How to sort elements in array without changing other elements indexes? [duplicate]

This question already has an answer here:

How to sort an array of odd numbers in ascending order, but keep even numbers at their position?

6 answers

I have this array:

var arr = [5, 3, 2, 8, 1, 4];

I'm trying to sort ONLY the elements that are odd values so I want this

output:

[1, 3, 2, 8, 5, 4]

As you can see the even elements don't change their position. Can anyone tell me what I'm missing? Here's my code:

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0)
 arr.push(val);
 
 return arr.sort();
 , );

 return oddElements;

console.log(myFunction([5, 3, 2, 8, 1, 4]));

I know I can use slice to add elements to array, but I'm stuck on how to get the indexes and put the elements in the array.

edited Jan 19 at 6:00

asked Jan 19 at 5:55

progx

328522

marked as duplicate by Nina Scholz javascript
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Jan 19 at 9:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment |

This question already has an answer here:

How to sort an array of odd numbers in ascending order, but keep even numbers at their position?

6 answers

I have this array:

var arr = [5, 3, 2, 8, 1, 4];

I'm trying to sort ONLY the elements that are odd values so I want this

output:

[1, 3, 2, 8, 5, 4]

As you can see the even elements don't change their position. Can anyone tell me what I'm missing? Here's my code:

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0)
 arr.push(val);
 
 return arr.sort();
 , );

 return oddElements;

console.log(myFunction([5, 3, 2, 8, 1, 4]));

I know I can use slice to add elements to array, but I'm stuck on how to get the indexes and put the elements in the array.

edited Jan 19 at 6:00

asked Jan 19 at 5:55

progx

328522

marked as duplicate by Nina Scholz javascript
Users with the javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.

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Jan 19 at 9:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment |

This question already has an answer here:

How to sort an array of odd numbers in ascending order, but keep even numbers at their position?

6 answers

I have this array:

var arr = [5, 3, 2, 8, 1, 4];

I'm trying to sort ONLY the elements that are odd values so I want this

output:

[1, 3, 2, 8, 5, 4]

As you can see the even elements don't change their position. Can anyone tell me what I'm missing? Here's my code:

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0)
 arr.push(val);
 
 return arr.sort();
 , );

 return oddElements;

console.log(myFunction([5, 3, 2, 8, 1, 4]));

I know I can use slice to add elements to array, but I'm stuck on how to get the indexes and put the elements in the array.

edited Jan 19 at 6:00

asked Jan 19 at 5:55

progx

328522

This question already has an answer here:

How to sort an array of odd numbers in ascending order, but keep even numbers at their position?

6 answers

I have this array:

var arr = [5, 3, 2, 8, 1, 4];

I'm trying to sort ONLY the elements that are odd values so I want this

output:

[1, 3, 2, 8, 5, 4]

As you can see the even elements don't change their position. Can anyone tell me what I'm missing? Here's my code:

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0)
 arr.push(val);
 
 return arr.sort();
 , );

 return oddElements;

console.log(myFunction([5, 3, 2, 8, 1, 4]));

I know I can use slice to add elements to array, but I'm stuck on how to get the indexes and put the elements in the array.

This question already has an answer here:

How to sort an array of odd numbers in ascending order, but keep even numbers at their position?

6 answers

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0)
 arr.push(val);
 
 return arr.sort();
 , );

 return oddElements;

console.log(myFunction([5, 3, 2, 8, 1, 4]));

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0)
 arr.push(val);
 
 return arr.sort();
 , );

 return oddElements;

console.log(myFunction([5, 3, 2, 8, 1, 4]));

javascript

edited Jan 19 at 6:00

asked Jan 19 at 5:55

progx

328522

edited Jan 19 at 6:00

asked Jan 19 at 5:55

progx

328522

edited Jan 19 at 6:00

asked Jan 19 at 5:55

progx

328522

asked Jan 19 at 5:55

progx

328522

asked Jan 19 at 5:55

progx

328522

marked as duplicate by Nina Scholz javascript
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Jan 19 at 9:21

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Jan 19 at 9:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment |

