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4

I am trying to arrange any kind of array input into a sorted array that also combines the equal pairs into an array inside the same array.

I do the following

const arrangeTheArray=(arr)=>
 
 //checking input here 
 if(arr.length<2)
 
 return arr;
 else 
 //sorting the array
 arr= arr.sort();
 
 //displaying the sorted array
 console.log(arr);
 for(let i=1; i<arr.length;i++)
 
 for(let j=0;j<i;j++)
 
 //here I am looping and comparing the values of array 
 if(arr[j]===arr[i])
 //putting the value
 arr[j]= [arr[i],arr[j]]; 
 
 
 
 //displaying the final output
 console.log(arr);

arrangeTheArray([0,2,2,1,1,6,3,1,0])

e.g array input : [0,2,2,1,1,6,3,1,0]
final out put: [[0,0],[1,1,1],[2,2],3,6]

javascript arrays

share|improve this question

edited Mar 24 at 15:29

Code Maniac

15.1k21035

asked Mar 24 at 15:19

jsLearnerjsLearner

574

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  your sorting goes wrong for numbers greater then 9, becaus ou are sorting strings and not numbers.
  
  – Nina Scholz
  Mar 24 at 16:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sometimes I really mess up with logic so by doing sort(a,b=>return a-b)method would it solve the problem
  
  – jsLearner
  Mar 24 at 16:29
  

  
  
  
  

add a comment | 

4

I am trying to arrange any kind of array input into a sorted array that also combines the equal pairs into an array inside the same array.

I do the following

const arrangeTheArray=(arr)=>
 
 //checking input here 
 if(arr.length<2)
 
 return arr;
 else 
 //sorting the array
 arr= arr.sort();
 
 //displaying the sorted array
 console.log(arr);
 for(let i=1; i<arr.length;i++)
 
 for(let j=0;j<i;j++)
 
 //here I am looping and comparing the values of array 
 if(arr[j]===arr[i])
 //putting the value
 arr[j]= [arr[i],arr[j]]; 
 
 
 
 //displaying the final output
 console.log(arr);

arrangeTheArray([0,2,2,1,1,6,3,1,0])

e.g array input : [0,2,2,1,1,6,3,1,0]
final out put: [[0,0],[1,1,1],[2,2],3,6]

javascript arrays

share|improve this question

edited Mar 24 at 15:29

Code Maniac

15.1k21035

asked Mar 24 at 15:19

jsLearnerjsLearner

574

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  your sorting goes wrong for numbers greater then 9, becaus ou are sorting strings and not numbers.
  
  – Nina Scholz
  Mar 24 at 16:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sometimes I really mess up with logic so by doing sort(a,b=>return a-b)method would it solve the problem
  
  – jsLearner
  Mar 24 at 16:29
  

  
  
  
  

add a comment | 

I am trying to arrange any kind of array input into a sorted array that also combines the equal pairs into an array inside the same array.

I do the following

const arrangeTheArray=(arr)=>
 
 //checking input here 
 if(arr.length<2)
 
 return arr;
 else 
 //sorting the array
 arr= arr.sort();
 
 //displaying the sorted array
 console.log(arr);
 for(let i=1; i<arr.length;i++)
 
 for(let j=0;j<i;j++)
 
 //here I am looping and comparing the values of array 
 if(arr[j]===arr[i])
 //putting the value
 arr[j]= [arr[i],arr[j]]; 
 
 
 
 //displaying the final output
 console.log(arr);

arrangeTheArray([0,2,2,1,1,6,3,1,0])

e.g array input : [0,2,2,1,1,6,3,1,0]
final out put: [[0,0],[1,1,1],[2,2],3,6]

javascript arrays

share|improve this question

edited Mar 24 at 15:29

Code Maniac

15.1k21035

asked Mar 24 at 15:19

jsLearnerjsLearner

574

const arrangeTheArray=(arr)=>
 
 //checking input here 
 if(arr.length<2)
 
 return arr;
 else 
 //sorting the array
 arr= arr.sort();
 
 //displaying the sorted array
 console.log(arr);
 for(let i=1; i<arr.length;i++)
 
 for(let j=0;j<i;j++)
 
 //here I am looping and comparing the values of array 
 if(arr[j]===arr[i])
 //putting the value
 arr[j]= [arr[i],arr[j]]; 
 
 
 
 //displaying the final output
 console.log(arr);

arrangeTheArray([0,2,2,1,1,6,3,1,0])

share|improve this question

edited Mar 24 at 15:29

Code Maniac

15.1k21035

asked Mar 24 at 15:19

jsLearnerjsLearner

574

share|improve this question

edited Mar 24 at 15:29

Code Maniac

15.1k21035

asked Mar 24 at 15:19

jsLearnerjsLearner

574

edited Mar 24 at 15:29

Code Maniac

15.1k21035

edited Mar 24 at 15:29

Code Maniac

15.1k21035

asked Mar 24 at 15:19

jsLearnerjsLearner

574

asked Mar 24 at 15:19

jsLearnerjsLearner

574

3 Answers
3

active

oldest

votes

1

You can use reduce and map.

