How to sort elements in array without changing other elements indexes?
9
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.
javascript
asked 1 hour ago
progxprogx
309521
add a comment |
9
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.
javascript
asked 1 hour ago
progxprogx
309521
add a comment |
9
9
9
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.
javascript
asked 1 hour ago
progxprogx
309521
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.
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
javascript
asked 1 hour ago
progxprogx
309521
asked 1 hour ago
progxprogx
309521
edited 1 hour ago
progx
asked 1 hour ago
progxprogx
309521
asked 1 hour ago
progxprogx
309521
asked 1 hour ago
progxprogx
309521
309521
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
4
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 temp = [...arr];
var oddSorted = temp.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));answered 51 mins ago
Amardeep BhowmickAmardeep Bhowmick
1,7521821
easy-to-follow answer. now, can you think of a way to do this without having to do the% 2 != 0calculation twice per element? :D
– user633183
46 mins ago
One thing that comes to my mind is checking whether the current element is included in theoddSortedarray, if yes then replace the current number with the first number from theoddSortedarray. This will also work as if the current element is included in theoddSortedarray 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
37 mins ago
add a comment |
3
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 ]));answered 1 hour ago
CertainPerformanceCertainPerformance
80.5k143865
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
1 hour ago
Ah, the 11 was throwing things off, since.sortsorts lexiographically - use a custom.sortfunction instead, see edit
– CertainPerformance
1 hour ago
add a comment |
0
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()
answered 1 hour ago
Tanmoy Krishna DasTanmoy Krishna Das
520212
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
1 hour ago
Okay... I'm taking a look at it
– Tanmoy Krishna Das
1 hour ago
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
1 hour ago
add a comment |
0
Fun problem, thanks for sharing! I think this is a straightforward approach -
const compare = (a, b) =>
a < b ? -1
: a > b ? 1
: 0
const main = (arr) =>
{ const odds =
const idxs =
for (const [ i, x ] of arr.entries())
if (x & 1)
(odds.push(x), idxs.push(i))
// odds = [ 5, 3, 11, 1 ]
// idxs = [ 0, 1, 4, 6 ]
odds.sort(compare)
// no need to sort idxs, always inserted in order
// odds = [ 1, 3, 5, 11 ]
// idxs = [ 0, 1, 4, 6 ]
for (const [ i, x ] of odds.entries())
arr[idxs[i]] = x
return arr
}
console.log(main([ 5, 3, 2, 8, 11, 4, 1 ]))
// [ 1, 3, 2, 8, 5, 4, 11 ]If you want main to be an immutable operation, change the end to
const result = [ ...arr ]
for (const [ i, x ] of odds.entries())
result[idxs[i]] = x
return result
answered 29 mins ago
user633183user633183
68.8k21137177
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
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 temp = [...arr];
var oddSorted = temp.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));answered 51 mins ago
Amardeep BhowmickAmardeep Bhowmick
1,7521821
easy-to-follow answer. now, can you think of a way to do this without having to do the% 2 != 0calculation twice per element? :D
– user633183
46 mins ago
One thing that comes to my mind is checking whether the current element is included in theoddSortedarray, if yes then replace the current number with the first number from theoddSortedarray. This will also work as if the current element is included in theoddSortedarray 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
37 mins ago
add a comment |
4
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 temp = [...arr];
var oddSorted = temp.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));answered 51 mins ago
Amardeep BhowmickAmardeep Bhowmick
1,7521821
easy-to-follow answer. now, can you think of a way to do this without having to do the% 2 != 0calculation twice per element? :D
– user633183
46 mins ago
One thing that comes to my mind is checking whether the current element is included in theoddSortedarray, if yes then replace the current number with the first number from theoddSortedarray. This will also work as if the current element is included in theoddSortedarray 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
37 mins ago
add a comment |
4
4
4
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 temp = [...arr];
var oddSorted = temp.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));answered 51 mins ago
Amardeep BhowmickAmardeep Bhowmick
1,7521821
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 temp = [...arr];
var oddSorted = temp.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 temp = [...arr];
var oddSorted = temp.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 temp = [...arr];
var oddSorted = temp.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));answered 51 mins ago
