# Solving Leetcode Interviews in Seconds with AI: Merge Sorted Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "88" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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    Follow up: Can you come up with an algorithm that runs in O(m + n) time? 

	# Explanation
	Here's the optimal approach to merge the sorted arrays `nums1` and `nums2` in-place within `nums1`:

*   **Two Pointers (from the end):** Use two pointers, one for `nums1` (starting at index `m-1`) and one for `nums2` (starting at index `n-1`).
*   **Compare and Place:** Compare the elements pointed to by the two pointers. Place the larger element at the end of `nums1` (index `m+n-1`). Decrement the appropriate pointer and the index in `nums1`.
*   **Handle Remaining Elements:** If there are any remaining elements in `nums2` after `nums1` is exhausted, copy them into the beginning of `nums1`.

*   **Runtime Complexity:** O(m + n), **Storage Complexity:** O(1)

	
	# Code
	```python
	def merge(nums1, m, nums2, n):
    """
    Merges two sorted arrays nums1 and nums2 into nums1 in-place.

    Args:
        nums1: The first array (modified in-place).
        m: The number of elements in nums1 that are valid.
        nums2: The second array.
        n: The number of elements in nums2.
    """
    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]
            i -= 1
        else:
            nums1[k] = nums2[j]
            j -= 1
        k -= 1

    # Copy remaining elements from nums2, if any
    while j >= 0:
        nums1[k] = nums2[j]
        j -= 1
        k -= 1
	```