LeetCode 2932: Maximum Strong Pair XOR I Solution

Master LeetCode problem 2932 (Maximum Strong Pair XOR I), a easy 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.

2932. Maximum Strong Pair XOR I

Easy

Array Hash Table Bit Manipulation Trie Sliding Window

Problem Explanation

Explanation:

To solve this problem, we need to find the two integers from the array that form a strong pair with the maximum XOR value. We can achieve this by iterating through all pairs of numbers and calculating their XOR value, while considering the condition for a strong pair.

Initialize a variable maxXor to store the maximum XOR value found.
Iterate through all pairs of numbers in the array.
For each pair (num1, num2), calculate their XOR value and check if it forms a strong pair.
Update maxXor if the current XOR value is greater than the previous maximum.
Return the final maxXor value. :

Solution Code

class Solution {
    public int findMaximumXOR(int[] nums) {
        int maxXor = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i; j < nums.length; j++) {
                int xor = nums[i] ^ nums[j];
                if (xor > maxXor && Math.abs(nums[i] - nums[j]) <= Math.min(nums[i], nums[j])) {
                    maxXor = xor;
                }
            }
        }
        return maxXor;
    }
}

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