How can I split my array into multiple smaller arrays?

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up vote
7
down vote

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My target 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?






javascript arrays







share|improve this question























  • I think you should set j=i in begin loop
    – Alexandr Kudryashov
    59 mins ago










  • @AlexandrKudryashov I have tried. It is not correct
    – TeodorKolev
    57 mins ago










  • set j=i and remove the if condition in the nested loop
    – rock star
    56 mins ago










  • init i with 0 and remove the = symbol in the corresponding condition
    – rock star
    56 mins ago






  • 1




    @rockstar nope, it is not correct
    – TeodorKolev
    55 mins ago














up vote
7
down vote

favorite
1












My target 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?






javascript arrays







share|improve this question























  • I think you should set j=i in begin loop
    – Alexandr Kudryashov
    59 mins ago










  • @AlexandrKudryashov I have tried. It is not correct
    – TeodorKolev
    57 mins ago










  • set j=i and remove the if condition in the nested loop
    – rock star
    56 mins ago










  • init i with 0 and remove the = symbol in the corresponding condition
    – rock star
    56 mins ago






  • 1




    @rockstar nope, it is not correct
    – TeodorKolev
    55 mins ago












up vote
7
down vote

favorite
1









up vote
7
down vote

favorite
1






1





My target 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?






javascript arrays







share|improve this question















My target 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




javascript arrays






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 52 mins ago









Jenny O'Reilly

10.1k73651




10.1k73651










asked 1 hour ago









TeodorKolev

95752043




95752043











  • I think you should set j=i in begin loop
    – Alexandr Kudryashov
    59 mins ago










  • @AlexandrKudryashov I have tried. It is not correct
    – TeodorKolev
    57 mins ago










  • set j=i and remove the if condition in the nested loop
    – rock star
    56 mins ago










  • init i with 0 and remove the = symbol in the corresponding condition
    – rock star
    56 mins ago






  • 1




    @rockstar nope, it is not correct
    – TeodorKolev
    55 mins ago
















  • I think you should set j=i in begin loop
    – Alexandr Kudryashov
    59 mins ago










  • @AlexandrKudryashov I have tried. It is not correct
    – TeodorKolev
    57 mins ago










  • set j=i and remove the if condition in the nested loop
    – rock star
    56 mins ago










  • init i with 0 and remove the = symbol in the corresponding condition
    – rock star
    56 mins ago






  • 1




    @rockstar nope, it is not correct
    – TeodorKolev
    55 mins ago















I think you should set j=i in begin loop
– Alexandr Kudryashov
59 mins ago




I think you should set j=i in begin loop
– Alexandr Kudryashov
59 mins ago












@AlexandrKudryashov I have tried. It is not correct
– TeodorKolev
57 mins ago




@AlexandrKudryashov I have tried. It is not correct
– TeodorKolev
57 mins ago












set j=i and remove the if condition in the nested loop
– rock star
56 mins ago




set j=i and remove the if condition in the nested loop
– rock star
56 mins ago












init i with 0 and remove the = symbol in the corresponding condition
– rock star
56 mins ago




init i with 0 and remove the = symbol in the corresponding condition
– rock star
56 mins ago




1




1




@rockstar nope, it is not correct
– TeodorKolev
55 mins ago




@rockstar nope, it is not correct
– TeodorKolev
55 mins ago












5 Answers
5






active

oldest

votes

















up vote
3
down vote



accepted










You have two issues in your code:



  1. 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

  2. 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);








share|improve this answer



























    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;








    share|improve this answer
















    • 1




      I was expecting answer from you :)
      – Ankit Agarwal
      54 mins ago










    • This do one big array. Target is to create small arrays
      – TeodorKolev
      54 mins ago










    • what do you want with small arrays? instead of pushing, you could display the sub array.
      – Nina Scholz
      50 mins ago

















    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);








    share|improve this answer






















    • Why do you need the tmp variable?
      – Jenny O'Reilly
      50 mins ago










    • you can do it without tmp variable=)
      – Alexandr Kudryashov
      49 mins ago

















    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






    share|improve this answer



























      up vote
      0
      down vote













      Use two iteration



      1. get slice array based on loop index.

