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var merge = function(nums1, m, nums2, n) {

    //contcating two array
     let array = nums2.concat(nums1)
    // sort the array
     array.sort((a,b) => a-b)
     // remove element > m+n length
 return array.slice(m+n-n)
};

This ^ function is returning -> [1,2,3,0,0,0] if i'm applying console then answer is same as expected -> [1,2,2,3,5,6]

Why is this happening ?

var merge = function(nums1, m, nums2, n) {

    //contcating two array
     let array = nums2.concat(nums1)
    // sort the array
     array.sort((a,b) => a-b)
     // remove element > m+n length
 return array.slice(m+n-n)
};

This ^ function is returning -> [1,2,3,0,0,0] if i'm applying console then answer is same as expected -> [1,2,2,3,5,6]

Why is this happening ?

• javascript
• merge

Share Improve this question Follow edited Jun 7, 2022 at 6:11 Hritik Sharma 2,01611 gold badge1111 silver badges2727 bronze badges asked Jun 7, 2022 at 5:46 ma_12ma_12 7511 silver badge66 bronze badges 5

• What's the input? num1, m, num2 and n? – zer00ne Commented Jun 7, 2022 at 5:50
• nums1 ->[1,2,3,0,0,0] nums2 ->[2,5,6] n , m -> 3 – ma_12 Commented Jun 7, 2022 at 5:56
• So m and n are both 3? The last line you are removing index 3+3-3 ...so slice(3) will return an array without the first 3 numbers...why? and why not just m if n-n? – zer00ne Commented Jun 7, 2022 at 6:11
• m + n - n = m !? do you mean array.slice(m+n,n) – Xupitan Commented Jun 7, 2022 at 6:16
• with 2 arrays : [1,2,3.0,0,0] and [2,5,6] . result is [1,2,2,3,5,6], must be array.slice(3) or array.slice(3, array.length) – Xupitan Commented Jun 7, 2022 at 6:51

Add a ment  | 

5 Answers 5

Sorted by: Reset to default 4

Remove slice function from the end of the function. slice(m+n-n) slices your sorted array and returns array from index m+1 to the last index.

var merge = function(nums1, m, nums2, n) {

    //contcating two array
     let array = nums2.concat(nums1)
    // sort the array
     array.sort((a,b) => a-b)
     // remove element > m+n length
 return array.slice(m+n-n);
};

console.log(merge([2,4,8,9],4,[0,4,6,9],4));

You can use the following function to merge and then sort the two arrays. Time plexity of this approach is O(nlogn)

function merge(arr1,arr2){
  return [...arr1,...arr2].sort();
}

console.log(merge([4,8,6],[1,3,9,10]));

The second approach runs in O(n) time.

 function merge(arr1,m,arr2,n){
          let result = [];
          let i=0 , j = 0 ;
          while(i<m && j<n){
            if(arr1[i]<arr2[j]){
              result.push(arr1[i]);
              i++;
            }else{
              result.push(arr2[j]);
              j++;
            }
          }
          while(i<m){
            result.push(arr1[i]);
            i++;
          }
          while(j<n){
            result.push(arr2[j]);
            j++;
          }
          return result;
     }

        console.log(merge([4,5,6],3,[1,3,8,10],4));

It is mentioned that Do not return anything, modify nums1 in-place instead., So the output is your nums1 array and not array which you have created.

Note: To resolve this issue you can modify the nums1 array, below is the sample code:

var merge = function(nums1, m, nums2, n) {
   const index = nums1.length - n;
    for(let i=index;i < nums1.length;i++) {
        nums1[i] = nums2[i-index];
    }
    nums1.sort((a,b)=> a-b);
    
};

I did not see any problem with your code and it works, I just added another code. I think the problem with your console.log.

var merge = function(nums1, m, nums2, n) {

    //contcating two array
     let array = nums2.concat(nums1)
    // sort the array
     array.sort((a,b) => a-b)
     // remove element > m+n length
 return array.slice(m+n-n)
};

console.log(merge([1,2,3,0,0,0],3,[2,5,6],3));

var merge = function(nums1, m, nums2, n) {
    return    nums1.slice(0, m)
    .concat(nums2.slice(0, n))
    .sort((i, j) => i - j);
       
       
    };

  console.log(  merge([1,2,3,0,0,0],3,[2,5,6],3))

By using simple for loop

let arr1 = [10, 20, 30, 45, 50]

let arr2 = [15, 25, 26, 35]

const mergeArray = (arr1, arr2) => {

  let arr3 = []
  let i = 0;
  let j = 0;

  for (let n = 0; n < arr1.length + arr2.length; n++) {
    if (arr1[i] <= arr2[j] || arr2.length < j + 1) {
      arr3.push(arr1[i])
      i++
    } else {
      arr3.push(arr2[j])
      j++
    }
  }
  return arr3
}
console.log(JSON.stringify(mergeArray(arr1, arr2)))

I tried to use a two-pointer algorithm but realized that I need 3 pointers in this case.

var merge = function(nums1, m, nums2, n) {
    let i = m - 1;  // Pointer for the last initialized element in nums1
    let j = n - 1;  // Pointer for the last element in nums2
    let k = m + n - 1;  // Pointer for the last position in nums1

    // Iterate backwards and fill nums1 from the end
    while (j >= 0) {
        if (i >= 0 && nums1[i] > nums2[j]) {
        nums1[k] = nums1[i];
        i--;
    } else {
        nums1[k] = nums2[j];
        j--;
    }
    k--;
}

return nums1;

};

var merge = function(nums1, m, nums2, n) {

    //contcating two array
     let array = nums2.concat(nums1)
    // sort the array
     array.sort((a,b) => a-b)
     // remove element > m+n length
 return array.slice(m+n-n)
};

This ^ function is returning -> [1,2,3,0,0,0] if i'm applying console then answer is same as expected -> [1,2,2,3,5,6]

Why is this happening ?

var merge = function(nums1, m, nums2, n) {

    //contcating two array
     let array = nums2.concat(nums1)
    // sort the array
     array.sort((a,b) => a-b)
     // remove element > m+n length
 return array.slice(m+n-n)
};

This ^ function is returning -> [1,2,3,0,0,0] if i'm applying console then answer is same as expected -> [1,2,2,3,5,6]

Why is this happening ?

