How to get all substrings (contiguous subsequences) of my JavaScript array?

up vote
9
down vote

favorite

My task is to split the given array into smaller arrays using JavaScript. For example [1, 2, 3, 4] should be split to [1] [1, 2] [1, 2, 3] [1, 2, 3, 4] [2] [2, 3] [2, 3, 4] [3] [3, 4] [4].

I am using this code:

let arr = [1, 2, 3, 4];

for (let i = 1; i <= arr.length; i++) 
 let a = ;
 for (let j = 0; j < arr.length; j++) 
 a.push(arr[j]);
 if (a.length === i) 
 break;
 
 
 console.log(a);

And I get the following result: [1] [1, 2] [1, 2, 3] [1, 2, 3, 4] undefined

What am I missing/doing wrong?

edited 13 mins ago

Bergi

346k54502823

asked 2 hours ago

TeodorKolev

96752043

I think you should set j=i in begin loop
â€“Â Alexandr Kudryashov
2 hours ago

@AlexandrKudryashov I have tried. It is not correct
â€“Â TeodorKolev
2 hours ago

set j=i and remove the if condition in the nested loop
â€“Â rock star
2 hours ago

init i with 0 and remove the = symbol in the corresponding condition
â€“Â rock star
2 hours ago

1

@rockstar nope, it is not correct
â€“Â TeodorKolev
2 hours ago

Â |Â
show 1 more comment

up vote
9
down vote

favorite

My task is to split the given array into smaller arrays using JavaScript. For example [1, 2, 3, 4] should be split to [1] [1, 2] [1, 2, 3] [1, 2, 3, 4] [2] [2, 3] [2, 3, 4] [3] [3, 4] [4].

I am using this code:

let arr = [1, 2, 3, 4];

for (let i = 1; i <= arr.length; i++) 
 let a = ;
 for (let j = 0; j < arr.length; j++) 
 a.push(arr[j]);
 if (a.length === i) 
 break;
 
 
 console.log(a);

And I get the following result: [1] [1, 2] [1, 2, 3] [1, 2, 3, 4] undefined

What am I missing/doing wrong?

edited 13 mins ago

Bergi

346k54502823

asked 2 hours ago

TeodorKolev

96752043

I think you should set j=i in begin loop
â€“Â Alexandr Kudryashov
2 hours ago

@AlexandrKudryashov I have tried. It is not correct
â€“Â TeodorKolev
2 hours ago

set j=i and remove the if condition in the nested loop
â€“Â rock star
2 hours ago

init i with 0 and remove the = symbol in the corresponding condition
â€“Â rock star
2 hours ago

1

@rockstar nope, it is not correct
â€“Â TeodorKolev
2 hours ago

Â |Â
show 1 more comment

up vote
9
down vote

favorite

My task is to split the given array into smaller arrays using JavaScript. For example [1, 2, 3, 4] should be split to [1] [1, 2] [1, 2, 3] [1, 2, 3, 4] [2] [2, 3] [2, 3, 4] [3] [3, 4] [4].

I am using this code:

let arr = [1, 2, 3, 4];

for (let i = 1; i <= arr.length; i++) 
 let a = ;
 for (let j = 0; j < arr.length; j++) 
 a.push(arr[j]);
 if (a.length === i) 
 break;
 
 
 console.log(a);

And I get the following result: [1] [1, 2] [1, 2, 3] [1, 2, 3, 4] undefined

What am I missing/doing wrong?

edited 13 mins ago

Bergi

346k54502823

asked 2 hours ago

TeodorKolev

96752043

My task is to split the given array into smaller arrays using JavaScript. For example [1, 2, 3, 4] should be split to [1] [1, 2] [1, 2, 3] [1, 2, 3, 4] [2] [2, 3] [2, 3, 4] [3] [3, 4] [4].

