Creating a wave of string in Javascript
7
I can't seem to figure it out how to make a wave from a string in Javascript.
Rules:
- The input will always be lower case string.
- Ignore whitespace.
Expected result:
wave("hello") => ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
wave (" h e y ") => [" H e y ", " h E y ", " h e Y "]
wave ("") =>
This is as far as I got. Current code will give me an answer ["hello", "hello", "hello", "hello", "hello"]. I'm thinking using second for loop and somehow capitalize each new letter but I'am stumped. Also I would appreciate if answer would avoid using loop inside loop O(n^2). Because of BIG O Scalability.
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
array.push(str);
}
for (let index = 0; index < str.length; index++) {
console.log(array);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
javascript arrays
edited Mar 4 at 12:24
vahdet
2,06431436
asked Mar 4 at 12:21
Arnas DičkusArnas Dičkus
8719
add a comment |
7
I can't seem to figure it out how to make a wave from a string in Javascript.
Rules:
- The input will always be lower case string.
- Ignore whitespace.
Expected result:
wave("hello") => ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
wave (" h e y ") => [" H e y ", " h E y ", " h e Y "]
wave ("") =>
This is as far as I got. Current code will give me an answer ["hello", "hello", "hello", "hello", "hello"]. I'm thinking using second for loop and somehow capitalize each new letter but I'am stumped. Also I would appreciate if answer would avoid using loop inside loop O(n^2). Because of BIG O Scalability.
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
array.push(str);
}
for (let index = 0; index < str.length; index++) {
console.log(array);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
javascript arrays
edited Mar 4 at 12:24
vahdet
2,06431436
asked Mar 4 at 12:21
Arnas DičkusArnas Dičkus
8719
Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
Mar 4 at 12:28
add a comment |
7
7
7
1
I can't seem to figure it out how to make a wave from a string in Javascript.
Rules:
- The input will always be lower case string.
- Ignore whitespace.
Expected result:
wave("hello") => ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
wave (" h e y ") => [" H e y ", " h E y ", " h e Y "]
wave ("") =>
This is as far as I got. Current code will give me an answer ["hello", "hello", "hello", "hello", "hello"]. I'm thinking using second for loop and somehow capitalize each new letter but I'am stumped. Also I would appreciate if answer would avoid using loop inside loop O(n^2). Because of BIG O Scalability.
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
array.push(str);
}
for (let index = 0; index < str.length; index++) {
console.log(array);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
javascript arrays
edited Mar 4 at 12:24
vahdet
2,06431436
asked Mar 4 at 12:21
Arnas DičkusArnas Dičkus
8719
I can't seem to figure it out how to make a wave from a string in Javascript.
Rules:
- The input will always be lower case string.
- Ignore whitespace.
Expected result:
wave("hello") => ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
wave (" h e y ") => [" H e y ", " h E y ", " h e Y "]
wave ("") =>
This is as far as I got. Current code will give me an answer ["hello", "hello", "hello", "hello", "hello"]. I'm thinking using second for loop and somehow capitalize each new letter but I'am stumped. Also I would appreciate if answer would avoid using loop inside loop O(n^2). Because of BIG O Scalability.
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
array.push(str);
}
for (let index = 0; index < str.length; index++) {
console.log(array);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
javascript arrays
javascript arrays
edited Mar 4 at 12:24
vahdet
2,06431436
asked Mar 4 at 12:21
Arnas DičkusArnas Dičkus
8719
edited Mar 4 at 12:24
vahdet
2,06431436
asked Mar 4 at 12:21
Arnas DičkusArnas Dičkus
8719
edited Mar 4 at 12:24
vahdet
2,06431436
edited Mar 4 at 12:24
vahdet
2,06431436
edited Mar 4 at 12:24
vahdet
2,06431436
2,06431436
asked Mar 4 at 12:21
Arnas DičkusArnas Dičkus
8719
asked Mar 4 at 12:21
Arnas DičkusArnas Dičkus
8719
asked Mar 4 at 12:21
Arnas DičkusArnas Dičkus
8719
8719
Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
Mar 4 at 12:28
add a comment |
Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
Mar 4 at 12:28
Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
Mar 4 at 12:28
Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
Mar 4 at 12:28
add a comment |
5 Answers
5
active
oldest
votes
8
You could take an outer loop for visiting the characters and if a non space character is found, create a new string with an uppercase letter at this position.