Here idea is

• First create a object with each digit as key and group the values by key.


• Now map on the grouped data, if the length of element greater than one pass element as it is, else pass the 0th index value

let combine = (arr) =>
 let groups = arr.reduce((op,inp)=>,)
 let final = Object.values(groups).map(e=> e.length > 1 ? e : e[0])
 return final


console.log(combine([0,2,2,1,1,6,3,1,0]))

const arrangeTheArray=(arr)=>
 if(arr.length<2)
 
 return arr;
 else 
 arr = arr.sort((a,b)=>a-b);
 
 let final = []
 for(let i=0;i<arr.length;i++)
 let current = arr[i]
 let j = i;
 let temp = []
 while(arr[j] === current)
 temp.push(arr[j])
 j++
 
 i = j-1
 temp = temp.length > 1 ? temp : temp[0]
 final.push(temp)
 
 console.log(final)

arrangeTheArray([0,2,2,1,1,6,3,1,0])

share|improve this answer

edited Mar 24 at 16:10

answered Mar 24 at 15:22

Code ManiacCode Maniac

15.1k21035

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well Thank you so much for your answer it is giving the desired output.I am facing little difficulties in understanding your code though. Can you tell me if I can achieve the same output using my code instead
  
  – jsLearner
  Mar 24 at 15:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jsLearner you can tell me where you're facing problem to understand will help you. wait let me update question with your version too
  
  – Code Maniac
  Mar 24 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I gotta read and understand reduce properly. I was trying to achieve that with just arrays and for loop. So far i was able to sort the array and even manager to combine few pairs but that does not give the correct answer for all the inputs
  
  – jsLearner
  Mar 24 at 15:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This works however the best answer should be the first way you mentioned
  
  – jsLearner
  Mar 24 at 16:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  map and filter i knew but could not think if we can use it here...gotta work on programming logic alot I am still learning
  
  – jsLearner
  Mar 24 at 16:18
  

  
  
  
  

 | 
show 2 more comments

1

You can use Array.reduce to accumulate the common elements in an object.

Then use Object.values and Array.from to process the nested arrays into arrays of common elements and distinct elements:

const arrangeTheArray = (arr) => 
 if (!Array.isArray(arr) && arr.length < 2)
 return arr;
 
 const pairs = arr.reduce((acc, ele) => 
 if(acc[ele])
 acc[ele].push(ele);
 else
 acc[ele] = [ele];
 
 return acc;
 , );
 return Array.from(Object.values(pairs), ele => ele.length > 1 ? ele : +ele.join())
 
 
 console.log(arrangeTheArray([0,2,2,1,1,6,3,1,0]));

share|improve this answer

edited Mar 24 at 15:51

answered Mar 24 at 15:32

Amardeep BhowmickAmardeep Bhowmick

6,54421231

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  thanks for your answer let me check
  
  – jsLearner
  Mar 24 at 15:59
  

  
  
  
  

add a comment | 

1

You could sort the array with a callback for numbers and reduce the array by checking the predecessor p and the actual value v and push either the value or an array of the last group and the value.

function arrangeTheArray(array) 
 return array
 .sort((a, b) => a - b)
 .reduce(
 (r, v, i, [i - 1]: p ) => r.concat([p !== v ? v : [].concat(r.pop(), v)]),
 []
 );


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

With a classsic while loop from the end, because the array shrinks.

function arrangeTheArray(array) 
 var i = array.length - 1;
 array.sort((a, b) => a - b);
 while (i--) 
 if (array[i] === (Array.isArray(array[i + 1]) ? array[i + 1][0] : array[i + 1])) 
 array[i] = [].concat(array[i], ...array.splice(i + 1, 1));
 
 
 return array;


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

share|improve this answer

edited Mar 24 at 16:22

answered Mar 24 at 15:33

Nina ScholzNina Scholz

208k16117190

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this looks good, thanks for the answer gotta read the reduce method once. But can we do it with for loop as well the one which I am trying ?
  
