MergeSortedArray

88. Merge Sorted Array
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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




Solution (Python):

#the reason we don't use len(nums2) - 1 as p2, for example, and vice versa for p1, is because we don't merge 0s we see in both arrays, some of which may be in nums2 input array, and n (given input) dosen't include the 0s!

class Solution:
    import math
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        """
        Do not return anything, modify nums1 in-place instead.
        """
        p1, p2 = m - 1, n - 1 #p1 and p2 represent the final indexes of each array 
        
        while p1 >= 0 and p2 >= 0: #a boundary check since we know that p1 and p2 are not too far to the right outside the array due to the previous lines
            if nums1[p1] > nums2[p2]: #nums1 is an array while p1 is an integer index
                nums1[p1 + p2 + 1] = nums1[p1]
                p1 -= 1 #moving 1 pointer backwards because we already merged that element into the final list
            else:
                nums1[p1 + p2 + 1] = nums2[p2]
                p2 -= 1 #moving 2 pointer backwards because we already merged that element into the final list (we merge either 1 or 2, not both. we are starting from the end of each list and doing this comparison)
        
        while p2 >= 0: #boundary check
            nums1[p2] = nums2[p2]
            p2 -= 1