function wave(string) {
var result = ,
i;
for (i = 0; i < string.length; i++) {
if (string[i] === ' ') continue;
result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join(''));
}
return result;
}
console.log(wave("hello")); // ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
console.log(wave(" h e y ")); // [" H e y ", " h E y ", " h e Y "]
console.log(wave("")); // .as-console-wrapper { max-height: 100% !important; top: 0; }answered Mar 4 at 12:32
Nina ScholzNina Scholz
190k15100174
add a comment |
1
This does it. However, if you have spaces in your string, it will output string without any "waved letter" (since also space is handled):
const wave = (str = '') => {
return str
.split('')
.map((letter, i, arr) => `${arr.slice(0, i)}${letter.toUpperCase()}${arr.slice(i + 1, arr.length)}`.replace(/,/g, ''));
}
console.log(wave('wave'));
console.log(wave('foo bar'));
console.log(wave());answered Mar 4 at 12:37
zvonazvona
12.6k14059
add a comment |
1
You can take each char in string in for loop and make it uppercase and the append with prefix and post fix string
var array=;
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
if (str[index] === ' ') continue;
waveChar=str.charAt(index).toUpperCase();
preStr=str.substring(0,index);
postStr=str.substring(index,str.length-1);
array.push(preStr+waveChar+postStr);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
console.log(array);answered Mar 4 at 12:40
Code_ModeCode_Mode
1,212816
add a comment |
0
I use tradicional js. This works on 99% off today browsers.
Where answer very pragmatic. I use array access for string ;
Magic is "String.fromCharCode(str.charCodeAt(x) ^ 32);"
Make it inverse always when we call this line.
var index = 0;
var mydata= "hello";
function wave (str){
var FINAL = ;
var newString = "";
for (var j =0; j < str.length; j++) {
for (var x =0; x < str.length; x++) {
var rez = "";
if (x == index) {
rez = String.fromCharCode(str.charCodeAt(x) ^ 32);
} else {
rez = str[x];
}
newString += rez ;
}
FINAL.push(newString);
newString = "";
index++;
}
return FINAL;
}
var rezArray = wave(mydata);
console.log(rezArray);
// Just for teory
console.log(Array.from([1, 2, 3], x => console.log(x)));answered Mar 4 at 13:08
Nikola LukicNikola Lukic
1,75721937
Your answer uses 2 for loops inside each other. Which is Bad for Scalability O(n^2). According to BIG O notations.BIG O Cheatsheet
– Arnas Dičkus
Mar 4 at 13:12
You must understand that " for (i = 0; i < string.length; i++) { if (string[i] === ' ') continue; result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); }" is ALSO loop in twice. ;) When you call this line : Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); you work with iterator.
– Nikola Lukic
Mar 4 at 13:18
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
Mar 4 at 13:19
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
Mar 4 at 13:24
2
"String.fromCharCode(str.charCodeAt(x) ^ 32)" is too much magic. It's a nice trick, but works only on ASCII characters. Don't do that. There's no reason not to use.toUpperCase().
– Bergi
Mar 4 at 18:14
|
show 3 more comments
0
I use modify of String prototype :
for implementation of replaceAt .
// Modify prototype
String.prototype.replaceAt=function(index, replacement) {
return this.substr(0, index) + replacement+ this.substr(index + replacement.length);
}
function WaveFunction(str) {
var base = str;
var R = ;
var n =str.length;
for (var i=0 ; i<n ;i++){
str = base;
var c=str.charAt(i);
var res = c.toUpperCase();
// console.log(res);
str = str.replaceAt(i, res);
R.push(str);
}
return R;
}
var REZ = WaveFunction("hello");
console.log(REZ);edited Mar 4 at 14:23
Nikola Lukic
1,75721937
answered Mar 4 at 12:31
Peter HassaballahPeter Hassaballah
825
1
Nice solution with replaceAt !
– Nikola Lukic
Mar 4 at 13:23
where isreplaceAtfrom?
– Nina Scholz
Mar 4 at 13:39
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
Mar 4 at 13:48
1
String.prototype.replaceAt=function(index, replacement) { return this.substr(0, index) + replacement+ this.substr(index + replacement.length); } I am sorry i forgot to add it to my answer :)
– Peter Hassaballah
Mar 4 at 14:22
1
Don't modify objects you don't own.
