Maximum Strong Pair XOR

Problem Statement

Given an array of integers, find the maximum XOR value of any two elements in the array.

The condition |x - y| <= min(x, y) means that for a pair of integers (x, y) to be considered a "strong pair", the absolute difference between x and y (|x - y|) must be less than or equal to the minimum of x and y (min(x, y)).

In simpler terms, it means that the absolute difference between x and y should not exceed the smaller of the two numbers.

For example:

• For the pair (3, 5), the absolute difference is |3 - 5| = 2, and the minimum of 3 and 5 is 3. Since 2 is less than or equal to 3, the pair (3, 5) is a strong pair.
• For the pair (8, 2), the absolute difference is |8 - 2| = 6, and the minimum of 8 and 2 is 2. Since 6 is greater than 2, the pair (8, 2) is not a strong pair.

Examples

Example 1:

Input: arr = [3, 10, 5, 25, 2, 8]

Output: Maximum XOR value: 28

Explanation: In the given array, the maximum XOR value of any two elements is achieved when 5 and 25 are selected.

5^25 = 28

Hence, the maximum XOR value is 28.

Example 2:

Input: arr[] = {0, 2, 5, 7}

Output: 7

Explanation: The maximum XOR value is achieved by 0 XOR 7 = 7

Theory

XOR, also known as exclusive OR, is a bitwise operation that outputs true only when the input differs (one is true, other is false) (both are different). The XOR operation is denote dby the symbol ^. For example:

0 XOR 0 = 0
0 XOR 1 = 1
1 XOR 0 = 1
1 XOR 1 = 0

Different Approaches

1 Brute Force Approach

Algorithm:

1. Initialize a variable maxXOR to store the maximum XOR value.
2. Iterate through all pairs of elements in the array.
3. For each pair of elements (arr[i], arr[j]), calculate their XOR value and update maxXOR if the XOR value is greater than the current maxXOR.
4. Return maxXOR.

Code Implementation:

#include <iostream>
#include <vector>
using namespace std;

int findMaximumXORPair(vector<int>& arr) {
    int n = arr.size();
    int maxXOR = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            maxXOR = max(maxXOR, arr[i] ^ arr[j]);
        }
    }
    return maxXOR;
}

int main() {
    vector<int> arr = {3, 10, 5, 25, 2, 8};
    cout << "Maximum XOR Pair: " << findMaximumXORPair(arr) << endl;
    return 0;
}

Complexity Analysis:

Time Complexity:

• The brute force approach involves checking all possible pairs of elements in the array, resulting in a time complexity of O(N^2), where N is the size of the array.

Space Complexity:

• The space complexity is O(1) as we are not using any extra space apart from the input array.