= 4*1 + 8 = 12 (product of adjacent balloons)
1 explodes score= 4 + 12= 16 (only 4 is left on the left side)
4 explodes score = 4 + 16 = 20 (no balloons left so add 4)
score =20
other combinations will result in lesser scores.
No sorting is allowed
But when I tried to apply this on this input n = 6
values = 7, 10, 8, 20, 50, 9.
In my opinion it should be = 1331
by using 20, 8, 10, 50, 7, 9 ballons.
I've used this nPn ways (1st balloon N ways, 2nd N-1 ways …last balloon 1 ways N*(N-1)(N-2)..2*1= N! to select balloons.
But from the geekforgeeks code this output is 1400. I'm unable to understand how can be this 1400? I know this is not pure mathematical problem because it's programming related problem.
But I'm confused how this combination is actually selected
algorithms permutations combinations programming
asked Nov 23 '18 at 8:22
Anzum AAnzum A
85
I was trying to understand this problem Burst Balloon . Mainly selecting the most efficient balloon to burst from the given condition. Given conditions are :
Below is the condition to calculate the score.
When Balloon $B_i$ Explodes then score will be a product of $B_{i-1}$ & $B_{i+1}$ (score ${}= B_{i-1}times B_{i+1}$).
When Balloon $B_i$ Explodes and there is only left Balloon present then score will be $B_{i-1}$.
When Balloon $B_i$ Explodes and there is only right Balloon present then score will be $B_{i+1}$.
When Balloon $B_i$ explodes and there is no left and right Balloon present then score will be $B_i$.
Example:
Input: B = {1, 2, 3, 4}
Output: 20
Explanation:
For max score:
3 explodes, score= 4*2=8 (product of adjacent balloons)
2 explodes score= 4*1 + 8 = 12 (product of adjacent balloons)
1 explodes score= 4 + 12= 16 (only 4 is left on the left side)
4 explodes score = 4 + 16 = 20 (no balloons left so add 4)
score =20
other combinations will result in lesser scores.
No sorting is allowed
But when I tried to apply this on this input n = 6
values = 7, 10, 8, 20, 50, 9.
In my opinion it should be = 1331
by using 20, 8, 10, 50, 7, 9 ballons.
I've used this nPn ways (1st balloon N ways, 2nd N-1 ways …last balloon 1 ways N*(N-1)(N-2)..2*1= N! to select balloons.
But from the geekforgeeks code this output is 1400. I'm unable to understand how can be this 1400? I know this is not pure mathematical problem because it's programming related problem.
But I'm confused how this combination is actually selected
algorithms permutations combinations programming
algorithms permutations combinations programming
asked Nov 23 '18 at 8:22
Anzum AAnzum A
85
asked Nov 23 '18 at 8:22
Anzum AAnzum A
85
edited Nov 24 '18 at 7:10
Anzum A
asked Nov 23 '18 at 8:22
Anzum AAnzum A
85
asked Nov 23 '18 at 8:22
Anzum AAnzum A
85
asked Nov 23 '18 at 8:22
Anzum AAnzum A
85
85
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 23 '18 at 8:37
add a comment |
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 23 '18 at 8:37
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 23 '18 at 8:37
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 23 '18 at 8:37
add a comment |
1 Answer
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Your last three moves are wrong. The position before these moves is ${7,50,9}$ and you burst $50$ scoring $63$, then $7$ scoring $9$, then $9$ scoring $9$, for $81$. But you can do better by bursting $7$ (scoring $50$) then $9$ (scoring $50$) then $50$ (scoring $50$) for $150$.
The point here is that it isn't always best to make the highest scoring move, because it may remove good options for future moves. (In other words, the greedy algorithm is not optimal.)
answered Nov 23 '18 at 8:30
Especially LimeEspecially Lime
21.7k22858
okay, I got this. This is a greedy problem
– Anzum A
Nov 23 '18 at 8:51
add a comment |
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Your last three moves are wrong. The position before these moves is ${7,50,9}$ and you burst $50$ scoring $63$, then $7$ scoring $9$, then $9$ scoring $9$, for $81$. But you can do better by bursting $7$ (scoring $50$) then $9$ (scoring $50$) then $50$ (scoring $50$) for $150$.
The point here is that it isn't always best to make the highest scoring move, because it may remove good options for future moves. (In other words, the greedy algorithm is not optimal.)
answered Nov 23 '18 at 8:30
Especially LimeEspecially Lime
21.7k22858
okay, I got this. This is a greedy problem
– Anzum A
Nov 23 '18 at 8:51
add a comment |
Your last three moves are wrong. The position before these moves is ${7,50,9}$ and you burst $50$ scoring $63$, then $7$ scoring $9$, then $9$ scoring $9$, for $81$. But you can do better by bursting $7$ (scoring $50$) then $9$ (scoring $50$) then $50$ (scoring $50$) for $150$.
The point here is that it isn't always best to make the highest scoring move, because it may remove good options for future moves. (In other words, the greedy algorithm is not optimal.)
answered Nov 23 '18 at 8:30
Especially LimeEspecially Lime
21.7k22858
okay, I got this. This is a greedy problem
– Anzum A
Nov 23 '18 at 8:51
add a comment |
Your last three moves are wrong. The position before these moves is ${7,50,9}$ and you burst $50$ scoring $63$, then $7$ scoring $9$, then $9$ scoring $9$, for $81$. But you can do better by bursting $7$ (scoring $50$) then $9$ (scoring $50$) then $50$ (scoring $50$) for $150$.
The point here is that it isn't always best to make the highest scoring move, because it may remove good options for future moves. (In other words, the greedy algorithm is not optimal.)
answered Nov 23 '18 at 8:30
Especially LimeEspecially Lime
21.7k22858
Your last three moves are wrong. The position before these moves is ${7,50,9}$ and you burst $50$ scoring $63$, then $7$ scoring $9$, then $9$ scoring $9$, for $81$. But you can do better by bursting $7$ (scoring $50$) then $9$ (scoring $50$) then $50$ (scoring $50$) for $150$.
The point here is that it isn't always best to make the highest scoring move, because it may remove good options for future moves. (In other words, the greedy algorithm is not optimal.)
answered Nov 23 '18 at 8:30
Especially LimeEspecially Lime
21.7k22858
answered Nov 23 '18 at 8:30
Especially LimeEspecially Lime
21.7k22858
answered Nov 23 '18 at 8:30
Especially LimeEspecially Lime
21.7k22858
answered Nov 23 '18 at 8:30
Especially LimeEspecially Lime
21.7k22858
21.7k22858
okay, I got this. This is a greedy problem
– Anzum A
Nov 23 '18 at 8:51
add a comment |
okay, I got this. This is a greedy problem
– Anzum A
Nov 23 '18 at 8:51
okay, I got this. This is a greedy problem
– Anzum A
Nov 23 '18 at 8:51
okay, I got this. This is a greedy problem
– Anzum A
Nov 23 '18 at 8:51
add a comment |
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 23 '18 at 8:37