  – jsLearner
  Mar 24 at 15:49
  

  
  
  
  

add a comment | 

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1

You can use reduce and map.

Here idea is

• First create a object with each digit as key and group the values by key.


• Now map on the grouped data, if the length of element greater than one pass element as it is, else pass the 0th index value

let combine = (arr) =>
 let groups = arr.reduce((op,inp)=>,)
 let final = Object.values(groups).map(e=> e.length > 1 ? e : e[0])
 return final


console.log(combine([0,2,2,1,1,6,3,1,0]))

const arrangeTheArray=(arr)=>
 if(arr.length<2)
 
 return arr;
 else 
 arr = arr.sort((a,b)=>a-b);
 
 let final = []
 for(let i=0;i<arr.length;i++)
 let current = arr[i]
 let j = i;
 let temp = []
 while(arr[j] === current)
 temp.push(arr[j])
 j++
 
 i = j-1
 temp = temp.length > 1 ? temp : temp[0]
 final.push(temp)
 
 console.log(final)

arrangeTheArray([0,2,2,1,1,6,3,1,0])

share|improve this answer

edited Mar 24 at 16:10

answered Mar 24 at 15:22

Code ManiacCode Maniac

15.1k21035

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well Thank you so much for your answer it is giving the desired output.I am facing little difficulties in understanding your code though. Can you tell me if I can achieve the same output using my code instead
  
  – jsLearner
  Mar 24 at 15:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jsLearner you can tell me where you're facing problem to understand will help you. wait let me update question with your version too
  
  – Code Maniac
  Mar 24 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I gotta read and understand reduce properly. I was trying to achieve that with just arrays and for loop. So far i was able to sort the array and even manager to combine few pairs but that does not give the correct answer for all the inputs
  
  – jsLearner
  Mar 24 at 15:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This works however the best answer should be the first way you mentioned
  
  – jsLearner
  Mar 24 at 16:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  map and filter i knew but could not think if we can use it here...gotta work on programming logic alot I am still learning
  
  – jsLearner
  Mar 24 at 16:18
  

  
  
  
  

 | 
show 2 more comments

1

You can use reduce and map.

Here idea is

• First create a object with each digit as key and group the values by key.


• Now map on the grouped data, if the length of element greater than one pass element as it is, else pass the 0th index value

let combine = (arr) =>
 let groups = arr.reduce((op,inp)=>,)
 let final = Object.values(groups).map(e=> e.length > 1 ? e : e[0])
 return final


console.log(combine([0,2,2,1,1,6,3,1,0]))

const arrangeTheArray=(arr)=>
 if(arr.length<2)
 
 return arr;
 else 
 arr = arr.sort((a,b)=>a-b);
 
 let final = []
 for(let i=0;i<arr.length;i++)
 let current = arr[i]
 let j = i;
 let temp = []
 while(arr[j] === current)
 temp.push(arr[j])
 j++
 
 i = j-1
 temp = temp.length > 1 ? temp : temp[0]
 final.push(temp)
 
 console.log(final)

arrangeTheArray([0,2,2,1,1,6,3,1,0])

share|improve this answer

edited Mar 24 at 16:10

answered Mar 24 at 15:22

Code ManiacCode Maniac

15.1k21035

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well Thank you so much for your answer it is giving the desired output.I am facing little difficulties in understanding your code though. Can you tell me if I can achieve the same output using my code instead
  
  – jsLearner
  Mar 24 at 15:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jsLearner you can tell me where you're facing problem to understand will help you. wait let me update question with your version too
  
  – Code Maniac
  Mar 24 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I gotta read and understand reduce properly. I was trying to achieve that with just arrays and for loop. So far i was able to sort the array and even manager to combine few pairs but that does not give the correct answer for all the inputs
  
  – jsLearner
  Mar 24 at 15:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This works however the best answer should be the first way you mentioned
  
  – jsLearner
  Mar 24 at 16:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  map and filter i knew but could not think if we can use it here...gotta work on programming logic alot I am still learning
  
  – jsLearner
  Mar 24 at 16:18
  

  
  
  
  

 | 
show 2 more comments

You can use reduce and map.

Here idea is

• First create a object with each digit as key and group the values by key.