– Emile Bergeron
Mar 4 at 14:51
|
show 1 more comment
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
8
You could take an outer loop for visiting the characters and if a non space character is found, create a new string with an uppercase letter at this position.
function wave(string) {
var result = ,
i;
for (i = 0; i < string.length; i++) {
if (string[i] === ' ') continue;
result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join(''));
}
return result;
}
console.log(wave("hello")); // ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
console.log(wave(" h e y ")); // [" H e y ", " h E y ", " h e Y "]
console.log(wave("")); // .as-console-wrapper { max-height: 100% !important; top: 0; }answered Mar 4 at 12:32
Nina ScholzNina Scholz
190k15100174
add a comment |
8
You could take an outer loop for visiting the characters and if a non space character is found, create a new string with an uppercase letter at this position.
function wave(string) {
var result = ,
i;
for (i = 0; i < string.length; i++) {
if (string[i] === ' ') continue;
result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join(''));
}
return result;
}
console.log(wave("hello")); // ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
console.log(wave(" h e y ")); // [" H e y ", " h E y ", " h e Y "]
console.log(wave("")); // .as-console-wrapper { max-height: 100% !important; top: 0; }answered Mar 4 at 12:32
Nina ScholzNina Scholz
190k15100174
add a comment |
8
8
8
You could take an outer loop for visiting the characters and if a non space character is found, create a new string with an uppercase letter at this position.
function wave(string) {
var result = ,
i;
for (i = 0; i < string.length; i++) {
if (string[i] === ' ') continue;
result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join(''));
}
return result;
}
console.log(wave("hello")); // ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
console.log(wave(" h e y ")); // [" H e y ", " h E y ", " h e Y "]
console.log(wave("")); // .as-console-wrapper { max-height: 100% !important; top: 0; }answered Mar 4 at 12:32
Nina ScholzNina Scholz
190k15100174
You could take an outer loop for visiting the characters and if a non space character is found, create a new string with an uppercase letter at this position.
function wave(string) {
var result = ,
i;
for (i = 0; i < string.length; i++) {
if (string[i] === ' ') continue;
result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join(''));
}
return result;
}
console.log(wave("hello")); // ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
console.log(wave(" h e y ")); // [" H e y ", " h E y ", " h e Y "]
console.log(wave("")); // .as-console-wrapper { max-height: 100% !important; top: 0; }function wave(string) {
var result = ,
i;
for (i = 0; i < string.length; i++) {
if (string[i] === ' ') continue;
result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join(''));
}
return result;
}
console.log(wave("hello")); // ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
console.log(wave(" h e y ")); // [" H e y ", " h E y ", " h e Y "]
console.log(wave("")); // .as-console-wrapper { max-height: 100% !important; top: 0; }function wave(string) {
var result = ,
i;
for (i = 0; i < string.length; i++) {
if (string[i] === ' ') continue;
result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join(''));
}
return result;
}
console.log(wave("hello")); // ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
console.log(wave(" h e y ")); // [" H e y ", " h E y ", " h e Y "]
console.log(wave("")); // .as-console-wrapper { max-height: 100% !important; top: 0; }answered Mar 4 at 12:32
Nina ScholzNina Scholz
190k15100174
answered Mar 4 at 12:32
Nina ScholzNina Scholz
190k15100174
answered Mar 4 at 12:32
Nina ScholzNina Scholz
190k15100174
answered Mar 4 at 12:32
Nina ScholzNina Scholz
190k15100174
190k15100174
add a comment |
add a comment |
1
This does it. However, if you have spaces in your string, it will output string without any "waved letter" (since also space is handled):
const wave = (str = '') => {
return str
.split('')
.map((letter, i, arr) => `${arr.slice(0, i)}${letter.toUpperCase()}${arr.slice(i + 1, arr.length)}`.replace(/,/g, ''));
}
console.log(wave('wave'));
console.log(wave('foo bar'));
console.log(wave());answered Mar 4 at 12:37
zvonazvona
12.6k14059
add a comment |
1
This does it. However, if you have spaces in your string, it will output string without any "waved letter" (since also space is handled):
const wave = (str = '') => {
return str
.split('')
.map((letter, i, arr) => `${arr.slice(0, i)}${letter.toUpperCase()}${arr.slice(i + 1, arr.length)}`.replace(/,/g, ''));