I am using this code:

let arr = [1, 2, 3, 4];

for (let i = 1; i <= arr.length; i++) 
 let a = ;
 for (let j = 0; j < arr.length; j++) 
 a.push(arr[j]);
 if (a.length === i) 
 break;
 
 
 console.log(a);

And I get the following result: [1] [1, 2] [1, 2, 3] [1, 2, 3, 4] undefined

What am I missing/doing wrong?

let arr = [1, 2, 3, 4];

for (let i = 1; i <= arr.length; i++) 
 let a = ;
 for (let j = 0; j < arr.length; j++) 
 a.push(arr[j]);
 if (a.length === i) 
 break;
 
 
 console.log(a);

let arr = [1, 2, 3, 4];

for (let i = 1; i <= arr.length; i++) 
 let a = ;
 for (let j = 0; j < arr.length; j++) 
 a.push(arr[j]);
 if (a.length === i) 
 break;
 
 
 console.log(a);

javascript arrays substring

edited 13 mins ago

Bergi

346k54502823

asked 2 hours ago

TeodorKolev

96752043

edited 13 mins ago

Bergi

346k54502823

asked 2 hours ago

TeodorKolev

96752043

edited 13 mins ago

Bergi

346k54502823

edited 13 mins ago

Bergi

346k54502823

edited 13 mins ago

Bergi

346k54502823

asked 2 hours ago

TeodorKolev

96752043

asked 2 hours ago

TeodorKolev

96752043

asked 2 hours ago

TeodorKolev

96752043

I think you should set j=i in begin loop
â€“Â Alexandr Kudryashov
2 hours ago

@AlexandrKudryashov I have tried. It is not correct
â€“Â TeodorKolev
2 hours ago

set j=i and remove the if condition in the nested loop
â€“Â rock star
2 hours ago

init i with 0 and remove the = symbol in the corresponding condition
â€“Â rock star
2 hours ago

1

@rockstar nope, it is not correct
â€“Â TeodorKolev
2 hours ago

Â |Â
show 1 more comment

I think you should set j=i in begin loop
â€“Â Alexandr Kudryashov
2 hours ago

@AlexandrKudryashov I have tried. It is not correct
â€“Â TeodorKolev
2 hours ago

set j=i and remove the if condition in the nested loop
â€“Â rock star
2 hours ago

init i with 0 and remove the = symbol in the corresponding condition
â€“Â rock star
2 hours ago

1

@rockstar nope, it is not correct
â€“Â TeodorKolev
2 hours ago

I think you should set j=i in begin loop
â€“Â Alexandr Kudryashov
2 hours ago

@AlexandrKudryashov I have tried. It is not correct
â€“Â TeodorKolev
2 hours ago

set j=i and remove the if condition in the nested loop
â€“Â rock star
2 hours ago

init i with 0 and remove the = symbol in the corresponding condition
â€“Â rock star
2 hours ago

@rockstar nope, it is not correct
â€“Â TeodorKolev
2 hours ago

Â |Â
show 1 more comment

6 Answers
6

active

oldest

votes

up vote
4
down vote

accepted

You have two issues in your code:

You need to have loop to initialize with the value of i for the inner loop so that it consider the next index for new iteration of i

You need to remove that break on the length which you have in inner loop.

let arr = [1, 2, 3, 4];
for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a.push(arr[j]);
 console.log(a);

answered 1 hour ago

Ankit Agarwal

20.8k41840

add a commentÂ |Â

up vote
7
down vote

For the inner array, you could just start with the index of the outer array.

var array = [1, 2, 3, 4],
 temp,
 i, j, l = array.length,
 result = ;
 
for (i = 0; i < l; i++) 
 for (j = i; j < l; j++) 
 result.push(array.slice(i, j + 1));
 

console.log(result.map(a => a.join(' ')));