• javascript
• merge

Share Improve this question Follow edited Jun 7, 2022 at 6:11 Hritik Sharma 2,01611 gold badge1111 silver badges2727 bronze badges asked Jun 7, 2022 at 5:46 ma_12ma_12 7511 silver badge66 bronze badges 5

• What's the input? num1, m, num2 and n? – zer00ne Commented Jun 7, 2022 at 5:50
• nums1 ->[1,2,3,0,0,0] nums2 ->[2,5,6] n , m -> 3 – ma_12 Commented Jun 7, 2022 at 5:56
• So m and n are both 3? The last line you are removing index 3+3-3 ...so slice(3) will return an array without the first 3 numbers...why? and why not just m if n-n? – zer00ne Commented Jun 7, 2022 at 6:11
• m + n - n = m !? do you mean array.slice(m+n,n) – Xupitan Commented Jun 7, 2022 at 6:16
• with 2 arrays : [1,2,3.0,0,0] and [2,5,6] . result is [1,2,2,3,5,6], must be array.slice(3) or array.slice(3, array.length) – Xupitan Commented Jun 7, 2022 at 6:51

Add a ment  | 

5 Answers 5

Sorted by: Reset to default 4

Remove slice function from the end of the function. slice(m+n-n) slices your sorted array and returns array from index m+1 to the last index.

var merge = function(nums1, m, nums2, n) {

    //contcating two array
     let array = nums2.concat(nums1)
    // sort the array
     array.sort((a,b) => a-b)
     // remove element > m+n length
 return array.slice(m+n-n);
};

console.log(merge([2,4,8,9],4,[0,4,6,9],4));

You can use the following function to merge and then sort the two arrays. Time plexity of this approach is O(nlogn)

function merge(arr1,arr2){
  return [...arr1,...arr2].sort();
}

console.log(merge([4,8,6],[1,3,9,10]));

The second approach runs in O(n) time.

 function merge(arr1,m,arr2,n){
          let result = [];
          let i=0 , j = 0 ;
          while(i<m && j<n){
            if(arr1[i]<arr2[j]){
              result.push(arr1[i]);
              i++;
            }else{
              result.push(arr2[j]);
              j++;
            }
          }
          while(i<m){
            result.push(arr1[i]);
            i++;
          }
          while(j<n){
            result.push(arr2[j]);
            j++;
          }
          return result;
     }

        console.log(merge([4,5,6],3,[1,3,8,10],4));

It is mentioned that Do not return anything, modify nums1 in-place instead., So the output is your nums1 array and not array which you have created.

Note: To resolve this issue you can modify the nums1 array, below is the sample code:

var merge = function(nums1, m, nums2, n) {
   const index = nums1.length - n;
    for(let i=index;i < nums1.length;i++) {
        nums1[i] = nums2[i-index];
    }
    nums1.sort((a,b)=> a-b);
    
};

I did not see any problem with your code and it works, I just added another code. I think the problem with your console.log.

var merge = function(nums1, m, nums2, n) {

    //contcating two array
     let array = nums2.concat(nums1)
    // sort the array
     array.sort((a,b) => a-b)
     // remove element > m+n length
 return array.slice(m+n-n)
};

console.log(merge([1,2,3,0,0,0],3,[2,5,6],3));

var merge = function(nums1, m, nums2, n) {
    return    nums1.slice(0, m)
    .concat(nums2.slice(0, n))
    .sort((i, j) => i - j);
       
       
    };

  console.log(  merge([1,2,3,0,0,0],3,[2,5,6],3))

By using simple for loop

let arr1 = [10, 20, 30, 45, 50]

let arr2 = [15, 25, 26, 35]

const mergeArray = (arr1, arr2) => {

  let arr3 = []
  let i = 0;
  let j = 0;

  for (let n = 0; n < arr1.length + arr2.length; n++) {
    if (arr1[i] <= arr2[j] || arr2.length < j + 1) {
      arr3.push(arr1[i])
      i++
    } else {
      arr3.push(arr2[j])
      j++
    }
  }
  return arr3
}
console.log(JSON.stringify(mergeArray(arr1, arr2)))

I tried to use a two-pointer algorithm but realized that I need 3 pointers in this case.

var merge = function(nums1, m, nums2, n) {
    let i = m - 1;  // Pointer for the last initialized element in nums1
    let j = n - 1;  // Pointer for the last element in nums2
    let k = m + n - 1;  // Pointer for the last position in nums1

    // Iterate backwards and fill nums1 from the end
    while (j >= 0) {
        if (i >= 0 && nums1[i] > nums2[j]) {
        nums1[k] = nums1[i];
        i--;
    } else {
        nums1[k] = nums2[j];
        j--;
    }
    k--;
}

return nums1;

};

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