Merge Sorted Arrays

88. Merge Sorted Array

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:

1. The number of elements initialized in nums1 and nums2 are m and n respectively.
2. You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2.

   Example:

   Input: 
   nums1 = [1,2,3,0,0,0], m = 3
   nums2 = [2,5,6],       n = 3

   Output: [1,2,2,3,5,6]

1. First Solution

• Starting at the end of the first array -> nums1
  • compare the last initialized elements in each of the two arrays
  • which one is bigger we will put in the last element of the first array -> nums1[nums1.length-1]
  1. First we start by comparing 6 and 3 -> which one is bigger? 6 -> nums1[nums1.length-1] => [1, 2, 3, 0, 0, 6] 1.1 We continue to compare 5 and 3 -> which one is bigger? 5 -> nums1[nums1.length-2] => [1, 2, 3, 0, 5, 6] 1.2 We continue to compare 2 and 3 -> which one is bigger? 3 -> nums1[nums1.length-3] => [1, 2, 3, 3, 5, 6] 1.3 We continue to compare 2 and 2 -> which one is bigger? no one -> nums1[nums1.length-4] => [1, 2, 2, 3, 5, 6]
  • m is the number of the initialized variables in nums1
  • n is the number of the initialized variables in nums2

    const merge = function (nums1, m, nums2, n) {
        let first = m - 1;
        let second = n - 1;

        for (let i = m + n - 1; i >= 0; i--) {
            if (second < 0) {
                break;
            }

            if (nums1[first] > nums2[second]) {
                nums1[i] = nums1[first];
                first--;
            } else { 
                nums1[i] = nums2[second];
                second--;
            }
        }
    };

2. Second Solution

Approach:

• Start from back of nums1 and iterate through both lists backwards, putting the larger of nums1[m] and nums2[n] into the last element of nums1 so far. Starting from the back allows us to do this in one loop.

• Once an element is copied over, it is in it’s “final spot” in the sorted array, and will not move again.

• The algorithm has a runtime of O(m+n) and space of O(1).

• The latter is true because even though a variable is initialized, it is not a function of the inputs m or n.

• Note: This could have been done without initializing ‘final’ at all, but it reads better with it.

    const merge = function (nums1, m, nums2, n) {
        let final = m + n - 1;
        m--;
        n--;

        while (m >= 0 || n >= 0) {
            // if no more nums2 then just copy remaining m elements
            if (n < 0 || nums1[m] > nums2[n]) {
                nums1[final] = nums1[m];
                m--;
            } else { // m < 0 || nums1[m] < nums2[n]
                nums1[final] = nums2[n];
                n--;
            }
            final--;
        }

        return nums1;
    };

Merge Sorted Arrays

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