• Now map on the grouped data, if the length of element greater than one pass element as it is, else pass the 0th index value

let combine = (arr) =>
 let groups = arr.reduce((op,inp)=>,)
 let final = Object.values(groups).map(e=> e.length > 1 ? e : e[0])
 return final


console.log(combine([0,2,2,1,1,6,3,1,0]))

const arrangeTheArray=(arr)=>
 if(arr.length<2)
 
 return arr;
 else 
 arr = arr.sort((a,b)=>a-b);
 
 let final = []
 for(let i=0;i<arr.length;i++)
 let current = arr[i]
 let j = i;
 let temp = []
 while(arr[j] === current)
 temp.push(arr[j])
 j++
 
 i = j-1
 temp = temp.length > 1 ? temp : temp[0]
 final.push(temp)
 
 console.log(final)

arrangeTheArray([0,2,2,1,1,6,3,1,0])

share|improve this answer

edited Mar 24 at 16:10

answered Mar 24 at 15:22

Code ManiacCode Maniac

15.1k21035

let combine = (arr) =>
 let groups = arr.reduce((op,inp)=>,)
 let final = Object.values(groups).map(e=> e.length > 1 ? e : e[0])
 return final


console.log(combine([0,2,2,1,1,6,3,1,0]))

const arrangeTheArray=(arr)=>
 if(arr.length<2)
 
 return arr;
 else 
 arr = arr.sort((a,b)=>a-b);
 
 let final = []
 for(let i=0;i<arr.length;i++)
 let current = arr[i]
 let j = i;
 let temp = []
 while(arr[j] === current)
 temp.push(arr[j])
 j++
 
 i = j-1
 temp = temp.length > 1 ? temp : temp[0]
 final.push(temp)
 
 console.log(final)

arrangeTheArray([0,2,2,1,1,6,3,1,0])

let combine = (arr) =>
 let groups = arr.reduce((op,inp)=>,)
 let final = Object.values(groups).map(e=> e.length > 1 ? e : e[0])
 return final


console.log(combine([0,2,2,1,1,6,3,1,0]))

let combine = (arr) =>
 let groups = arr.reduce((op,inp)=>,)
 let final = Object.values(groups).map(e=> e.length > 1 ? e : e[0])
 return final


console.log(combine([0,2,2,1,1,6,3,1,0]))

const arrangeTheArray=(arr)=>
 if(arr.length<2)
 
 return arr;
 else 
 arr = arr.sort((a,b)=>a-b);
 
 let final = []
 for(let i=0;i<arr.length;i++)
 let current = arr[i]
 let j = i;
 let temp = []
 while(arr[j] === current)
 temp.push(arr[j])
 j++
 
 i = j-1
 temp = temp.length > 1 ? temp : temp[0]
 final.push(temp)
 
 console.log(final)

arrangeTheArray([0,2,2,1,1,6,3,1,0])

const arrangeTheArray=(arr)=>
 if(arr.length<2)
 
 return arr;
 else 
 arr = arr.sort((a,b)=>a-b);
 
 let final = []
 for(let i=0;i<arr.length;i++)
 let current = arr[i]
 let j = i;
 let temp = []
 while(arr[j] === current)
 temp.push(arr[j])
 j++
 
 i = j-1
 temp = temp.length > 1 ? temp : temp[0]
 final.push(temp)
 
 console.log(final)

arrangeTheArray([0,2,2,1,1,6,3,1,0])

share|improve this answer

edited Mar 24 at 16:10

answered Mar 24 at 15:22

Code ManiacCode Maniac

15.1k21035

answered Mar 24 at 15:22

Code ManiacCode Maniac

15.1k21035

answered Mar 24 at 15:22

Code ManiacCode Maniac

15.1k21035

1

You can use Array.reduce to accumulate the common elements in an object.

Then use Object.values and Array.from to process the nested arrays into arrays of common elements and distinct elements:

const arrangeTheArray = (arr) => 
 if (!Array.isArray(arr) && arr.length < 2)
 return arr;
 
 const pairs = arr.reduce((acc, ele) => 
 if(acc[ele])
 acc[ele].push(ele);
 else
 acc[ele] = [ele];
 
 return acc;
 , );
 return Array.from(Object.values(pairs), ele => ele.length > 1 ? ele : +ele.join())
 
 
 console.log(arrangeTheArray([0,2,2,1,1,6,3,1,0]));

share|improve this answer

edited Mar 24 at 15:51

answered Mar 24 at 15:32

Amardeep BhowmickAmardeep Bhowmick

6,54421231

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  thanks for your answer let me check
  
  – jsLearner
  Mar 24 at 15:59
  

  
  
  
  

add a comment | 

1

You can use Array.reduce to accumulate the common elements in an object.

