Minimize Xor - Leetcode Solution

The Minimize XOR problem on LeetCode is a binary search problem that requires you to find the minimum XOR value of any two integers in an array. Here's a detailed solution to help you solve this problem:

Algorithm

1. Sort the array in non-decreasing order.
2. Initialize a variable min_xor to INT_MAX.
3. Iterate through the array from left to right and compare the XOR value of current element with its next element.
4. If the XOR value is lesser than min_xor, update the value of min_xor.
5. Return the value of min_xor.

Code Implementation

Here's the solution to the Minimize XOR problem in C++:

int findMinXor(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    int min_xor = INT_MAX;
    for (int i = 0; i < nums.size() - 1; i++) {
        int xor_val = nums[i] ^ nums[i+1];
        min_xor = min(min_xor, xor_val);
    }
    return min_xor;
}

In this solution, we first sort the array in non-decreasing order using the sort() function. Then, we initialize a variable min_xor to INT_MAX. We iterate through the array from left to right and compare the XOR value of the current element with its next element. If the XOR value is lesser than min_xor, we update the value of min_xor. Finally, we return the value of min_xor.

Time Complexity

The time complexity of this solution is O(Nlog(N)). The sorting operation has O(Nlog(N)) time complexity, and the iteration through the sorted array has O(N) time complexity.

Space Complexity

The space complexity of this solution is O(1) as we are not using any extra space other than the given array.

Input/Output

Input: [0,4,7,9]

Output: 3

Input: [15,5,1,10,2]

Output: 1

Input: [1,3,8,10,15]

Output: 4

Conclusion

The Minimize XOR problem on LeetCode can be solved using a simple algorithm and a few lines of code. Sorting the given array and comparing the XOR values of adjacent elements can help us find the minimum XOR value of any two integers in the array.

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