.as-console-wrapper max-height: 100% !important; top: 0;

answered 2 hours ago

Nina Scholz

158k1277136

1

I was expecting answer from you :)
â€“Â Ankit Agarwal
2 hours ago

This do one big array. Target is to create small arrays
â€“Â TeodorKolev
2 hours ago

what do you want with small arrays? instead of pushing, you could display the sub array.
â€“Â Nina Scholz
1 hour ago

add a commentÂ |Â

up vote
2
down vote

Try this

 let arr = [1, 2, 3, 4];
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 var tmp=i;
 for (let j = tmp; j < arr.length; j++) 
 a.push(arr[j]); 
 console.log(a);

edited 2 hours ago

Sudhir Ojha

1,024313

answered 2 hours ago

Alexandr Kudryashov

559213

Why do you need the tmp variable?
â€“Â Jenny O'Reilly
1 hour ago

you can do it without tmp variable=)
â€“Â Alexandr Kudryashov
1 hour ago

add a commentÂ |Â

up vote
0
down vote

i have prepare stackblitz for this case.

let source = [1,2,3,4];
const output = ;
const arrayMultiplier = (source) => 
 const eachValueArray = ;
 source.forEach((item, index) => 
 // Will push new array who will be sliced source array.
 eachValueArray.push(source.slice(0, source.length - index));
 );
 //We reverse array to have right order.
 return eachValueArray.reverse();
;

for(let i = 0; i <= source.length; i++) 
 output.push(...arrayMultiplier(source));
 source.shift(); // Will recraft source array by removing first index.

//Don't forget last item.
output.push(source);
console.log(output);

Is not the most shorten solution but do the job

answered 1 hour ago

Yanis-git

1,3171217

Avoid forEach + push, use map instead
â€“Â Bergi
18 mins ago

add a commentÂ |Â

up vote
0
down vote

Use two iteration

get slice array based on loop index.

use sliced array and combine array element.

 var arr = [1, 2, 3, 4];
 let newArra =;
 arr.map((x,i)=> 
 let remainArr = arr.slice(i);
 return remainArr.forEach((y, r) => newArra.push(arr.slice(i).slice(0, r+1)))
 )
 newArra.forEach(x=> console.log(x))

edited 1 hour ago

answered 1 hour ago

Anoop

18.8k94468

add a commentÂ |Â

up vote
0
down vote

If you don't want to mutate your array.

 let arr = [1, 2, 3, 4];
 let res = ;
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a = [...a, arr[j]];
 res = [...res, a];
 
 
 console.log(res);

edited 32 mins ago

Yanis-git

1,3171217

answered 48 mins ago

Varun Arya

New contributor

add a commentÂ |Â

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

up vote
4
down vote

accepted

You have two issues in your code:

You need to have loop to initialize with the value of i for the inner loop so that it consider the next index for new iteration of i

You need to remove that break on the length which you have in inner loop.

let arr = [1, 2, 3, 4];
for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a.push(arr[j]);
 console.log(a);

answered 1 hour ago

Ankit Agarwal

20.8k41840

add a commentÂ |Â

up vote
4
down vote

accepted

You have two issues in your code:

You need to have loop to initialize with the value of i for the inner loop so that it consider the next index for new iteration of i

You need to remove that break on the length which you have in inner loop.

let arr = [1, 2, 3, 4];
for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a.push(arr[j]);
 console.log(a);

answered 1 hour ago

Ankit Agarwal

20.8k41840

add a commentÂ |Â

up vote
4
down vote

accepted

You have two issues in your code:

You need to have loop to initialize with the value of i for the inner loop so that it consider the next index for new iteration of i

You need to remove that break on the length which you have in inner loop.

let arr = [1, 2, 3, 4];
for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a.push(arr[j]);
 console.log(a);

answered 1 hour ago

Ankit Agarwal

20.8k41840

You have two issues in your code:

You need to have loop to initialize with the value of i for the inner loop so that it consider the next index for new iteration of i

You need to remove that break on the length which you have in inner loop.