}
console.log(wave('wave'));
console.log(wave('foo bar'));
console.log(wave());answered Mar 4 at 12:37
zvonazvona
12.6k14059
add a comment |
1
1
1
This does it. However, if you have spaces in your string, it will output string without any "waved letter" (since also space is handled):
const wave = (str = '') => {
return str
.split('')
.map((letter, i, arr) => `${arr.slice(0, i)}${letter.toUpperCase()}${arr.slice(i + 1, arr.length)}`.replace(/,/g, ''));
}
console.log(wave('wave'));
console.log(wave('foo bar'));
console.log(wave());answered Mar 4 at 12:37
zvonazvona
12.6k14059
This does it. However, if you have spaces in your string, it will output string without any "waved letter" (since also space is handled):
const wave = (str = '') => {
return str
.split('')
.map((letter, i, arr) => `${arr.slice(0, i)}${letter.toUpperCase()}${arr.slice(i + 1, arr.length)}`.replace(/,/g, ''));
}
console.log(wave('wave'));
console.log(wave('foo bar'));
console.log(wave());const wave = (str = '') => {
return str
.split('')
.map((letter, i, arr) => `${arr.slice(0, i)}${letter.toUpperCase()}${arr.slice(i + 1, arr.length)}`.replace(/,/g, ''));
}
console.log(wave('wave'));
console.log(wave('foo bar'));
console.log(wave());const wave = (str = '') => {
return str
.split('')
.map((letter, i, arr) => `${arr.slice(0, i)}${letter.toUpperCase()}${arr.slice(i + 1, arr.length)}`.replace(/,/g, ''));
}
console.log(wave('wave'));
console.log(wave('foo bar'));
console.log(wave());answered Mar 4 at 12:37
zvonazvona
12.6k14059
answered Mar 4 at 12:37
zvonazvona
12.6k14059
answered Mar 4 at 12:37
zvonazvona
12.6k14059
answered Mar 4 at 12:37
zvonazvona
12.6k14059
12.6k14059
add a comment |
add a comment |
1
You can take each char in string in for loop and make it uppercase and the append with prefix and post fix string
var array=;
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
if (str[index] === ' ') continue;
waveChar=str.charAt(index).toUpperCase();
preStr=str.substring(0,index);
postStr=str.substring(index,str.length-1);
array.push(preStr+waveChar+postStr);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
console.log(array);answered Mar 4 at 12:40
Code_ModeCode_Mode
1,212816
add a comment |
1
You can take each char in string in for loop and make it uppercase and the append with prefix and post fix string
var array=;
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
if (str[index] === ' ') continue;
waveChar=str.charAt(index).toUpperCase();
preStr=str.substring(0,index);
postStr=str.substring(index,str.length-1);
array.push(preStr+waveChar+postStr);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
console.log(array);answered Mar 4 at 12:40
Code_ModeCode_Mode
1,212816
add a comment |
1
1
1
You can take each char in string in for loop and make it uppercase and the append with prefix and post fix string
var array=;
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
if (str[index] === ' ') continue;
waveChar=str.charAt(index).toUpperCase();
preStr=str.substring(0,index);
postStr=str.substring(index,str.length-1);
array.push(preStr+waveChar+postStr);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
console.log(array);answered Mar 4 at 12:40
Code_ModeCode_Mode
1,212816
You can take each char in string in for loop and make it uppercase and the append with prefix and post fix string
var array=;
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
if (str[index] === ' ') continue;
waveChar=str.charAt(index).toUpperCase();
preStr=str.substring(0,index);
postStr=str.substring(index,str.length-1);
array.push(preStr+waveChar+postStr);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
console.log(array);var array=;
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
if (str[index] === ' ') continue;
waveChar=str.charAt(index).toUpperCase();
preStr=str.substring(0,index);
postStr=str.substring(index,str.length-1);
array.push(preStr+waveChar+postStr);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
console.log(array);var array=;
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
if (str[index] === ' ') continue;
waveChar=str.charAt(index).toUpperCase();
preStr=str.substring(0,index);
postStr=str.substring(index,str.length-1);
array.push(preStr+waveChar+postStr);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
console.log(array);answered Mar 4 at 12:40
Code_ModeCode_Mode
1,212816
answered Mar 4 at 12:40
Code_ModeCode_Mode
1,212816
answered Mar 4 at 12:40
Code_ModeCode_Mode
1,212816
answered Mar 4 at 12:40
Code_ModeCode_Mode
1,212816
1,212816
add a comment |
add a comment |
0
I use tradicional js. This works on 99% off today browsers.