3 Answers
3

active

oldest

votes

One option is to keep track of the indicies of the odd numbers in the original array, and after .reduceing and sorting, then iterate through the original odd indicies and reassign, taking from the sorted odd array:

function oddSort(array) 
 const oddIndicies = ;
 const newArr = array.slice();
 const sortedOdd = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 oddIndicies.push(index);
 
 return arr;
 , )
 .sort((a, b) => a - b);
 while (oddIndicies.length > 0) 
 newArr[oddIndicies.shift()] = sortedOdd.shift();
 
 return newArr;


console.log(oddSort([5, 3, 2, 8, 1, 4]));
console.log(oddSort([5, 3, 2, 8, 1, 4, 11 ]));

edited Jan 19 at 6:09

answered Jan 19 at 6:00

CertainPerformance

84.6k154169

your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:06

Ah, the 11 was throwing things off, since .sort sorts lexiographically - use a custom .sort function instead, see edit

– CertainPerformance
Jan 19 at 6:10

Thanks a lot bro!

– progx
Jan 19 at 21:21

add a comment |

First sort only the odd numbers and put it in an array oddSorted. Then map through each element in the original array and check if the current element is odd, if odd replace it with the corresponding sorted number from the oddSorted array.

function sortOddElements(arr)
 var oddSorted = arr.filter(ele => ele %2 != 0).sort((a, b) => a - b);
 var evenNotSorted = arr.map((ele, idx) => 
 if(ele % 2 != 0)
 return oddSorted.shift(); 
 
 return ele;
 );
 return evenNotSorted;

var arr = [5, 3, 2, 8, 1, 4];
console.log(sortOddElements(arr));
arr = [5, 3, 2, 8, 1, 4, 11 ];
console.log(sortOddElements(arr));

edited Jan 19 at 8:43

answered Jan 19 at 6:30

Amardeep Bhowmick

2,56511022

easy-to-follow answer. now, can you think of a way to do this without having to do the % 2 != 0 calculation twice per element? :D

– user633183
Jan 19 at 6:35

One thing that comes to my mind is checking whether the current element is included in the oddSorted array, if yes then replace the current number with the first number from the oddSorted array. This will also work as if the current element is included in the oddSorted array then it implies that the number is odd and hence there is no need to check it again. But do you think this will improve the performance?

– Amardeep Bhowmick
Jan 19 at 6:44

You don't need to clone arr. Filter will return a new array anyway.

– kremerd
Jan 19 at 8:32

add a comment |

I modified your code a little bit to fulfill your objective. Take a look below

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 
 return arr.sort(function(a, b)return a - b);
 , );

 var index = 0;
 var finalElements = ;
 for(var i=0; i<array.length; i++) 
 var element = array[i];
 if(element %2 !==0) 
 finalElements.push(oddElements[index]);
 index++;
 else 
 finalElements.push(element);
 
 
 return finalElements;

console.log(myFunction([5, 3, 2, 8, 1, 4, 11]));

Remember, the default sort function sorts the values alphabetically. That's why you can't just use arr.sort()

edited Jan 19 at 6:17

answered Jan 19 at 6:08

Tanmoy Krishna Das

625414

Nope, your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:09

Okay... I'm taking a look at it

– Tanmoy Krishna Das
Jan 19 at 6:10

The problem was with the sort function. The default sort function sorts the values alphabetically. That's why you can't just use arr.sort(). I modified the code to reflect your needs.

– Tanmoy Krishna Das
Jan 19 at 6:18

add a comment |

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

function oddSort(array) 
 const oddIndicies = ;
 const newArr = array.slice();
 const sortedOdd = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 oddIndicies.push(index);
 
 return arr;
 , )
 .sort((a, b) => a - b);
 while (oddIndicies.length > 0) 
 newArr[oddIndicies.shift()] = sortedOdd.shift();
 
 return newArr;


console.log(oddSort([5, 3, 2, 8, 1, 4]));
console.log(oddSort([5, 3, 2, 8, 1, 4, 11 ]));

edited Jan 19 at 6:09

answered Jan 19 at 6:00

CertainPerformance

84.6k154169

your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:06

Ah, the 11 was throwing things off, since .sort sorts lexiographically - use a custom .sort function instead, see edit

– CertainPerformance
Jan 19 at 6:10

Thanks a lot bro!

