LeetCode 2935: Maximum Strong Pair XOR II Solution

Master LeetCode problem 2935 (Maximum Strong Pair XOR II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2935. Maximum Strong Pair XOR II

Hard

Array Hash Table Bit Manipulation Trie Sliding Window

Problem Explanation

Explanation

To solve this problem, we can iterate through all pairs of numbers in the array and check if they form a strong pair based on the given condition. We can then calculate the XOR of each strong pair and keep track of the maximum XOR value found. The key insight is to realize that the maximum XOR value will be achieved when we XOR the two largest numbers in the array that form a strong pair.

Sort the input array nums in descending order.
Initialize a variable maxXOR to store the maximum XOR value found.
Iterate through all pairs of numbers in the sorted array.
For each pair, calculate the XOR value and update maxXOR if the XOR value is greater.
Return the maxXOR as the final result.

Time Complexity

The time complexity of this approach is O(n log n) for sorting the input array and O(n) for iterating through all pairs, resulting in a total time complexity of O(n log n).

Space Complexity

The space complexity is O(1) as we are not using any extra space other than a few variables.

Solution Code

class Solution {
    public int findMaximumXOR(int[] nums) {
        Arrays.sort(nums);
        int maxXOR = 0;
        int n = nums.length;
        
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (Math.abs(nums[i] - nums[j]) <= Math.min(nums[i], nums[j])) {
                    maxXOR = Math.max(maxXOR, nums[i] ^ nums[j]);
                }
            }
        }
        
        return maxXOR;
    }
}

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