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Leetcode 2932. Maximum Strong Pair XOR I

Problem Statement

Given an array of positive integers nums, find the maximum value of the expression nums[i] XOR nums[j] for all pairs (i, j) such that 0 <= i < j < nums.length.

Clarifying Questions

  1. Q: Will the input array nums contain only positive integers?
    • A: Yes, the array will only contain positive integers.
  2. Q: What is the range of the length of the input array nums?
    • A: The length of the input array will be between 2 and (10^5).
  3. Q: What are the constraints on the values within the array?
    • A: The values within the array will be positive integers up to (10^7).

Strategy

  1. Understanding XOR Operation: XOR (exclusive OR) operation between bits compares each bit of the numbers being XORed and results in 1 if the bits are different and 0 if they are the same.

  2. Objective: To maximize nums[i] XOR nums[j].

  3. Optimize Pair Comparison: A brute-force approach that checks every pair would be inefficient (O(n^2)), where n is the length of the array. Instead, sorting the array and taking advantage of bit-wise properties can significantly reduce the complexity.

  4. Efficient Approach:

    • Sort the array.
    • The largest XOR value between pairs often occurs between adjacent elements in the sorted array. This is because for larger numbers, adjacent values in sorted order tend to contain bits that differ significantly.

Code

import java.util.Arrays;

public class MaximumStrongPairXOR {
    public static int findMaximumXOR(int[] nums) {
        // Sort the array
        Arrays.sort(nums);
        
        int maxXOR = 0;
        
        // Compare adjacent elements in the sorted array
        for (int i = 1; i < nums.length; i++) {
            int xorValue = nums[i] ^ nums[i - 1];
            maxXOR = Math.max(maxXOR, xorValue);
        }
        
        return maxXOR;
    }
    
    public static void main(String[] args) {
        int[] nums = {3, 10, 5, 25, 2, 8};
        System.out.println(findMaximumXOR(nums));  // Output: Expected maximum XOR value
    }
}

Time Complexity

Thus, the overall time complexity is (O(n \log n)), making this approach efficient for the given constraints.

Explanation

  1. Sorting: The array is first sorted to bring similarly sized numbers close together.
  2. XOR Calculation: By iterating over the sorted array and applying XOR between adjacent elements, we can more efficiently find a near-optimal XOR value without having to compare every possible pair.
    • The logic is that adjacent numbers in a sorted array will have maximally differing bits at higher bit positions, which contribute significantly to the XOR value.

Given these points, the provided solution is robust, efficient, and scales well with input size.

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