Merge Two 2D Arrays by Summing Values

I start off with 2 pointers, i1 = 0 and i2 = 0, which are used to iterate through nums1 and nums2 respectively.

Creating ans as our resultant array, since both nums1 and nums2 are in ascending order, if:

• nums1[i1][0] == nums2[i2][0], then:
  • we know both nums1[i1][0] and nums2[i1][0] are the elements with the smallest id out of all remaining elements
  • we can add the id nums1[i1][0] as well as the sum of both values, nums1[i1][1] + nums2[i2][1] to the ans array
  • finally, increment both i1 and i2
• nums1[i1][0] < nums2[i2][0], then:
  • we know nums1[i1][0] has the smallest id out of all remaining elements
  • therefore, we can add the id nums1[i1][0] and its corresponding value, nums1[i1][1] to the ans array
  • finally, increment i1
• nums1[i1][0] > nums2[i2][0], then:
  • we know nums2[i2][0] has the smallest id out of all remaining elements
  • therefore, we can add the id nums2[i2][0] and its corresponding value, nums2[i2][1] to the ans array
  • finally, increment i2

We can repeat the conditional until we reach the end of nums1 or nums2 (or both).

Finally, to complete ans, we can just append all the remaining elements in either nums1 or nums2 before simply returning ans.

code

class Solution {
public:
    vector<vector<int>> mergeArrays(vector<vector<int>>& nums1, vector<vector<int>>& nums2) {
        int i1 = 0;
        int i2 = 0;

        vector<vector<int>> ans;
        while (i1 < nums1.size() && i2 < nums2.size()) {
            if (nums1[i1][0] == nums2[i2][0]) {
                ans.push_back({nums1[i1][0], nums1[i1][1] + nums2[i2][1]});
                ++i1;
                ++i2;
            } else if (nums1[i1][0] <= nums2[i2][0])  {
                ans.push_back({nums1[i1][0], nums1[i1][1]});
                ++i1;
            } else {
                ans.push_back({nums2[i2][0], nums2[i2][1]});
                ++i2;
            }
        }

        while (i1 < nums1.size()) {
            ans.push_back({nums1[i1][0], nums1[i1][1]});
            ++i1;
        }

        while (i2 < nums2.size()) {
            ans.push_back({nums2[i2][0], nums2[i2][1]});
            ++i2;
        }

        return ans;
    }
};

complexity

• Time: O(n)
• Space: O(n)