let arr = [1, 2, 3, 4];
for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a.push(arr[j]);
 console.log(a);

let arr = [1, 2, 3, 4];
for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a.push(arr[j]);
 console.log(a);

let arr = [1, 2, 3, 4];
for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a.push(arr[j]);
 console.log(a);

answered 1 hour ago

Ankit Agarwal

20.8k41840

answered 1 hour ago

Ankit Agarwal

20.8k41840

answered 1 hour ago

Ankit Agarwal

20.8k41840

answered 1 hour ago

Ankit Agarwal

20.8k41840

add a commentÂ |Â

up vote
7
down vote

For the inner array, you could just start with the index of the outer array.

var array = [1, 2, 3, 4],
 temp,
 i, j, l = array.length,
 result = ;
 
for (i = 0; i < l; i++) 
 for (j = i; j < l; j++) 
 result.push(array.slice(i, j + 1));
 

console.log(result.map(a => a.join(' ')));

.as-console-wrapper max-height: 100% !important; top: 0;

answered 2 hours ago

Nina Scholz

158k1277136

1

I was expecting answer from you :)
â€“Â Ankit Agarwal
2 hours ago

This do one big array. Target is to create small arrays
â€“Â TeodorKolev
2 hours ago

what do you want with small arrays? instead of pushing, you could display the sub array.
â€“Â Nina Scholz
1 hour ago

add a commentÂ |Â

up vote
7
down vote

For the inner array, you could just start with the index of the outer array.

var array = [1, 2, 3, 4],
 temp,
 i, j, l = array.length,
 result = ;
 
for (i = 0; i < l; i++) 
 for (j = i; j < l; j++) 
 result.push(array.slice(i, j + 1));
 

console.log(result.map(a => a.join(' ')));

.as-console-wrapper max-height: 100% !important; top: 0;

answered 2 hours ago

Nina Scholz

158k1277136

1

I was expecting answer from you :)
â€“Â Ankit Agarwal
2 hours ago

This do one big array. Target is to create small arrays
â€“Â TeodorKolev
2 hours ago

what do you want with small arrays? instead of pushing, you could display the sub array.
â€“Â Nina Scholz
1 hour ago

add a commentÂ |Â

up vote
7
down vote

For the inner array, you could just start with the index of the outer array.

var array = [1, 2, 3, 4],
 temp,
 i, j, l = array.length,
 result = ;
 
for (i = 0; i < l; i++) 
 for (j = i; j < l; j++) 
 result.push(array.slice(i, j + 1));
 

console.log(result.map(a => a.join(' ')));

.as-console-wrapper max-height: 100% !important; top: 0;

answered 2 hours ago

Nina Scholz

158k1277136

For the inner array, you could just start with the index of the outer array.

var array = [1, 2, 3, 4],
 temp,
 i, j, l = array.length,
 result = ;
 
for (i = 0; i < l; i++) 
 for (j = i; j < l; j++) 
 result.push(array.slice(i, j + 1));
 

console.log(result.map(a => a.join(' ')));

.as-console-wrapper max-height: 100% !important; top: 0;

var array = [1, 2, 3, 4],
 temp,
 i, j, l = array.length,
 result = ;
 
for (i = 0; i < l; i++) 
 for (j = i; j < l; j++) 
 result.push(array.slice(i, j + 1));
 

console.log(result.map(a => a.join(' ')));

.as-console-wrapper max-height: 100% !important; top: 0;

var array = [1, 2, 3, 4],
 temp,
 i, j, l = array.length,
 result = ;
 
for (i = 0; i < l; i++) 
 for (j = i; j < l; j++) 
 result.push(array.slice(i, j + 1));
 

console.log(result.map(a => a.join(' ')));

.as-console-wrapper max-height: 100% !important; top: 0;

answered 2 hours ago

Nina Scholz

158k1277136

answered 2 hours ago

Nina Scholz

158k1277136

answered 2 hours ago

Nina Scholz

158k1277136

answered 2 hours ago

Nina Scholz

158k1277136

1

I was expecting answer from you :)
â€“Â Ankit Agarwal
2 hours ago

This do one big array. Target is to create small arrays
â€“Â TeodorKolev
2 hours ago

what do you want with small arrays? instead of pushing, you could display the sub array.
â€“Â Nina Scholz
1 hour ago

add a commentÂ |Â

1

I was expecting answer from you :)
â€“Â Ankit Agarwal
2 hours ago

This do one big array. Target is to create small arrays
â€“Â TeodorKolev
2 hours ago

what do you want with small arrays? instead of pushing, you could display the sub array.
â€“Â Nina Scholz
1 hour ago

