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

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# 題目敘述

You are given a 0-indexed integer array nums and a positive integer k .

You can apply the following operation on the array any number of times:

• Choose any element of the array and flip a bit in its binary representation. Flipping a bit means changing a 0 to 1 or vice versa.

Return the minimum number of operations required to make the bitwise XOR of all elements of the final array equal to k .

Note that you can flip leading zero bits in the binary representation of elements. For example, for the number (101)2 you can flip the fourth bit and obtain (1101)2 .

# Example 1

Input: nums = [2,1,3,4], k = 1
Output: 2
Explanation: We can do the following operations:

• Choose element 2 which is 3 == (011)2, we flip the first bit and we obtain (010)2 == 2. nums becomes [2,1,2,4].
• Choose element 0 which is 2 == (010)2, we flip the third bit and we obtain (110)2 = 6. nums becomes [6,1,2,4].
  The XOR of elements of the final array is (6 XOR 1 XOR 2 XOR 4) == 1 == k.
  It can be shown that we cannot make the XOR equal to k in less than 2 operations.

# Example 2

Input: nums = [2,0,2,0], k = 0
Output: 0
Explanation: The XOR of elements of the array is (2 XOR 0 XOR 2 XOR 0) == 0 == k. So no operation is needed.

# 解題思路

# Solution

class Solution {

    public int minOperations(int[] nums, int k) {

        int bitwise = 0;

        for(int num : nums){

            bitwise ^= num;

        }

        bitwise ^= k;

        String binary = Integer.toBinaryString(bitwise);

        int ans = 0;

        for(int i = 0; i < binary.length(); i++){

            if(binary.charAt(i) == '1'){

                ans += 1;

            }

        }

        return ans;

    }

}