Introduction

In this blog post, we will explore how to solve the LeetCode problem "870" 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 both of the same length. The advantage of nums1 with respect to nums2 is the number of indices i for which nums1[i] > nums2[i]. Return any permutation of nums1 that maximizes its advantage with respect to nums2. Example 1: Input: nums1 = [2,7,11,15], nums2 = [1,10,4,11] Output: [2,11,7,15] Example 2: Input: nums1 = [12,24,8,32], nums2 = [13,25,32,11] Output: [24,32,8,12] Constraints: 1 <= nums1.length <= 105 nums2.length == nums1.length 0 <= nums1[i], nums2[i] <= 109

Explanation

Here's the approach, runtime analysis, and Python code to solve the problem:

• Greedy Assignment: Sort both nums1 and nums2. Iterate through nums2 (in sorted order). For each element in nums2, find the smallest element in nums1 that is greater than it. If such an element exists, assign it; otherwise, assign the smallest remaining element from nums1 (since we can't win this round, we minimize the loss).

• Index Tracking: Since we need to return a permutation of nums1 corresponding to the original order of nums2, we need to track the original indices of nums2 while sorting. We do this using an auxiliary array of indices.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(n) to store indices and result array.

Code

    def advantageCount(nums1, nums2):
    """
    Returns any permutation of nums1 that maximizes its advantage with respect to nums2.
    """
    n = len(nums1)
    nums1.sort()
    idx = sorted(range(n), key=lambda i: nums2[i])  # Sort indices based on nums2 values
    nums2_sorted = [nums2[i] for i in idx]  # Construct a sorted view of nums2 based on sorted indices
    result = [0] * n
    left = 0
    right = n - 1

    for num1 in nums1:
        if num1 > nums2_sorted[left]:
            result[idx[left]] = num1
            left += 1
        else:
            result[idx[right]] = num1
            right -= 1

    return result