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, 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

    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