Then use Object.values and Array.from to process the nested arrays into arrays of common elements and distinct elements:

const arrangeTheArray = (arr) => 
 if (!Array.isArray(arr) && arr.length < 2)
 return arr;
 
 const pairs = arr.reduce((acc, ele) => 
 if(acc[ele])
 acc[ele].push(ele);
 else
 acc[ele] = [ele];
 
 return acc;
 , );
 return Array.from(Object.values(pairs), ele => ele.length > 1 ? ele : +ele.join())
 
 
 console.log(arrangeTheArray([0,2,2,1,1,6,3,1,0]));

share|improve this answer

edited Mar 24 at 15:51

answered Mar 24 at 15:32

Amardeep BhowmickAmardeep Bhowmick

6,54421231

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  thanks for your answer let me check
  
  – jsLearner
  Mar 24 at 15:59
  

  
  
  
  

add a comment | 

You can use Array.reduce to accumulate the common elements in an object.

Then use Object.values and Array.from to process the nested arrays into arrays of common elements and distinct elements:

const arrangeTheArray = (arr) => 
 if (!Array.isArray(arr) && arr.length < 2)
 return arr;
 
 const pairs = arr.reduce((acc, ele) => 
 if(acc[ele])
 acc[ele].push(ele);
 else
 acc[ele] = [ele];
 
 return acc;
 , );
 return Array.from(Object.values(pairs), ele => ele.length > 1 ? ele : +ele.join())
 
 
 console.log(arrangeTheArray([0,2,2,1,1,6,3,1,0]));

share|improve this answer

edited Mar 24 at 15:51

answered Mar 24 at 15:32

Amardeep BhowmickAmardeep Bhowmick

6,54421231

const arrangeTheArray = (arr) => 
 if (!Array.isArray(arr) && arr.length < 2)
 return arr;
 
 const pairs = arr.reduce((acc, ele) => 
 if(acc[ele])
 acc[ele].push(ele);
 else
 acc[ele] = [ele];
 
 return acc;
 , );
 return Array.from(Object.values(pairs), ele => ele.length > 1 ? ele : +ele.join())
 
 
 console.log(arrangeTheArray([0,2,2,1,1,6,3,1,0]));

const arrangeTheArray = (arr) => 
 if (!Array.isArray(arr) && arr.length < 2)
 return arr;
 
 const pairs = arr.reduce((acc, ele) => 
 if(acc[ele])
 acc[ele].push(ele);
 else
 acc[ele] = [ele];
 
 return acc;
 , );
 return Array.from(Object.values(pairs), ele => ele.length > 1 ? ele : +ele.join())
 
 
 console.log(arrangeTheArray([0,2,2,1,1,6,3,1,0]));

const arrangeTheArray = (arr) => 
 if (!Array.isArray(arr) && arr.length < 2)
 return arr;
 
 const pairs = arr.reduce((acc, ele) => 
 if(acc[ele])
 acc[ele].push(ele);
 else
 acc[ele] = [ele];
 
 return acc;
 , );
 return Array.from(Object.values(pairs), ele => ele.length > 1 ? ele : +ele.join())
 
 
 console.log(arrangeTheArray([0,2,2,1,1,6,3,1,0]));

share|improve this answer

edited Mar 24 at 15:51

answered Mar 24 at 15:32

Amardeep BhowmickAmardeep Bhowmick

6,54421231

answered Mar 24 at 15:32

Amardeep BhowmickAmardeep Bhowmick

6,54421231

answered Mar 24 at 15:32

Amardeep BhowmickAmardeep Bhowmick

6,54421231

1

You could sort the array with a callback for numbers and reduce the array by checking the predecessor p and the actual value v and push either the value or an array of the last group and the value.

function arrangeTheArray(array) 
 return array
 .sort((a, b) => a - b)
 .reduce(
 (r, v, i, [i - 1]: p ) => r.concat([p !== v ? v : [].concat(r.pop(), v)]),
 []
 );


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

With a classsic while loop from the end, because the array shrinks.

function arrangeTheArray(array) 
 var i = array.length - 1;
 array.sort((a, b) => a - b);
 while (i--) 
 if (array[i] === (Array.isArray(array[i + 1]) ? array[i + 1][0] : array[i + 1])) 
 array[i] = [].concat(array[i], ...array.splice(i + 1, 1));
 
 
 return array;


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

share|improve this answer

edited Mar 24 at 16:22

answered Mar 24 at 15:33

Nina ScholzNina Scholz

208k16117190

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this looks good, thanks for the answer gotta read the reduce method once. But can we do it with for loop as well the one which I am trying ?
  
