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

[mah-shamim]

Topics: Array, Bit Manipulation, Biweekly Contest 121

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)(1101)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

Hint:

1. Calculate the bitwise XOR of all elements of the original array and compare it to k in their binary representation.
2. For each different bit between the bitwise XOR of elements of the original array and k we have to flip exactly one bit of an element in nums to make that bit equal.

[mah-shamim]

We are given an array nums and an integer k. We can flip any bit in any element any number of times. Each flip counts as one operation. The goal is to make the XOR of all elements in the array equal to k with the minimum number of operations.

Approach:

The problem requires transforming the array such that the XOR of all elements equals k by flipping bits in the elements' binary representations. Each flip operation changes one bit in one element and counts as one operation. The solution involves determining the minimum number of such operations needed.

The key insight is that the XOR of all elements in the array initially computes to a value, say current_xor. The goal is to change current_xor to k. Since flipping a bit in any element flips the corresponding bit in the overall XOR, the number of operations required is exactly the number of bits where current_xor and k differ.

Let's implement this solution in PHP: 2997. Minimum Number of Operations to Make Array XOR Equal to K

<?php
/**
 * @param Integer[] $nums
 * @param Integer $k
 * @return Integer
 */
function minOperations(array $nums, int $k): int
{
    $xors = array_reduce($nums, function($carry, $item) {
        return $carry ^ $item;
    }, 0);
    return substr_count(decbin($k ^ $xors), '1');
}

// Test cases
echo minOperations([2, 1, 3, 4], 1);  // Output: 2
echo "\n";
echo minOperations([2, 0, 2, 0], 0);  // Output: 0
?>

Explanation:

1. Compute Initial XOR: Calculate the XOR of all elements in the array. This gives the current XOR value, current_xor.
2. Find Differing Bits: Compute the XOR between current_xor and k. The result, diff, will have bits set to 1 exactly where current_xor and k differ.
3. Count Set Bits: The number of set bits (1s) in diff represents the number of bits that need to be flipped in the array elements to make the overall XOR equal to k. Each bit flip requires one operation, so the count of set bits is the answer.

Time Complexity

• Time Complexity: O(n), where n is the number of elements in nums. This is because we traverse the array once to compute the XOR.
• Space Complexity: O(1), as we use a constant amount of extra space regardless of the input size. The binary string representation of the XOR result has a length logarithmic in the value of the numbers, which is manageable.