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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 positions i for which nums1[i] > nums2[i].

Return any permutation of nums1 that maximizes its advantage with respect to nums2.

Clarifying Questions

  1. What is the length range of nums1 and nums2?
    • The arrays can have length from 1 to 10^5.
  2. What is the range of values for elements in nums1 and nums2?
    • The elements range from 0 to 10^9.
  3. Do the arrays contain duplicate elements?
    • Yes, the arrays can contain duplicate elements.
  4. What if multiple permutations give the same advantage?
    • Any permutation that maximizes the advantage is acceptable.

Strategy

To maximize the advantage of nums1 over nums2:

  1. Sort both arrays: Given the sorted version, you can use a two-pointer technique to maximize the counts of nums1[i] > nums2[i].
  2. Greedy Allocation: Use a two-pointer approach to assign the smallest possible nums1 that can beat nums2’s current largest remaining element. If no such nums1 is found, assign it to the smallest remaining nums2 (these can be paired with any nums1).

Approach

  1. Sort nums2 and keep track of their original indices.
  2. Sort nums1.
  3. Use a deque (double-ended queue) to allocate elements from nums1:
    • Match the smallest element in nums1 that is still greater than the current element in nums2.
    • If no such element exists, use the smallest element in nums1 with the smallest element in nums2 (i.e., nums1 cannot win here).

Code

Here’s the Java code implementing the strategy:

import java.util.Arrays;
import java.util.PriorityQueue;

public class AdvantageShuffle {
    public int[] advantageCount(int[] nums1, int[] nums2) {
        // Length of the arrays
        int n = nums1.length;

        // Sort nums1 to facilitate sorting logic
        Arrays.sort(nums1);
        
        // Generate pairs of (value, index) for nums2, then sort by value
        int[][] nums2WithIndex = new int[n][2];
        for (int i = 0; i < n; i++) {
            nums2WithIndex[i][0] = nums2[i];
            nums2WithIndex[i][1] = i;
        }
        Arrays.sort(nums2WithIndex, (a, b) -> Integer.compare(a[0], b[0]));
        
        // Result array to hold the permutation of nums1
        int[] result = new int[n];
        
        // Two pointers to allocate numbers
        int left = 0, right = n - 1;

        // Use a for loop to process sorted nums1
        for (int num : nums1) {
            if (num > nums2WithIndex[left][0]) {
                // If the current number can take advantage, place it in the result
                result[nums2WithIndex[left][1]] = num;
                left++;
            } else {
                // Otherwise, place it against the largest remaining number in nums2
                result[nums2WithIndex[right][1]] = num;
                right--;
            }
        }
        
        return result;
    }

    public static void main(String[] args) {
        AdvantageShuffle solution = new AdvantageShuffle();
        int[] nums1 = {2,7,11,15};
        int[] nums2 = {1,10,4,11};
        System.out.println(Arrays.toString(solution.advantageCount(nums1, nums2))); // Expected output [2,11,7,15] or any permutation with maximum advantage.
    }
}

Time Complexity

This approach ensures that we use a greedy method to maximize the advantage by making optimal pairing strategies.

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