– progx
Jan 19 at 21:21

add a comment |

function oddSort(array) 
 const oddIndicies = ;
 const newArr = array.slice();
 const sortedOdd = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 oddIndicies.push(index);
 
 return arr;
 , )
 .sort((a, b) => a - b);
 while (oddIndicies.length > 0) 
 newArr[oddIndicies.shift()] = sortedOdd.shift();
 
 return newArr;


console.log(oddSort([5, 3, 2, 8, 1, 4]));
console.log(oddSort([5, 3, 2, 8, 1, 4, 11 ]));

edited Jan 19 at 6:09

answered Jan 19 at 6:00

CertainPerformance

84.6k154169

your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:06

Ah, the 11 was throwing things off, since .sort sorts lexiographically - use a custom .sort function instead, see edit

– CertainPerformance
Jan 19 at 6:10

Thanks a lot bro!

– progx
Jan 19 at 21:21

add a comment |

function oddSort(array) 
 const oddIndicies = ;
 const newArr = array.slice();
 const sortedOdd = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 oddIndicies.push(index);
 
 return arr;
 , )
 .sort((a, b) => a - b);
 while (oddIndicies.length > 0) 
 newArr[oddIndicies.shift()] = sortedOdd.shift();
 
 return newArr;


console.log(oddSort([5, 3, 2, 8, 1, 4]));
console.log(oddSort([5, 3, 2, 8, 1, 4, 11 ]));

edited Jan 19 at 6:09

answered Jan 19 at 6:00

CertainPerformance

84.6k154169

function oddSort(array) 
 const oddIndicies = ;
 const newArr = array.slice();
 const sortedOdd = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 oddIndicies.push(index);
 
 return arr;
 , )
 .sort((a, b) => a - b);
 while (oddIndicies.length > 0) 
 newArr[oddIndicies.shift()] = sortedOdd.shift();
 
 return newArr;


console.log(oddSort([5, 3, 2, 8, 1, 4]));
console.log(oddSort([5, 3, 2, 8, 1, 4, 11 ]));

function oddSort(array) 
 const oddIndicies = ;
 const newArr = array.slice();
 const sortedOdd = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 oddIndicies.push(index);
 
 return arr;
 , )
 .sort((a, b) => a - b);
 while (oddIndicies.length > 0) 
 newArr[oddIndicies.shift()] = sortedOdd.shift();
 
 return newArr;


console.log(oddSort([5, 3, 2, 8, 1, 4]));
console.log(oddSort([5, 3, 2, 8, 1, 4, 11 ]));

function oddSort(array) 
 const oddIndicies = ;
 const newArr = array.slice();
 const sortedOdd = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 oddIndicies.push(index);
 
 return arr;
 , )
 .sort((a, b) => a - b);
 while (oddIndicies.length > 0) 
 newArr[oddIndicies.shift()] = sortedOdd.shift();
 
 return newArr;


console.log(oddSort([5, 3, 2, 8, 1, 4]));
console.log(oddSort([5, 3, 2, 8, 1, 4, 11 ]));

edited Jan 19 at 6:09

answered Jan 19 at 6:00

CertainPerformance

84.6k154169

edited Jan 19 at 6:09

answered Jan 19 at 6:00

CertainPerformance

84.6k154169

answered Jan 19 at 6:00

CertainPerformance

84.6k154169

answered Jan 19 at 6:00

CertainPerformance

84.6k154169

your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:06

Ah, the 11 was throwing things off, since .sort sorts lexiographically - use a custom .sort function instead, see edit

– CertainPerformance
Jan 19 at 6:10

Thanks a lot bro!

– progx
Jan 19 at 21:21

add a comment |

your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:06

Ah, the 11 was throwing things off, since .sort sorts lexiographically - use a custom .sort function instead, see edit

– CertainPerformance
Jan 19 at 6:10

Thanks a lot bro!

– progx
Jan 19 at 21:21

your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:06

Ah, the 11 was throwing things off, since .sort sorts lexiographically - use a custom .sort function instead, see edit

– CertainPerformance
Jan 19 at 6:10

Thanks a lot bro!