Where answer very pragmatic. I use array access for string ;
Magic is "String.fromCharCode(str.charCodeAt(x) ^ 32);"
Make it inverse always when we call this line.
var index = 0;
var mydata= "hello";
function wave (str){
var FINAL = ;
var newString = "";
for (var j =0; j < str.length; j++) {
for (var x =0; x < str.length; x++) {
var rez = "";
if (x == index) {
rez = String.fromCharCode(str.charCodeAt(x) ^ 32);
} else {
rez = str[x];
}
newString += rez ;
}
FINAL.push(newString);
newString = "";
index++;
}
return FINAL;
}
var rezArray = wave(mydata);
console.log(rezArray);
// Just for teory
console.log(Array.from([1, 2, 3], x => console.log(x)));answered Mar 4 at 13:08
Nikola LukicNikola Lukic
1,75721937
Your answer uses 2 for loops inside each other. Which is Bad for Scalability O(n^2). According to BIG O notations.BIG O Cheatsheet
– Arnas Dičkus
Mar 4 at 13:12
You must understand that " for (i = 0; i < string.length; i++) { if (string[i] === ' ') continue; result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); }" is ALSO loop in twice. ;) When you call this line : Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); you work with iterator.
– Nikola Lukic
Mar 4 at 13:18
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
Mar 4 at 13:19
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
Mar 4 at 13:24
2
"String.fromCharCode(str.charCodeAt(x) ^ 32)" is too much magic. It's a nice trick, but works only on ASCII characters. Don't do that. There's no reason not to use.toUpperCase().
– Bergi
Mar 4 at 18:14
|
show 3 more comments
0
I use tradicional js. This works on 99% off today browsers.
Where answer very pragmatic. I use array access for string ;
Magic is "String.fromCharCode(str.charCodeAt(x) ^ 32);"
Make it inverse always when we call this line.
var index = 0;
var mydata= "hello";
function wave (str){
var FINAL = ;
var newString = "";
for (var j =0; j < str.length; j++) {
for (var x =0; x < str.length; x++) {
var rez = "";
if (x == index) {
rez = String.fromCharCode(str.charCodeAt(x) ^ 32);
} else {
rez = str[x];
}
newString += rez ;
}
FINAL.push(newString);
newString = "";
index++;
}
return FINAL;
}
var rezArray = wave(mydata);
console.log(rezArray);
// Just for teory
console.log(Array.from([1, 2, 3], x => console.log(x)));answered Mar 4 at 13:08
Nikola LukicNikola Lukic
1,75721937
Your answer uses 2 for loops inside each other. Which is Bad for Scalability O(n^2). According to BIG O notations.BIG O Cheatsheet
– Arnas Dičkus
Mar 4 at 13:12
You must understand that " for (i = 0; i < string.length; i++) { if (string[i] === ' ') continue; result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); }" is ALSO loop in twice. ;) When you call this line : Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); you work with iterator.
– Nikola Lukic
Mar 4 at 13:18
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
Mar 4 at 13:19
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
Mar 4 at 13:24
2
"String.fromCharCode(str.charCodeAt(x) ^ 32)" is too much magic. It's a nice trick, but works only on ASCII characters. Don't do that. There's no reason not to use.toUpperCase().
– Bergi
Mar 4 at 18:14
|
show 3 more comments
0
0
0
I use tradicional js. This works on 99% off today browsers.
Where answer very pragmatic. I use array access for string ;
Magic is "String.fromCharCode(str.charCodeAt(x) ^ 32);"
Make it inverse always when we call this line.
var index = 0;
var mydata= "hello";
function wave (str){
var FINAL = ;
var newString = "";
for (var j =0; j < str.length; j++) {
for (var x =0; x < str.length; x++) {
var rez = "";
if (x == index) {
rez = String.fromCharCode(str.charCodeAt(x) ^ 32);
} else {
rez = str[x];
}
newString += rez ;
}
FINAL.push(newString);
newString = "";
index++;
}
return FINAL;
}
var rezArray = wave(mydata);
console.log(rezArray);
// Just for teory
console.log(Array.from([1, 2, 3], x => console.log(x)));answered Mar 4 at 13:08
Nikola LukicNikola Lukic
1,75721937
I use tradicional js. This works on 99% off today browsers.
