88. Merge Sorted Array

Problem Statement

You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.

Merge nums1 and nums2 into a single array sorted in non-decreasing order.

The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

Example 1:

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]
Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.

Example 2:

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].

Example 3:

Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.

Constraints:

nums1.length == m + n
nums2.length == n
0 <= m, n <= 200
1 <= m + n <= 200
-109 <= nums1[i], nums2[j] <= 109

Intuition

Approach:

One can do in extra space, sorting the array

But we want no extra space and O(m+n)

So we start the pointers from behind i,j
We want to exhaust j pointer of second array, J exhausted means, All went to correct 
Position

Links

https://leetcode.com/problems/merge-sorted-array/description/

Video Links

Approach 1:

Pointers

C++

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int i=m-1, j=n-1, k=m+n-1;

        while(i>=0 and j>=0){
            if(nums1[i] > nums2[j])
                nums1[k--] = nums1[i--];
            else
                nums1[k--] = nums2[j--];
        }

        while(j>=0)
            nums1[k--] = nums2[j--];
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++