– progx
Jan 19 at 21:21

add a comment |

function sortOddElements(arr)
 var oddSorted = arr.filter(ele => ele %2 != 0).sort((a, b) => a - b);
 var evenNotSorted = arr.map((ele, idx) => 
 if(ele % 2 != 0)
 return oddSorted.shift(); 
 
 return ele;
 );
 return evenNotSorted;

var arr = [5, 3, 2, 8, 1, 4];
console.log(sortOddElements(arr));
arr = [5, 3, 2, 8, 1, 4, 11 ];
console.log(sortOddElements(arr));

edited Jan 19 at 8:43

answered Jan 19 at 6:30

Amardeep Bhowmick

2,56511022

easy-to-follow answer. now, can you think of a way to do this without having to do the % 2 != 0 calculation twice per element? :D

– user633183
Jan 19 at 6:35

One thing that comes to my mind is checking whether the current element is included in the oddSorted array, if yes then replace the current number with the first number from the oddSorted array. This will also work as if the current element is included in the oddSorted array then it implies that the number is odd and hence there is no need to check it again. But do you think this will improve the performance?

– Amardeep Bhowmick
Jan 19 at 6:44

You don't need to clone arr. Filter will return a new array anyway.

– kremerd
Jan 19 at 8:32

add a comment |

function sortOddElements(arr)
 var oddSorted = arr.filter(ele => ele %2 != 0).sort((a, b) => a - b);
 var evenNotSorted = arr.map((ele, idx) => 
 if(ele % 2 != 0)
 return oddSorted.shift(); 
 
 return ele;
 );
 return evenNotSorted;

var arr = [5, 3, 2, 8, 1, 4];
console.log(sortOddElements(arr));
arr = [5, 3, 2, 8, 1, 4, 11 ];
console.log(sortOddElements(arr));

edited Jan 19 at 8:43

answered Jan 19 at 6:30

Amardeep Bhowmick

2,56511022

easy-to-follow answer. now, can you think of a way to do this without having to do the % 2 != 0 calculation twice per element? :D

– user633183
Jan 19 at 6:35

One thing that comes to my mind is checking whether the current element is included in the oddSorted array, if yes then replace the current number with the first number from the oddSorted array. This will also work as if the current element is included in the oddSorted array then it implies that the number is odd and hence there is no need to check it again. But do you think this will improve the performance?

– Amardeep Bhowmick
Jan 19 at 6:44

You don't need to clone arr. Filter will return a new array anyway.

– kremerd
Jan 19 at 8:32

add a comment |

function sortOddElements(arr)
 var oddSorted = arr.filter(ele => ele %2 != 0).sort((a, b) => a - b);
 var evenNotSorted = arr.map((ele, idx) => 
 if(ele % 2 != 0)
 return oddSorted.shift(); 
 
 return ele;
 );
 return evenNotSorted;

var arr = [5, 3, 2, 8, 1, 4];
console.log(sortOddElements(arr));
arr = [5, 3, 2, 8, 1, 4, 11 ];
console.log(sortOddElements(arr));

edited Jan 19 at 8:43

answered Jan 19 at 6:30

Amardeep Bhowmick

2,56511022

function sortOddElements(arr)
 var oddSorted = arr.filter(ele => ele %2 != 0).sort((a, b) => a - b);
 var evenNotSorted = arr.map((ele, idx) => 
 if(ele % 2 != 0)
 return oddSorted.shift(); 
 
 return ele;
 );
 return evenNotSorted;

var arr = [5, 3, 2, 8, 1, 4];
console.log(sortOddElements(arr));
arr = [5, 3, 2, 8, 1, 4, 11 ];
console.log(sortOddElements(arr));

function sortOddElements(arr)
 var oddSorted = arr.filter(ele => ele %2 != 0).sort((a, b) => a - b);
 var evenNotSorted = arr.map((ele, idx) => 
 if(ele % 2 != 0)
 return oddSorted.shift(); 
 
 return ele;
 );
 return evenNotSorted;

var arr = [5, 3, 2, 8, 1, 4];
console.log(sortOddElements(arr));
arr = [5, 3, 2, 8, 1, 4, 11 ];
console.log(sortOddElements(arr));

function sortOddElements(arr)
 var oddSorted = arr.filter(ele => ele %2 != 0).sort((a, b) => a - b);
 var evenNotSorted = arr.map((ele, idx) => 
 if(ele % 2 != 0)
 return oddSorted.shift(); 
 
