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

Description

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.

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 106
  • 0 <= k <= 106

Solutions

Solution 1: Bit Manipulation

We can perform a bitwise XOR operation on all elements in the array $nums$. The number of bits that differ from the binary representation of $k$ in the result is the minimum number of operations.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

  • class Solution {
    public int minOperations(int[] nums, int k) {
        for (int x : nums) {
            k ^= x;
        }
        return Integer.bitCount(k);
    }
}

  • class Solution {
public:
    int minOperations(vector<int>& nums, int k) {
        for (int x : nums) {
            k ^= x;
        }
        return __builtin_popcount(k);
    }
};

  • class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        return reduce(xor, nums, k).bit_count()


  • func minOperations(nums []int, k int) (ans int) {
	for _, x := range nums {
		k ^= x
	}
	return bits.OnesCount(uint(k))
}

  • function minOperations(nums: number[], k: number): number {
    for (const x of nums) {
        k ^= x;
    }
    return bitCount(k);
}

function bitCount(i: number): number {
    i = i - ((i >>> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
    i = (i + (i >>> 4)) & 0x0f0f0f0f;
    i = i + (i >>> 8);
    i = i + (i >>> 16);
    return i & 0x3f;
}

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