Where answer very pragmatic. I use array access for string ;
Magic is "String.fromCharCode(str.charCodeAt(x) ^ 32);"
Make it inverse always when we call this line.
var index = 0;
var mydata= "hello";
function wave (str){
var FINAL = ;
var newString = "";
for (var j =0; j < str.length; j++) {
for (var x =0; x < str.length; x++) {
var rez = "";
if (x == index) {
rez = String.fromCharCode(str.charCodeAt(x) ^ 32);
} else {
rez = str[x];
}
newString += rez ;
}
FINAL.push(newString);
newString = "";
index++;
}
return FINAL;
}
var rezArray = wave(mydata);
console.log(rezArray);
// Just for teory
console.log(Array.from([1, 2, 3], x => console.log(x)));var index = 0;
var mydata= "hello";
function wave (str){
var FINAL = ;
var newString = "";
for (var j =0; j < str.length; j++) {
for (var x =0; x < str.length; x++) {
var rez = "";
if (x == index) {
rez = String.fromCharCode(str.charCodeAt(x) ^ 32);
} else {
rez = str[x];
}
newString += rez ;
}
FINAL.push(newString);
newString = "";
index++;
}
return FINAL;
}
var rezArray = wave(mydata);
console.log(rezArray);
// Just for teory
console.log(Array.from([1, 2, 3], x => console.log(x)));var index = 0;
var mydata= "hello";
function wave (str){
var FINAL = ;
var newString = "";
for (var j =0; j < str.length; j++) {
for (var x =0; x < str.length; x++) {
var rez = "";
if (x == index) {
rez = String.fromCharCode(str.charCodeAt(x) ^ 32);
} else {
rez = str[x];
}
newString += rez ;
}
FINAL.push(newString);
newString = "";
index++;
}
return FINAL;
}
var rezArray = wave(mydata);
console.log(rezArray);
// Just for teory
console.log(Array.from([1, 2, 3], x => console.log(x)));answered Mar 4 at 13:08
Nikola LukicNikola Lukic
1,75721937
edited Mar 4 at 13:18
answered Mar 4 at 13:08
Nikola LukicNikola Lukic
1,75721937
answered Mar 4 at 13:08
Nikola LukicNikola Lukic
1,75721937
answered Mar 4 at 13:08
Nikola LukicNikola Lukic
1,75721937
1,75721937
Your answer uses 2 for loops inside each other. Which is Bad for Scalability O(n^2). According to BIG O notations.BIG O Cheatsheet
– Arnas Dičkus
Mar 4 at 13:12
You must understand that " for (i = 0; i < string.length; i++) { if (string[i] === ' ') continue; result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); }" is ALSO loop in twice. ;) When you call this line : Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); you work with iterator.
– Nikola Lukic
Mar 4 at 13:18
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
Mar 4 at 13:19
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
Mar 4 at 13:24
2
"String.fromCharCode(str.charCodeAt(x) ^ 32)" is too much magic. It's a nice trick, but works only on ASCII characters. Don't do that. There's no reason not to use.toUpperCase().
– Bergi
Mar 4 at 18:14
|
show 3 more comments
Your answer uses 2 for loops inside each other. Which is Bad for Scalability O(n^2). According to BIG O notations.BIG O Cheatsheet
– Arnas Dičkus
Mar 4 at 13:12
You must understand that " for (i = 0; i < string.length; i++) { if (string[i] === ' ') continue; result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); }" is ALSO loop in twice. ;) When you call this line : Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); you work with iterator.
– Nikola Lukic
Mar 4 at 13:18
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
Mar 4 at 13:19
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
Mar 4 at 13:24
2
"String.fromCharCode(str.charCodeAt(x) ^ 32)" is too much magic. It's a nice trick, but works only on ASCII characters. Don't do that. There's no reason not to use.toUpperCase().
– Bergi
Mar 4 at 18:14
Your answer uses 2 for loops inside each other. Which is Bad for Scalability O(n^2). According to BIG O notations.BIG O Cheatsheet
– Arnas Dičkus
Mar 4 at 13:12
Your answer uses 2 for loops inside each other. Which is Bad for Scalability O(n^2). According to BIG O notations.BIG O Cheatsheet
– Arnas Dičkus
Mar 4 at 13:12
You must understand that " for (i = 0; i < string.length; i++) { if (string[i] === ' ') continue; result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); }" is ALSO loop in twice. ;) When you call this line : Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); you work with iterator.
