LeetCode 2997: Minimum Number of Operations to Make Array XOR Equal to K Solution

Master LeetCode problem 2997 (Minimum Number of Operations to Make Array XOR Equal to K), a medium 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.

2997. Minimum Number of Operations to Make Array XOR Equal to K

Medium

Array Bit Manipulation

Problem Explanation

Explanation:

To solve this problem, we need to understand that XOR operation is commutative and associative. We can iterate over all elements of the array and XOR them to get the initial XOR value. Then, we can iterate over each bit of the XOR value from the most significant bit (MSB) to the least significant bit (LSB). For each bit, we check if it is set in the desired XOR value k. If it is not set, we need to set it in the array elements to make the final XOR equal to k. To set a bit, we need to flip it in at least one element. Therefore, for each bit position that needs to be set, we count the number of elements that have that bit set and the number of elements that have that bit unset. The minimum number of operations required to set that bit is the minimum of these two counts. We repeat this process for all bit positions from MSB to LSB.

Time Complexity: O(n), where n is the number of elements in the array nums. Space Complexity: O(1)

Solution Code

class Solution {
    public int minOperations(int[] nums, int k) {
        int xor = 0;
        for (int num : nums) {
            xor ^= num;
        }
        
        int operations = 0;
        for (int i = 20; i >= 0; i--) {
            int bit = (k >> i) & 1;
            if (((xor >> i) & 1) != bit) {
                operations++;
                int cnt1 = 0, cnt0 = 0;
                for (int num : nums) {
                    if (((num >> i) & 1) == 1) {
                        cnt1++;
                    } else {
                        cnt0++;
                    }
                }
                if (cnt1 < cnt0) {
                    cnt0 = cnt1;
                }
                xor ^= (cnt1 > 0 ? (1 << i) : 0);
            }
        }
        
        return operations;
    }
}

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