88.Merge_Sorted_Array_(Easy).md

88. Merge Sorted Array (Easy)

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

Question:

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.

Walk-through:

Use three ptrs i, j for nums1, nums2, k ptr for nums1's end. We accessing the two nums backward, if nums1[i] <= nums2[j], we set nums1[k] = nums1[j], then decrement j, k ptrs.

By the end, we check if two ptrs i, j and at index 0 or not. If not, we modify the rest with nums1[k] with either nums1 or nums2.

My Solution:

class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        """
        Do not return anything, modify nums1 in-place instead.
        """
        # Modify nums1 to be merged with nums1 and nums2
        # Use three pointers, i,j for nums1, nums2, k for the end of nums1
        # Comparing nums1, nums2 backward, copy larger pointer to k, decrement that pointer and k pointer

        # TC: O(m+n), SC: O(1)
        i, j = m-1, n-1     # pointers for nums1, nums2
        k = m + n - 1   # pointers to the end of nums1
        while i >= 0 and j >= 0:
            # Left >= right, decrement i, k
            if nums1[i] >= nums2[j]:
                nums1[k] = nums1[i]
                i -= 1
            else:
                nums1[k] = nums2[j]
                j -= 1
            k -= 1
        # Add the rest from nums1 or nums2 into nums1
        while i >= 0:
            nums1[k] = nums1[i]
            i -= 1
            k -= 1
        while j >= 0:
            nums1[k] = nums2[j]
            j -= 1
            k -= 1

Time Complexity: $O(m + n)$, m is the length of nums1, n is the length of nums2.
Space Complexity: $O(1)$