– Nikola Lukic
Mar 4 at 13:18
You must understand that " for (i = 0; i < string.length; i++) { if (string[i] === ' ') continue; result.push(Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); }" is ALSO loop in twice. ;) When you call this line : Array.from(string, (c, j) => i === j ? c.toUpperCase() : c).join('')); you work with iterator.
– Nikola Lukic
Mar 4 at 13:18
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
Mar 4 at 13:19
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
Mar 4 at 13:19
1
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
Mar 4 at 13:24
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
Mar 4 at 13:24
2
2
"
String.fromCharCode(str.charCodeAt(x) ^ 32)" is too much magic. It's a nice trick, but works only on ASCII characters. Don't do that. There's no reason not to use .toUpperCase().– Bergi
Mar 4 at 18:14
"
String.fromCharCode(str.charCodeAt(x) ^ 32)" is too much magic. It's a nice trick, but works only on ASCII characters. Don't do that. There's no reason not to use .toUpperCase().– Bergi
Mar 4 at 18:14
|
show 3 more comments
0
I use modify of String prototype :
for implementation of replaceAt .
// Modify prototype
String.prototype.replaceAt=function(index, replacement) {
return this.substr(0, index) + replacement+ this.substr(index + replacement.length);
}
function WaveFunction(str) {
var base = str;
var R = ;
var n =str.length;
for (var i=0 ; i<n ;i++){
str = base;
var c=str.charAt(i);
var res = c.toUpperCase();
// console.log(res);
str = str.replaceAt(i, res);
R.push(str);
}
return R;
}
var REZ = WaveFunction("hello");
console.log(REZ);edited Mar 4 at 14:23
Nikola Lukic
1,75721937
answered Mar 4 at 12:31
Peter HassaballahPeter Hassaballah
825
1
Nice solution with replaceAt !
– Nikola Lukic
Mar 4 at 13:23
where isreplaceAtfrom?
– Nina Scholz
Mar 4 at 13:39
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
Mar 4 at 13:48
1
String.prototype.replaceAt=function(index, replacement) { return this.substr(0, index) + replacement+ this.substr(index + replacement.length); } I am sorry i forgot to add it to my answer :)
– Peter Hassaballah
Mar 4 at 14:22
1
Don't modify objects you don't own.
– Emile Bergeron
Mar 4 at 14:51
|
show 1 more comment
0
I use modify of String prototype :
for implementation of replaceAt .
// Modify prototype
String.prototype.replaceAt=function(index, replacement) {
return this.substr(0, index) + replacement+ this.substr(index + replacement.length);
}
function WaveFunction(str) {
var base = str;
var R = ;
var n =str.length;
for (var i=0 ; i<n ;i++){
str = base;
var c=str.charAt(i);
var res = c.toUpperCase();
// console.log(res);
str = str.replaceAt(i, res);
R.push(str);
}
return R;
}
var REZ = WaveFunction("hello");
console.log(REZ);edited Mar 4 at 14:23
Nikola Lukic
1,75721937
answered Mar 4 at 12:31
Peter HassaballahPeter Hassaballah
825
1
Nice solution with replaceAt !
– Nikola Lukic
Mar 4 at 13:23
where isreplaceAtfrom?
– Nina Scholz
Mar 4 at 13:39
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
Mar 4 at 13:48
1
String.prototype.replaceAt=function(index, replacement) { return this.substr(0, index) + replacement+ this.substr(index + replacement.length); } I am sorry i forgot to add it to my answer :)
– Peter Hassaballah
Mar 4 at 14:22
1
Don't modify objects you don't own.
– Emile Bergeron
Mar 4 at 14:51
|
show 1 more comment
0
0
0
I use modify of String prototype :
for implementation of replaceAt .
// Modify prototype
String.prototype.replaceAt=function(index, replacement) {
return this.substr(0, index) + replacement+ this.substr(index + replacement.length);
}
function WaveFunction(str) {
var base = str;
var R = ;
var n =str.length;
for (var i=0 ; i<n ;i++){
str = base;
var c=str.charAt(i);
var res = c.toUpperCase();
// console.log(res);
str = str.replaceAt(i, res);
R.push(str);
}
return R;
}
var REZ = WaveFunction("hello");
console.log(REZ);edited Mar 4 at 14:23
Nikola Lukic
1,75721937
answered Mar 4 at 12:31
Peter HassaballahPeter Hassaballah
825
I use modify of String prototype :
for implementation of replaceAt .