      2. 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))








      share|improve this answer






















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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        3
        down vote



        accepted










        You have two issues in your code:



        1. 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

        2. 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);








        share|improve this answer
























          up vote
          3
          down vote



          accepted










          You have two issues in your code:



          1. 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

          2. 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);








          share|improve this answer






















            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            You have two issues in your code:



            1. 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

            2. 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);








            share|improve this answer












            You have two issues in your code:



            1. 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

            2. 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);






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 50 mins ago









            Ankit Agarwal

            20.8k31840




            20.8k31840






















                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;








                share|improve this answer
















                • 1




                  I was expecting answer from you :)
                  – Ankit Agarwal
                  54 mins ago










                • This do one big array. Target is to create small arrays
                  – TeodorKolev
                  54 mins ago










                • what do you want with small arrays? instead of pushing, you could display the sub array.
                  – Nina Scholz
                  50 mins ago














                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;








                share|improve this answer
















                • 1




                  I was expecting answer from you :)
                  – Ankit Agarwal
                  54 mins ago










                • This do one big array. Target is to create small arrays
                  – TeodorKolev
                  54 mins ago










                • what do you want with small arrays? instead of pushing, you could display the sub array.
                  – Nina Scholz
                  50 mins ago












                up vote
                7
                down vote










                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;








                share|improve this answer












                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;






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 56 mins ago









                Nina Scholz

                158k1277136




                158k1277136







                • 1




                  I was expecting answer from you :)
                  – Ankit Agarwal
                  54 mins ago










                • This do one big array. Target is to create small arrays
                  – TeodorKolev
                  54 mins ago










                • what do you want with small arrays? instead of pushing, you could display the sub array.
                  – Nina Scholz
                  50 mins ago












                • 1




                  I was expecting answer from you :)
                  – Ankit Agarwal
                  54 mins ago










                • This do one big array. Target is to create small arrays
                  – TeodorKolev
                  54 mins ago










                • what do you want with small arrays? instead of pushing, you could display the sub array.
                  – Nina Scholz
                  50 mins ago







                1




                1




                I was expecting answer from you :)
                – Ankit Agarwal
                54 mins ago




                I was expecting answer from you :)
                – Ankit Agarwal
                54 mins ago












                This do one big array. Target is to create small arrays
                – TeodorKolev
                54 mins ago




                This do one big array. Target is to create small arrays
                – TeodorKolev
                54 mins ago












                what do you want with small arrays? instead of pushing, you could display the sub array.
                – Nina Scholz
                50 mins ago




                what do you want with small arrays? instead of pushing, you could display the sub array.
                – Nina Scholz
                50 mins ago










                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);








                share|improve this answer






















                • Why do you need the tmp variable?
                  – Jenny O'Reilly
                  50 mins ago










                • you can do it without tmp variable=)
                  – Alexandr Kudryashov
                  49 mins ago














                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);








                share|improve this answer






















                • Why do you need the tmp variable?
                  – Jenny O'Reilly
                  50 mins ago










                • you can do it without tmp variable=)
                  – Alexandr Kudryashov
                  49 mins ago












                up vote
                2
                down vote










                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);








                share|improve this answer














                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);






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 52 mins ago









                Sudhir Ojha

                1,024313




                1,024313










                answered 53 mins ago









                Alexandr Kudryashov

                559213




                559213











                • Why do you need the tmp variable?
                  – Jenny O'Reilly
                  50 mins ago










                • you can do it without tmp variable=)
                  – Alexandr Kudryashov
                  49 mins ago
















                • Why do you need the tmp variable?
                  – Jenny O'Reilly
                  50 mins ago










                • you can do it without tmp variable=)
                  – Alexandr Kudryashov
                  49 mins ago















                Why do you need the tmp variable?
                – Jenny O'Reilly
                50 mins ago




                Why do you need the tmp variable?
                – Jenny O'Reilly
                50 mins ago












                you can do it without tmp variable=)
                – Alexandr Kudryashov
                49 mins ago




                you can do it without tmp variable=)
                – Alexandr Kudryashov
                49 mins ago










                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






                share|improve this answer
























                  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






                  share|improve this answer






















                    up vote
                    0
                    down vote










                    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






                    share|improve this answer












                    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







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 37 mins ago









                    Yanis-git

                    1,3151217




                    1,3151217




















                        up vote
                        0
                        down vote













                        Use two iteration



                        1. get slice array based on loop index.

                        2. 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))








                        share|improve this answer


























                          up vote
                          0
                          down vote













                          Use two iteration



                          1. get slice array based on loop index.

                          2. 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))








                          share|improve this answer
























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Use two iteration



                            1. get slice array based on loop index.

                            2. 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))








                            share|improve this answer














                            Use two iteration



                            1. get slice array based on loop index.

                            2. 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))






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 31 mins ago

























                            answered 43 mins ago









                            Anoop

                            18.8k94468




                            18.8k94468



























                                 

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