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Creating a wave of string in Javascript
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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(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
javascript arrays
edited yesterday
vahdet
1,99131430
asked yesterday
Arnas DičkusArnas Dičkus
7518
add a comment |
7
I can't seem to figure it out how to make a wave from a string in Javascript.
Rules:
- The input will always be lower case string.
- Ignore whitespace.
Expected result:
wave("hello") => ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
wave (" h e y ") => [" H e y ", " h E y ", " h e Y "]
wave ("") => []
This is as far as I got. Current code will give me an answer ["hello", "hello", "hello", "hello", "hello"]. I'm thinking using second for loop and somehow capitalize each new letter but I'am stumped. Also I would appreciate if answer would avoid using loop inside loop O(n^2). Because of BIG O Scalability.
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
array.push(str);
}
for (let index = 0; index < str.length; index++) {
console.log(array);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
javascript arrays
edited yesterday
vahdet
1,99131430
asked yesterday
Arnas DičkusArnas Dičkus
7518
Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
yesterday
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(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
javascript arrays
edited yesterday
vahdet
1,99131430
asked yesterday
Arnas DičkusArnas Dičkus
7518
I can't seem to figure it out how to make a wave from a string in Javascript.
Rules:
- The input will always be lower case string.
- Ignore whitespace.
Expected result:
wave("hello") => ["Hello", "hEllo", "heLlo", "helLo", "hellO"]
wave (" h e y ") => [" H e y ", " h E y ", " h e Y "]
wave ("") => []
This is as far as I got. Current code will give me an answer ["hello", "hello", "hello", "hello", "hello"]. I'm thinking using second for loop and somehow capitalize each new letter but I'am stumped. Also I would appreciate if answer would avoid using loop inside loop O(n^2). Because of BIG O Scalability.
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
array.push(str);
}
for (let index = 0; index < str.length; index++) {
console.log(array);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
javascript arrays
javascript arrays
edited yesterday
vahdet
1,99131430
asked yesterday
Arnas DičkusArnas Dičkus
7518
edited yesterday
vahdet
1,99131430
asked yesterday
Arnas DičkusArnas Dičkus
7518
edited yesterday
vahdet
1,99131430
edited yesterday
vahdet
1,99131430
edited yesterday
vahdet
1,99131430
1,99131430
asked yesterday
Arnas DičkusArnas Dičkus
7518
asked yesterday
Arnas DičkusArnas Dičkus
7518
asked yesterday
Arnas DičkusArnas Dičkus
7518
7518
Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
yesterday
add a comment |
Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
yesterday
Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
yesterday
Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
yesterday
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 yesterday
Nina ScholzNina Scholz
189k1598173
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()}${arr.slice(i + 1, arr.length)}`.replace(/,/g, ''));
}
console.log(wave('wave'));
console.log(wave('foo bar'));
console.log(wave());answered yesterday
zvonazvona
12.6k14059
add a comment |
1
You can take each char in string in for loop and make it uppercase and the append with prefix and post fix string
var array=[];
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
if (str[index] === ' ') continue;
waveChar=str.charAt(index).toUpperCase();
preStr=str.substring(0,index);
postStr=str.substring(index,str.length-1);
array.push(preStr+waveChar+postStr);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
console.log(array);answered yesterday
Code_ModeCode_Mode
1,192715
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 yesterday
Nikola LukicNikola Lukic
1,75121835
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
yesterday
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
yesterday
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
yesterday
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
yesterday
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
23 hours ago
|
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 yesterday
Nikola Lukic
1,75121835
answered yesterday
Peter HassaballahPeter Hassaballah
825
1
Nice solution with replaceAt !
– Nikola Lukic
yesterday
where isreplaceAtfrom?
– Nina Scholz
yesterday
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
yesterday
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
yesterday
1
Don't modify objects you don't own.