Amardeep BhowmickAmardeep Bhowmick
1,7521821
edited 43 mins ago
answered 51 mins ago
Amardeep BhowmickAmardeep Bhowmick
1,7521821
answered 51 mins ago
Amardeep BhowmickAmardeep Bhowmick
1,7521821
answered 51 mins ago
Amardeep BhowmickAmardeep Bhowmick
1,7521821
1,7521821
easy-to-follow answer. now, can you think of a way to do this without having to do the% 2 != 0calculation twice per element? :D
– user633183
46 mins ago
One thing that comes to my mind is checking whether the current element is included in theoddSortedarray, if yes then replace the current number with the first number from theoddSortedarray. This will also work as if the current element is included in theoddSortedarray 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
37 mins ago
add a comment |
easy-to-follow answer. now, can you think of a way to do this without having to do the% 2 != 0calculation twice per element? :D
– user633183
46 mins ago
One thing that comes to my mind is checking whether the current element is included in theoddSortedarray, if yes then replace the current number with the first number from theoddSortedarray. This will also work as if the current element is included in theoddSortedarray 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
37 mins ago
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
46 mins ago
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
46 mins ago
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
37 mins ago
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
37 mins ago
add a comment |
3
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 ]));answered 1 hour ago
CertainPerformanceCertainPerformance
80.5k143865
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
1 hour ago
Ah, the 11 was throwing things off, since.sortsorts lexiographically - use a custom.sortfunction instead, see edit
– CertainPerformance
1 hour ago
add a comment |
3
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 ]));answered 1 hour ago
CertainPerformanceCertainPerformance
80.5k143865
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
1 hour ago
Ah, the 11 was throwing things off, since.sortsorts lexiographically - use a custom.sortfunction instead, see edit
– CertainPerformance
1 hour ago
add a comment |
3
3
3
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 ]));answered 1 hour ago
CertainPerformanceCertainPerformance
80.5k143865
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 ]));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 ]));answered 1 hour ago
CertainPerformanceCertainPerformance
80.5k143865
edited 1 hour ago
answered 1 hour ago
CertainPerformanceCertainPerformance
80.5k143865
answered 1 hour ago
CertainPerformanceCertainPerformance
80.5k143865
answered 1 hour ago
CertainPerformanceCertainPerformance
80.5k143865
80.5k143865
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
1 hour ago
Ah, the 11 was throwing things off, since.sortsorts lexiographically - use a custom.sortfunction instead, see edit
– CertainPerformance
1 hour ago
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
1 hour ago
Ah, the 11 was throwing things off, since.sortsorts lexiographically - use a custom.sortfunction instead, see edit
– CertainPerformance
1 hour ago
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
1 hour ago
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
1 hour ago
Ah, the 11 was throwing things off, since
.sort sorts lexiographically - use a custom .sort function instead, see edit– CertainPerformance
1 hour ago
Ah, the 11 was throwing things off, since
.sort sorts lexiographically - use a custom .sort function instead, see edit– CertainPerformance
1 hour ago
add a comment |
0
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()
answered 1 hour ago
Tanmoy Krishna DasTanmoy Krishna Das
520212
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
1 hour ago
Okay... I'm taking a look at it
– Tanmoy Krishna Das
1 hour ago
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
1 hour ago
add a comment |
0
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()
answered 1 hour ago
Tanmoy Krishna DasTanmoy Krishna Das
520212
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
1 hour ago
Okay... I'm taking a look at it
– Tanmoy Krishna Das
1 hour ago
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
1 hour ago
add a comment |
0
0
0
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()
answered 1 hour ago
Tanmoy Krishna DasTanmoy Krishna Das
520212
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]));answered 1 hour ago
Tanmoy Krishna DasTanmoy Krishna Das
520212
edited 1 hour ago
answered 1 hour ago
Tanmoy Krishna DasTanmoy Krishna Das
520212
answered 1 hour ago
Tanmoy Krishna DasTanmoy Krishna Das
520212
answered 1 hour ago
Tanmoy Krishna DasTanmoy Krishna Das
520212
520212
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
1 hour ago
Okay... I'm taking a look at it
– Tanmoy Krishna Das
1 hour ago
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
1 hour ago