// Modify prototype
String.prototype.replaceAt=function(index, replacement) {
return this.substr(0, index) + replacement+ this.substr(index + replacement.length);
}
function WaveFunction(str) {
var base = str;
var R = ;
var n =str.length;
for (var i=0 ; i<n ;i++){
str = base;
var c=str.charAt(i);
var res = c.toUpperCase();
// console.log(res);
str = str.replaceAt(i, res);
R.push(str);
}
return R;
}
var REZ = WaveFunction("hello");
console.log(REZ);// Modify prototype
String.prototype.replaceAt=function(index, replacement) {
return this.substr(0, index) + replacement+ this.substr(index + replacement.length);
}
function WaveFunction(str) {
var base = str;
var R = ;
var n =str.length;
for (var i=0 ; i<n ;i++){
str = base;
var c=str.charAt(i);
var res = c.toUpperCase();
// console.log(res);
str = str.replaceAt(i, res);
R.push(str);
}
return R;
}
var REZ = WaveFunction("hello");
console.log(REZ);// Modify prototype
String.prototype.replaceAt=function(index, replacement) {
return this.substr(0, index) + replacement+ this.substr(index + replacement.length);
}
function WaveFunction(str) {
var base = str;
var R = ;
var n =str.length;
for (var i=0 ; i<n ;i++){
str = base;
var c=str.charAt(i);
var res = c.toUpperCase();
// console.log(res);
str = str.replaceAt(i, res);
R.push(str);
}
return R;
}
var REZ = WaveFunction("hello");
console.log(REZ);edited Mar 4 at 14:23
Nikola Lukic
1,75721937
answered Mar 4 at 12:31
Peter HassaballahPeter Hassaballah
825
edited Mar 4 at 14:23
Nikola Lukic
1,75721937
edited Mar 4 at 14:23
Nikola Lukic
1,75721937
edited Mar 4 at 14:23
Nikola Lukic
1,75721937
1,75721937
answered Mar 4 at 12:31
Peter HassaballahPeter Hassaballah
825
answered Mar 4 at 12:31
Peter HassaballahPeter Hassaballah
825
answered Mar 4 at 12:31
Peter HassaballahPeter Hassaballah
825
825
1
Nice solution with replaceAt !
– Nikola Lukic
Mar 4 at 13:23
where isreplaceAtfrom?
– Nina Scholz
Mar 4 at 13:39
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
Mar 4 at 13:48
1
String.prototype.replaceAt=function(index, replacement) { return this.substr(0, index) + replacement+ this.substr(index + replacement.length); } I am sorry i forgot to add it to my answer :)
– Peter Hassaballah
Mar 4 at 14:22
1
Don't modify objects you don't own.
– Emile Bergeron
Mar 4 at 14:51
|
show 1 more comment
1
Nice solution with replaceAt !
– Nikola Lukic
Mar 4 at 13:23
where isreplaceAtfrom?
– Nina Scholz
Mar 4 at 13:39
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
Mar 4 at 13:48
1
String.prototype.replaceAt=function(index, replacement) { return this.substr(0, index) + replacement+ this.substr(index + replacement.length); } I am sorry i forgot to add it to my answer :)
– Peter Hassaballah
Mar 4 at 14:22
1
Don't modify objects you don't own.
– Emile Bergeron
Mar 4 at 14:51
1
1
Nice solution with replaceAt !
– Nikola Lukic
Mar 4 at 13:23
Nice solution with replaceAt !
– Nikola Lukic
Mar 4 at 13:23
where is
replaceAt from?– Nina Scholz
Mar 4 at 13:39
where is
replaceAt from?– Nina Scholz
Mar 4 at 13:39
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
Mar 4 at 13:48
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
Mar 4 at 13:48
1
1
String.prototype.replaceAt=function(index, replacement) { return this.substr(0, index) + replacement+ this.substr(index + replacement.length); } I am sorry i forgot to add it to my answer :)
– Peter Hassaballah
Mar 4 at 14:22
String.prototype.replaceAt=function(index, replacement) { return this.substr(0, index) + replacement+ this.substr(index + replacement.length); } I am sorry i forgot to add it to my answer :)
– Peter Hassaballah
Mar 4 at 14:22
1
1
Don't modify objects you don't own.
– Emile Bergeron
Mar 4 at 14:51
Don't modify objects you don't own.
– Emile Bergeron
Mar 4 at 14:51
|
show 1 more comment
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Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
Mar 4 at 12:28