I was expecting answer from you :)
â€“Â Ankit Agarwal
2 hours ago

This do one big array. Target is to create small arrays
â€“Â TeodorKolev
2 hours ago

what do you want with small arrays? instead of pushing, you could display the sub array.
â€“Â Nina Scholz
1 hour ago

add a commentÂ |Â

up vote
2
down vote

Try this

 let arr = [1, 2, 3, 4];
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 var tmp=i;
 for (let j = tmp; j < arr.length; j++) 
 a.push(arr[j]); 
 console.log(a);

edited 2 hours ago

Sudhir Ojha

1,024313

answered 2 hours ago

Alexandr Kudryashov

559213

Why do you need the tmp variable?
â€“Â Jenny O'Reilly
1 hour ago

you can do it without tmp variable=)
â€“Â Alexandr Kudryashov
1 hour ago

add a commentÂ |Â

up vote
2
down vote

Try this

 let arr = [1, 2, 3, 4];
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 var tmp=i;
 for (let j = tmp; j < arr.length; j++) 
 a.push(arr[j]); 
 console.log(a);

edited 2 hours ago

Sudhir Ojha

1,024313

answered 2 hours ago

Alexandr Kudryashov

559213

Why do you need the tmp variable?
â€“Â Jenny O'Reilly
1 hour ago

you can do it without tmp variable=)
â€“Â Alexandr Kudryashov
1 hour ago

add a commentÂ |Â

up vote
2
down vote

Try this

 let arr = [1, 2, 3, 4];
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 var tmp=i;
 for (let j = tmp; j < arr.length; j++) 
 a.push(arr[j]); 
 console.log(a);

edited 2 hours ago

Sudhir Ojha

1,024313

answered 2 hours ago

Alexandr Kudryashov

559213

Try this

 let arr = [1, 2, 3, 4];
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 var tmp=i;
 for (let j = tmp; j < arr.length; j++) 
 a.push(arr[j]); 
 console.log(a);

 let arr = [1, 2, 3, 4];
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 var tmp=i;
 for (let j = tmp; j < arr.length; j++) 
 a.push(arr[j]); 
 console.log(a);

 let arr = [1, 2, 3, 4];
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 var tmp=i;
 for (let j = tmp; j < arr.length; j++) 
 a.push(arr[j]); 
 console.log(a);

edited 2 hours ago

Sudhir Ojha

1,024313

answered 2 hours ago

Alexandr Kudryashov

559213

edited 2 hours ago

Sudhir Ojha

1,024313

edited 2 hours ago

Sudhir Ojha

1,024313

edited 2 hours ago

Sudhir Ojha

1,024313

answered 2 hours ago

Alexandr Kudryashov

559213

answered 2 hours ago

Alexandr Kudryashov

559213

answered 2 hours ago

Alexandr Kudryashov

559213

Why do you need the tmp variable?
â€“Â Jenny O'Reilly
1 hour ago

you can do it without tmp variable=)
â€“Â Alexandr Kudryashov
1 hour ago

add a commentÂ |Â

Why do you need the tmp variable?
â€“Â Jenny O'Reilly
1 hour ago

you can do it without tmp variable=)
â€“Â Alexandr Kudryashov
1 hour ago

Why do you need the tmp variable?
â€“Â Jenny O'Reilly
1 hour ago

you can do it without tmp variable=)
â€“Â Alexandr Kudryashov
1 hour ago

add a commentÂ |Â

up vote
0
down vote

i have prepare stackblitz for this case.

let source = [1,2,3,4];
const output = ;
const arrayMultiplier = (source) => 
 const eachValueArray = ;
 source.forEach((item, index) => 
 // Will push new array who will be sliced source array.
 eachValueArray.push(source.slice(0, source.length - index));
 );
 //We reverse array to have right order.
 return eachValueArray.reverse();
;

for(let i = 0; i <= source.length; i++) 
 output.push(...arrayMultiplier(source));
 source.shift(); // Will recraft source array by removing first index.