  – jsLearner
  Mar 24 at 15:49
  

  
  
  
  

add a comment | 

1

You could sort the array with a callback for numbers and reduce the array by checking the predecessor p and the actual value v and push either the value or an array of the last group and the value.

function arrangeTheArray(array) 
 return array
 .sort((a, b) => a - b)
 .reduce(
 (r, v, i, [i - 1]: p ) => r.concat([p !== v ? v : [].concat(r.pop(), v)]),
 []
 );


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

With a classsic while loop from the end, because the array shrinks.

function arrangeTheArray(array) 
 var i = array.length - 1;
 array.sort((a, b) => a - b);
 while (i--) 
 if (array[i] === (Array.isArray(array[i + 1]) ? array[i + 1][0] : array[i + 1])) 
 array[i] = [].concat(array[i], ...array.splice(i + 1, 1));
 
 
 return array;


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

share|improve this answer

edited Mar 24 at 16:22

answered Mar 24 at 15:33

Nina ScholzNina Scholz

208k16117190

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this looks good, thanks for the answer gotta read the reduce method once. But can we do it with for loop as well the one which I am trying ?
  
  – jsLearner
  Mar 24 at 15:49
  

  
  
  
  

add a comment | 

You could sort the array with a callback for numbers and reduce the array by checking the predecessor p and the actual value v and push either the value or an array of the last group and the value.

function arrangeTheArray(array) 
 return array
 .sort((a, b) => a - b)
 .reduce(
 (r, v, i, [i - 1]: p ) => r.concat([p !== v ? v : [].concat(r.pop(), v)]),
 []
 );


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

With a classsic while loop from the end, because the array shrinks.

function arrangeTheArray(array) 
 var i = array.length - 1;
 array.sort((a, b) => a - b);
 while (i--) 
 if (array[i] === (Array.isArray(array[i + 1]) ? array[i + 1][0] : array[i + 1])) 
 array[i] = [].concat(array[i], ...array.splice(i + 1, 1));
 
 
 return array;


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

share|improve this answer

edited Mar 24 at 16:22

answered Mar 24 at 15:33

Nina ScholzNina Scholz

208k16117190

function arrangeTheArray(array) 
 return array
 .sort((a, b) => a - b)
 .reduce(
 (r, v, i, [i - 1]: p ) => r.concat([p !== v ? v : [].concat(r.pop(), v)]),
 []
 );


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

function arrangeTheArray(array) 
 var i = array.length - 1;
 array.sort((a, b) => a - b);
 while (i--) 
 if (array[i] === (Array.isArray(array[i + 1]) ? array[i + 1][0] : array[i + 1])) 
 array[i] = [].concat(array[i], ...array.splice(i + 1, 1));
 
 
 return array;


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

function arrangeTheArray(array) 
 return array
 .sort((a, b) => a - b)
 .reduce(
 (r, v, i, [i - 1]: p ) => r.concat([p !== v ? v : [].concat(r.pop(), v)]),
 []
 );


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

function arrangeTheArray(array) 
 return array
 .sort((a, b) => a - b)
 .reduce(
 (r, v, i, [i - 1]: p ) => r.concat([p !== v ? v : [].concat(r.pop(), v)]),
 []
 );


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

function arrangeTheArray(array) 
 var i = array.length - 1;
 array.sort((a, b) => a - b);
 while (i--) 
 if (array[i] === (Array.isArray(array[i + 1]) ? array[i + 1][0] : array[i + 1])) 
 array[i] = [].concat(array[i], ...array.splice(i + 1, 1));
 
 
 return array;


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

function arrangeTheArray(array) 
 var i = array.length - 1;
 array.sort((a, b) => a - b);
 while (i--) 
 if (array[i] === (Array.isArray(array[i + 1]) ? array[i + 1][0] : array[i + 1])) 
 array[i] = [].concat(array[i], ...array.splice(i + 1, 1));
 
 
 return array;


console.log(arrangeTheArray([0, 2, 2, 1, 1, 6, 3, 1, 0]));

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

share|improve this answer

edited Mar 24 at 16:22

answered Mar 24 at 15:33

Nina ScholzNina Scholz

208k16117190

answered Mar 24 at 15:33

Nina ScholzNina Scholz

208k16117190

answered Mar 24 at 15:33

Nina ScholzNina Scholz

208k16117190