– Emile Bergeron
yesterday
|
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 yesterday
Nina ScholzNina Scholz
189k1598173
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 yesterday
Nina ScholzNina Scholz
189k1598173
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 yesterday
Nina ScholzNina Scholz
189k1598173
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 yesterday
Nina ScholzNina Scholz
189k1598173
answered yesterday
Nina ScholzNina Scholz
189k1598173
answered yesterday
Nina ScholzNina Scholz
189k1598173
answered yesterday
Nina ScholzNina Scholz
189k1598173
189k1598173
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()}${arr.slice(i + 1, arr.length)}`.replace(/,/g, ''));
}
console.log(wave('wave'));
console.log(wave('foo bar'));
console.log(wave());answered yesterday
zvonazvona
12.6k14059
add a comment |
1
This does it. However, if you have spaces in your string, it will output string without any "waved letter" (since also space is handled):
const wave = (str = '') => {
return str
.split('')
.map((letter, i, arr) => `${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 yesterday
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()}${arr.slice(i + 1, arr.length)}`.replace(/,/g, ''));
}
console.log(wave('wave'));
console.log(wave('foo bar'));
console.log(wave());answered yesterday
zvonazvona
12.6k14059
This does it. However, if you have spaces in your string, it will output string without any "waved letter" (since also space is handled):
const wave = (str = '') => {
return str
.split('')
.map((letter, i, arr) => `${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 yesterday
zvonazvona
12.6k14059
answered yesterday
zvonazvona
12.6k14059
answered yesterday
zvonazvona
12.6k14059
answered yesterday
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(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
console.log(array);answered yesterday
Code_ModeCode_Mode
1,192715
add a comment |
1
You can take each char in string in for loop and make it uppercase and the append with prefix and post fix string
var array=[];
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
if (str[index] === ' ') continue;
waveChar=str.charAt(index).toUpperCase();
preStr=str.substring(0,index);
postStr=str.substring(index,str.length-1);
array.push(preStr+waveChar+postStr);
}
}else{
alert(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
console.log(array);answered yesterday
Code_ModeCode_Mode
1,192715
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(`${str} is either not a string or not lowercase`);
}
}
wave("hello");
console.log(array);answered yesterday
Code_ModeCode_Mode
1,192715
You can take each char in string in for loop and make it uppercase and the append with prefix and post fix string
var array=[];
const wave = (str) => {
if(typeof str === 'string' && str === str.toLowerCase()){
for (let index = 0; index < str.length; index++) {
if (str[index] === ' ') continue;
waveChar=str.charAt(index).toUpperCase();
preStr=str.substring(0,index);
postStr=str.substring(index,str.length-1);
array.push(preStr+waveChar+postStr);
}
}else{
alert(`${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 yesterday
Code_ModeCode_Mode
1,192715
answered yesterday
Code_ModeCode_Mode
1,192715
answered yesterday
Code_ModeCode_Mode
1,192715
answered yesterday
Code_ModeCode_Mode
1,192715
1,192715
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 yesterday
Nikola LukicNikola Lukic
1,75121835
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
yesterday
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
yesterday
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
yesterday
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
yesterday
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
23 hours ago
|
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 yesterday
Nikola LukicNikola Lukic
1,75121835
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
yesterday
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
yesterday
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
yesterday
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
yesterday
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
23 hours ago
|
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 yesterday
Nikola LukicNikola Lukic
1,75121835
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 yesterday
Nikola LukicNikola Lukic
1,75121835
edited yesterday
answered yesterday
Nikola LukicNikola Lukic
1,75121835
answered yesterday
Nikola LukicNikola Lukic
1,75121835
answered yesterday
Nikola LukicNikola Lukic
1,75121835
1,75121835
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
yesterday
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
yesterday
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
yesterday
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
yesterday
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
23 hours ago
|
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
yesterday
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
yesterday
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
yesterday
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
yesterday
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
23 hours ago
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
yesterday
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
yesterday
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
yesterday
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
yesterday
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
yesterday
What you think what is it : console.log(Array.from([1, 2, 3], x => console.log(x)));
– Nikola Lukic
yesterday
1
1
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
yesterday
Thanks, for explanation I wasn't aware of this.
– Arnas Dičkus
yesterday
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
23 hours ago
"
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
23 hours ago
|
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 yesterday
Nikola Lukic
1,75121835
answered yesterday
Peter HassaballahPeter Hassaballah
825
1
Nice solution with replaceAt !
– Nikola Lukic
yesterday
where isreplaceAtfrom?
– Nina Scholz
yesterday
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
yesterday
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
yesterday
1
Don't modify objects you don't own.
– Emile Bergeron
yesterday
|
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 yesterday
Nikola Lukic
1,75121835
answered yesterday
Peter HassaballahPeter Hassaballah
825
1
Nice solution with replaceAt !
– Nikola Lukic
yesterday
where isreplaceAtfrom?
– Nina Scholz
yesterday
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
yesterday
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
yesterday
1
Don't modify objects you don't own.
– Emile Bergeron
yesterday
|
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 yesterday
Nikola Lukic
1,75121835
answered yesterday
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 yesterday
Nikola Lukic
1,75121835
answered yesterday
Peter HassaballahPeter Hassaballah
825
edited yesterday
Nikola Lukic
1,75121835
edited yesterday
Nikola Lukic
1,75121835
edited yesterday
Nikola Lukic
1,75121835
1,75121835
answered yesterday
Peter HassaballahPeter Hassaballah
825
answered yesterday
Peter HassaballahPeter Hassaballah
825
answered yesterday
Peter HassaballahPeter Hassaballah
825
825
1
Nice solution with replaceAt !
– Nikola Lukic
yesterday
where isreplaceAtfrom?
– Nina Scholz
yesterday
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
yesterday
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
yesterday
1
Don't modify objects you don't own.
– Emile Bergeron
yesterday
|
show 1 more comment
1
Nice solution with replaceAt !
– Nikola Lukic
yesterday
where isreplaceAtfrom?
– Nina Scholz
yesterday
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
yesterday
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
yesterday
1
Don't modify objects you don't own.
– Emile Bergeron
yesterday
1
1
Nice solution with replaceAt !
– Nikola Lukic
yesterday
Nice solution with replaceAt !
– Nikola Lukic
yesterday
where is
replaceAt from?– Nina Scholz
yesterday
where is
replaceAt from?– Nina Scholz
yesterday
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
yesterday
@NikolaLukic I can seem to get it work I get error str.replaceAt is not a function.
– Arnas Dičkus
yesterday
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
yesterday
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
yesterday
1
1
Don't modify objects you don't own.
– Emile Bergeron
yesterday
Don't modify objects you don't own.
– Emile Bergeron
yesterday
|
show 1 more comment
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Shouldn't you convert some characters to upper case?
– Aycan Yaşıt
yesterday