//Don't forget last item.
output.push(source);
console.log(output);

Is not the most shorten solution but do the job

answered 1 hour ago

Yanis-git

1,3171217

Avoid forEach + push, use map instead
â€“Â Bergi
18 mins ago

add a commentÂ |Â

up vote
0
down vote

i have prepare stackblitz for this case.

let source = [1,2,3,4];
const output = ;
const arrayMultiplier = (source) => 
 const eachValueArray = ;
 source.forEach((item, index) => 
 // Will push new array who will be sliced source array.
 eachValueArray.push(source.slice(0, source.length - index));
 );
 //We reverse array to have right order.
 return eachValueArray.reverse();
;

for(let i = 0; i <= source.length; i++) 
 output.push(...arrayMultiplier(source));
 source.shift(); // Will recraft source array by removing first index.

//Don't forget last item.
output.push(source);
console.log(output);

Is not the most shorten solution but do the job

answered 1 hour ago

Yanis-git

1,3171217

Avoid forEach + push, use map instead
â€“Â Bergi
18 mins ago

add a commentÂ |Â

up vote
0
down vote

i have prepare stackblitz for this case.

let source = [1,2,3,4];
const output = ;
const arrayMultiplier = (source) => 
 const eachValueArray = ;
 source.forEach((item, index) => 
 // Will push new array who will be sliced source array.
 eachValueArray.push(source.slice(0, source.length - index));
 );
 //We reverse array to have right order.
 return eachValueArray.reverse();
;

for(let i = 0; i <= source.length; i++) 
 output.push(...arrayMultiplier(source));
 source.shift(); // Will recraft source array by removing first index.

//Don't forget last item.
output.push(source);
console.log(output);

Is not the most shorten solution but do the job

answered 1 hour ago

Yanis-git

1,3171217

i have prepare stackblitz for this case.

let source = [1,2,3,4];
const output = ;
const arrayMultiplier = (source) => 
 const eachValueArray = ;
 source.forEach((item, index) => 
 // Will push new array who will be sliced source array.
 eachValueArray.push(source.slice(0, source.length - index));
 );
 //We reverse array to have right order.
 return eachValueArray.reverse();
;

for(let i = 0; i <= source.length; i++) 
 output.push(...arrayMultiplier(source));
 source.shift(); // Will recraft source array by removing first index.

//Don't forget last item.
output.push(source);
console.log(output);

Is not the most shorten solution but do the job

answered 1 hour ago

Yanis-git

1,3171217

answered 1 hour ago

Yanis-git

1,3171217

answered 1 hour ago

Yanis-git

1,3171217

answered 1 hour ago

Yanis-git

1,3171217

Avoid forEach + push, use map instead
â€“Â Bergi
18 mins ago

add a commentÂ |Â

Avoid forEach + push, use map instead
â€“Â Bergi
18 mins ago

Avoid forEach + push, use map instead
â€“Â Bergi
18 mins ago

add a commentÂ |Â

up vote
0
down vote

Use two iteration

get slice array based on loop index.

use sliced array and combine array element.

 var arr = [1, 2, 3, 4];
 let newArra =;
 arr.map((x,i)=> 
 let remainArr = arr.slice(i);
 return remainArr.forEach((y, r) => newArra.push(arr.slice(i).slice(0, r+1)))
 )
 newArra.forEach(x=> console.log(x))

edited 1 hour ago

answered 1 hour ago

Anoop

18.8k94468

add a commentÂ |Â

up vote
0
down vote

Use two iteration

get slice array based on loop index.

use sliced array and combine array element.