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
1 hour ago
Okay... I'm taking a look at it
– Tanmoy Krishna Das
1 hour ago
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
1 hour ago
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
1 hour ago
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
1 hour ago
Okay... I'm taking a look at it
– Tanmoy Krishna Das
1 hour ago
Okay... I'm taking a look at it
– Tanmoy Krishna Das
1 hour ago
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
1 hour ago
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
1 hour ago
add a comment |
0
Fun problem, thanks for sharing! I think this is a straightforward approach -
const compare = (a, b) =>
a < b ? -1
: a > b ? 1
: 0
const main = (arr) =>
{ const odds =
const idxs =
for (const [ i, x ] of arr.entries())
if (x & 1)
(odds.push(x), idxs.push(i))
// odds = [ 5, 3, 11, 1 ]
// idxs = [ 0, 1, 4, 6 ]
odds.sort(compare)
// no need to sort idxs, always inserted in order
// odds = [ 1, 3, 5, 11 ]
// idxs = [ 0, 1, 4, 6 ]
for (const [ i, x ] of odds.entries())
arr[idxs[i]] = x
return arr
}
console.log(main([ 5, 3, 2, 8, 11, 4, 1 ]))
// [ 1, 3, 2, 8, 5, 4, 11 ]If you want main to be an immutable operation, change the end to
const result = [ ...arr ]
for (const [ i, x ] of odds.entries())
result[idxs[i]] = x
return result
answered 29 mins ago
user633183user633183
68.8k21137177
add a comment |
0
Fun problem, thanks for sharing! I think this is a straightforward approach -
const compare = (a, b) =>
a < b ? -1
: a > b ? 1
: 0
const main = (arr) =>
{ const odds =
const idxs =
for (const [ i, x ] of arr.entries())
if (x & 1)
(odds.push(x), idxs.push(i))
// odds = [ 5, 3, 11, 1 ]
// idxs = [ 0, 1, 4, 6 ]
odds.sort(compare)
// no need to sort idxs, always inserted in order
// odds = [ 1, 3, 5, 11 ]
// idxs = [ 0, 1, 4, 6 ]
for (const [ i, x ] of odds.entries())
arr[idxs[i]] = x
return arr
}
console.log(main([ 5, 3, 2, 8, 11, 4, 1 ]))
// [ 1, 3, 2, 8, 5, 4, 11 ]If you want main to be an immutable operation, change the end to
const result = [ ...arr ]
for (const [ i, x ] of odds.entries())
result[idxs[i]] = x
return result
answered 29 mins ago
user633183user633183
68.8k21137177
add a comment |
0
0
0
Fun problem, thanks for sharing! I think this is a straightforward approach -
const compare = (a, b) =>
a < b ? -1
: a > b ? 1
: 0
const main = (arr) =>
{ const odds =
const idxs =
for (const [ i, x ] of arr.entries())
if (x & 1)
(odds.push(x), idxs.push(i))
// odds = [ 5, 3, 11, 1 ]
// idxs = [ 0, 1, 4, 6 ]
odds.sort(compare)
// no need to sort idxs, always inserted in order
// odds = [ 1, 3, 5, 11 ]
// idxs = [ 0, 1, 4, 6 ]
for (const [ i, x ] of odds.entries())
arr[idxs[i]] = x
return arr
}
console.log(main([ 5, 3, 2, 8, 11, 4, 1 ]))
// [ 1, 3, 2, 8, 5, 4, 11 ]If you want main to be an immutable operation, change the end to
const result = [ ...arr ]
for (const [ i, x ] of odds.entries())
result[idxs[i]] = x
return result
answered 29 mins ago
user633183user633183
68.8k21137177
Fun problem, thanks for sharing! I think this is a straightforward approach -
const compare = (a, b) =>
a < b ? -1
: a > b ? 1
: 0
const main = (arr) =>
{ const odds =
const idxs =
for (const [ i, x ] of arr.entries())
if (x & 1)
(odds.push(x), idxs.push(i))
// odds = [ 5, 3, 11, 1 ]
// idxs = [ 0, 1, 4, 6 ]
odds.sort(compare)
// no need to sort idxs, always inserted in order
// odds = [ 1, 3, 5, 11 ]
// idxs = [ 0, 1, 4, 6 ]
for (const [ i, x ] of odds.entries())
arr[idxs[i]] = x
return arr
}
console.log(main([ 5, 3, 2, 8, 11, 4, 1 ]))
// [ 1, 3, 2, 8, 5, 4, 11 ]If you want main to be an immutable operation, change the end to
const result = [ ...arr ]
for (const [ i, x ] of odds.entries())
result[idxs[i]] = x
return result
const compare = (a, b) =>
a < b ? -1
: a > b ? 1
: 0
const main = (arr) =>
{ const odds =
const idxs =
for (const [ i, x ] of arr.entries())
if (x & 1)
(odds.push(x), idxs.push(i))
// odds = [ 5, 3, 11, 1 ]
// idxs = [ 0, 1, 4, 6 ]
odds.sort(compare)
// no need to sort idxs, always inserted in order
// odds = [ 1, 3, 5, 11 ]
// idxs = [ 0, 1, 4, 6 ]
for (const [ i, x ] of odds.entries())
arr[idxs[i]] = x
return arr
}
console.log(main([ 5, 3, 2, 8, 11, 4, 1 ]))
// [ 1, 3, 2, 8, 5, 4, 11 ]const compare = (a, b) =>
a < b ? -1
: a > b ? 1
: 0
const main = (arr) =>
{ const odds =
const idxs =
for (const [ i, x ] of arr.entries())
if (x & 1)
(odds.push(x), idxs.push(i))
// odds = [ 5, 3, 11, 1 ]
// idxs = [ 0, 1, 4, 6 ]
odds.sort(compare)
// no need to sort idxs, always inserted in order
// odds = [ 1, 3, 5, 11 ]
// idxs = [ 0, 1, 4, 6 ]
for (const [ i, x ] of odds.entries())
arr[idxs[i]] = x
return arr
}
console.log(main([ 5, 3, 2, 8, 11, 4, 1 ]))
// [ 1, 3, 2, 8, 5, 4, 11 ]answered 29 mins ago
user633183user633183
68.8k21137177
edited 23 mins ago
answered 29 mins ago
user633183user633183
68.8k21137177
answered 29 mins ago
user633183user633183
68.8k21137177
answered 29 mins ago
user633183user633183
68.8k21137177
68.8k21137177
add a comment |
add a comment |
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