 return ele;
 );
 return evenNotSorted;

var arr = [5, 3, 2, 8, 1, 4];
console.log(sortOddElements(arr));
arr = [5, 3, 2, 8, 1, 4, 11 ];
console.log(sortOddElements(arr));

edited Jan 19 at 8:43

answered Jan 19 at 6:30

Amardeep Bhowmick

2,56511022

edited Jan 19 at 8:43

answered Jan 19 at 6:30

Amardeep Bhowmick

2,56511022

answered Jan 19 at 6:30

Amardeep Bhowmick

2,56511022

answered Jan 19 at 6:30

Amardeep Bhowmick

2,56511022

easy-to-follow answer. now, can you think of a way to do this without having to do the % 2 != 0 calculation twice per element? :D

– user633183
Jan 19 at 6:35

One thing that comes to my mind is checking whether the current element is included in the oddSorted array, if yes then replace the current number with the first number from the oddSorted array. This will also work as if the current element is included in the oddSorted array then it implies that the number is odd and hence there is no need to check it again. But do you think this will improve the performance?

– Amardeep Bhowmick
Jan 19 at 6:44

You don't need to clone arr. Filter will return a new array anyway.

– kremerd
Jan 19 at 8:32

add a comment |

easy-to-follow answer. now, can you think of a way to do this without having to do the % 2 != 0 calculation twice per element? :D

– user633183
Jan 19 at 6:35

One thing that comes to my mind is checking whether the current element is included in the oddSorted array, if yes then replace the current number with the first number from the oddSorted array. This will also work as if the current element is included in the oddSorted array then it implies that the number is odd and hence there is no need to check it again. But do you think this will improve the performance?

– Amardeep Bhowmick
Jan 19 at 6:44

You don't need to clone arr. Filter will return a new array anyway.

– kremerd
Jan 19 at 8:32

easy-to-follow answer. now, can you think of a way to do this without having to do the % 2 != 0 calculation twice per element? :D

– user633183
Jan 19 at 6:35

One thing that comes to my mind is checking whether the current element is included in the oddSorted array, if yes then replace the current number with the first number from the oddSorted array. This will also work as if the current element is included in the oddSorted array then it implies that the number is odd and hence there is no need to check it again. But do you think this will improve the performance?

– Amardeep Bhowmick
Jan 19 at 6:44

You don't need to clone arr. Filter will return a new array anyway.

– kremerd
Jan 19 at 8:32

add a comment |

I modified your code a little bit to fulfill your objective. Take a look below

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 
 return arr.sort(function(a, b)return a - b);
 , );

 var index = 0;
 var finalElements = ;
 for(var i=0; i<array.length; i++) 
 var element = array[i];
 if(element %2 !==0) 
 finalElements.push(oddElements[index]);
 index++;
 else 
 finalElements.push(element);
 
 
 return finalElements;

console.log(myFunction([5, 3, 2, 8, 1, 4, 11]));

Remember, the default sort function sorts the values alphabetically. That's why you can't just use arr.sort()

edited Jan 19 at 6:17

answered Jan 19 at 6:08

Tanmoy Krishna Das

625414

Nope, your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:09

Okay... I'm taking a look at it

– Tanmoy Krishna Das
Jan 19 at 6:10

The problem was with the sort function. The default sort function sorts the values alphabetically. That's why you can't just use arr.sort(). I modified the code to reflect your needs.

– Tanmoy Krishna Das
Jan 19 at 6:18

add a comment |

I modified your code a little bit to fulfill your objective. Take a look below

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 
 return arr.sort(function(a, b)return a - b);
 , );

 var index = 0;
 var finalElements = ;
 for(var i=0; i<array.length; i++) 
 var element = array[i];
 if(element %2 !==0) 
 finalElements.push(oddElements[index]);
 index++;
 else 
 finalElements.push(element);
 
 
 return finalElements;

console.log(myFunction([5, 3, 2, 8, 1, 4, 11]));

Remember, the default sort function sorts the values alphabetically. That's why you can't just use arr.sort()

edited Jan 19 at 6:17

answered Jan 19 at 6:08

Tanmoy Krishna Das

625414

Nope, your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:09

Okay... I'm taking a look at it

– Tanmoy Krishna Das
Jan 19 at 6:10

The problem was with the sort function. The default sort function sorts the values alphabetically. That's why you can't just use arr.sort(). I modified the code to reflect your needs.