 var arr = [1, 2, 3, 4];
 let newArra =;
 arr.map((x,i)=> 
 let remainArr = arr.slice(i);
 return remainArr.forEach((y, r) => newArra.push(arr.slice(i).slice(0, r+1)))
 )
 newArra.forEach(x=> console.log(x))

edited 1 hour ago

answered 1 hour ago

Anoop

18.8k94468

add a commentÂ |Â

up vote
0
down vote

Use two iteration

get slice array based on loop index.

use sliced array and combine array element.

 var arr = [1, 2, 3, 4];
 let newArra =;
 arr.map((x,i)=> 
 let remainArr = arr.slice(i);
 return remainArr.forEach((y, r) => newArra.push(arr.slice(i).slice(0, r+1)))
 )
 newArra.forEach(x=> console.log(x))

edited 1 hour ago

answered 1 hour ago

Anoop

18.8k94468

Use two iteration

get slice array based on loop index.

use sliced array and combine array element.

 var arr = [1, 2, 3, 4];
 let newArra =;
 arr.map((x,i)=> 
 let remainArr = arr.slice(i);
 return remainArr.forEach((y, r) => newArra.push(arr.slice(i).slice(0, r+1)))
 )
 newArra.forEach(x=> console.log(x))

 var arr = [1, 2, 3, 4];
 let newArra =;
 arr.map((x,i)=> 
 let remainArr = arr.slice(i);
 return remainArr.forEach((y, r) => newArra.push(arr.slice(i).slice(0, r+1)))
 )
 newArra.forEach(x=> console.log(x))

 var arr = [1, 2, 3, 4];
 let newArra =;
 arr.map((x,i)=> 
 let remainArr = arr.slice(i);
 return remainArr.forEach((y, r) => newArra.push(arr.slice(i).slice(0, r+1)))
 )
 newArra.forEach(x=> console.log(x))

edited 1 hour ago

answered 1 hour ago

Anoop

18.8k94468

edited 1 hour ago

answered 1 hour ago

Anoop

18.8k94468

answered 1 hour ago

Anoop

18.8k94468

answered 1 hour ago

Anoop

18.8k94468

add a commentÂ |Â

up vote
0
down vote

If you don't want to mutate your array.

 let arr = [1, 2, 3, 4];
 let res = ;
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a = [...a, arr[j]];
 res = [...res, a];
 
 
 console.log(res);

edited 32 mins ago

Yanis-git

1,3171217

answered 48 mins ago

Varun Arya

New contributor

add a commentÂ |Â

up vote
0
down vote

If you don't want to mutate your array.

 let arr = [1, 2, 3, 4];
 let res = ;
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a = [...a, arr[j]];
 res = [...res, a];
 
 
 console.log(res);

edited 32 mins ago

Yanis-git

1,3171217

answered 48 mins ago

Varun Arya

New contributor

add a commentÂ |Â

up vote
0
down vote

If you don't want to mutate your array.

 let arr = [1, 2, 3, 4];
 let res = ;
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a = [...a, arr[j]];
 res = [...res, a];
 
 
 console.log(res);

edited 32 mins ago

Yanis-git

1,3171217

answered 48 mins ago

Varun Arya

New contributor

If you don't want to mutate your array.

 let arr = [1, 2, 3, 4];
 let res = ;
 for (let i = 0; i <= arr.length; i++) 
 let a = ;
 for (let j = i; j < arr.length; j++) 
 a = [...a, arr[j]];
 res = [...res, a];
 
 
 console.log(res);

edited 32 mins ago

Yanis-git

1,3171217

answered 48 mins ago

Varun Arya

New contributor

edited 32 mins ago

Yanis-git

1,3171217

edited 32 mins ago

Yanis-git

1,3171217

edited 32 mins ago

Yanis-git

1,3171217

answered 48 mins ago

Varun Arya

New contributor

answered 48 mins ago

Varun Arya

answered 48 mins ago

Varun Arya

New contributor

Varun Arya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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