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Leetcode 870. Advantage Shuffle

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.

Clarifying Questions

  1. Are the elements in nums1 and nums2 unique?
    • The problem statement does not specify whether the elements are unique, so we have to assume that there can be duplicates.
  2. Is there a certain range for the integer values within the arrays?
    • The range is not specified, so let’s consider the generic integer range in C++.

Strategy

The problem can be efficiently solved using a greedy approach with sorting:

  1. Sort nums1 in non-decreasing order.
  2. Sort nums2 maintaining the original indices.
  3. Use a two-pointer technique to determine which elements in nums1 can give an advantage over elements in nums2.
  4. Traverse through the sorted nums2, and for each element, try to find the smallest element in nums1 that is greater.
  5. Use two additional pointers or lists to keep track of the elements in nums1 that can be used and the ones that cannot provide an advantage.
  6. Construct the result array by combining elements that provide an advantage and the remaining elements.

Code

Here’s the C++ code implementing the above strategy:

#include <vector>
#include <algorithm>
#include <queue>

using namespace std;

vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
    int n = nums1.size();
    
    // Sort nums1
    sort(nums1.begin(), nums1.end());
    
    // Create a pair of (value, index) for nums2
    vector<pair<int, int>> sorted_nums2(n);
    for (int i = 0; i < n; i++) {
        sorted_nums2[i] = {nums2[i], i};
    }
    
    // Sort nums2 based on the value
    sort(sorted_nums2.begin(), sorted_nums2.end());
    
    // Result array
    vector<int> result(n);
    // Two pointers for nums1
    int left = 0, right = n - 1;
    
    // Try to maximize advantage
    for (int i = 0; i < n; i++) {
        if (nums1[left] > sorted_nums2[i].first) {
            result[sorted_nums2[i].second] = nums1[left++];
        } else {
            result[sorted_nums2[i].second] = nums1[right--];
        }
    }
    
    return result;
}

Time Complexity

  1. Sorting nums1: (O(n \log n))
  2. Creating the pair and sorting nums2: (O(n \log n))
  3. Traversing through nums2 with two pointers: (O(n))

The overall time complexity is dominated by the sorting steps: [ O(n \log n) + O(n \log n) + O(n) = O(n \log n) ]

This is efficient for the given problem constraints and provides an optimal solution.

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