– Tanmoy Krishna Das
Jan 19 at 6:18

add a comment |

I modified your code a little bit to fulfill your objective. Take a look below

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 
 return arr.sort(function(a, b)return a - b);
 , );

 var index = 0;
 var finalElements = ;
 for(var i=0; i<array.length; i++) 
 var element = array[i];
 if(element %2 !==0) 
 finalElements.push(oddElements[index]);
 index++;
 else 
 finalElements.push(element);
 
 
 return finalElements;

console.log(myFunction([5, 3, 2, 8, 1, 4, 11]));

Remember, the default sort function sorts the values alphabetically. That's why you can't just use arr.sort()

edited Jan 19 at 6:17

answered Jan 19 at 6:08

Tanmoy Krishna Das

625414

I modified your code a little bit to fulfill your objective. Take a look below

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 
 return arr.sort(function(a, b)return a - b);
 , );

 var index = 0;
 var finalElements = ;
 for(var i=0; i<array.length; i++) 
 var element = array[i];
 if(element %2 !==0) 
 finalElements.push(oddElements[index]);
 index++;
 else 
 finalElements.push(element);
 
 
 return finalElements;

console.log(myFunction([5, 3, 2, 8, 1, 4, 11]));

Remember, the default sort function sorts the values alphabetically. That's why you can't just use arr.sort()

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 
 return arr.sort(function(a, b)return a - b);
 , );

 var index = 0;
 var finalElements = ;
 for(var i=0; i<array.length; i++) 
 var element = array[i];
 if(element %2 !==0) 
 finalElements.push(oddElements[index]);
 index++;
 else 
 finalElements.push(element);
 
 
 return finalElements;

console.log(myFunction([5, 3, 2, 8, 1, 4, 11]));

function myFunction(array) 

 var oddElements = array.reduce((arr, val, index) => 
 if (val % 2 !== 0) 
 arr.push(val);
 
 return arr.sort(function(a, b)return a - b);
 , );

 var index = 0;
 var finalElements = ;
 for(var i=0; i<array.length; i++) 
 var element = array[i];
 if(element %2 !==0) 
 finalElements.push(oddElements[index]);
 index++;
 else 
 finalElements.push(element);
 
 
 return finalElements;

console.log(myFunction([5, 3, 2, 8, 1, 4, 11]));

edited Jan 19 at 6:17

answered Jan 19 at 6:08

Tanmoy Krishna Das

625414

edited Jan 19 at 6:17

answered Jan 19 at 6:08

Tanmoy Krishna Das

625414

answered Jan 19 at 6:08

Tanmoy Krishna Das

625414

answered Jan 19 at 6:08

Tanmoy Krishna Das

625414

Nope, your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:09

Okay... I'm taking a look at it

– Tanmoy Krishna Das
Jan 19 at 6:10

The problem was with the sort function. The default sort function sorts the values alphabetically. That's why you can't just use arr.sort(). I modified the code to reflect your needs.

– Tanmoy Krishna Das
Jan 19 at 6:18

add a comment |

Nope, your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:09

Okay... I'm taking a look at it

– Tanmoy Krishna Das
Jan 19 at 6:10

The problem was with the sort function. The default sort function sorts the values alphabetically. That's why you can't just use arr.sort(). I modified the code to reflect your needs.

– Tanmoy Krishna Das
Jan 19 at 6:18

Nope, your solution doesn't work when I pass this array: [5, 3, 2, 8, 1, 4, 11 ], it supposes to return [1, 3, 2, 8, 5, 4, 11], but it returns [1, 11, 2, 8, 3, 4, 5]

– progx
Jan 19 at 6:09

Okay... I'm taking a look at it

– Tanmoy Krishna Das
Jan 19 at 6:10

The problem was with the sort function. The default sort function sorts the values alphabetically. That's why you can't just use arr.sort(). I modified the code to reflect your needs.

– Tanmoy Krishna Das
Jan 19 at 6:18

add a comment |

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