Introduction

In this blog post, we will explore how to solve the LeetCode problem "2997" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's a breakdown of the solution approach, followed by the Python code:

• Calculate Initial XOR: Compute the bitwise XOR of all the numbers in the input array nums.
• Compare with Target: Compare the initial XOR result with the target value k.

• Count Different Bits: The number of differing bits between the initial XOR and k represents the minimum number of flips needed. This is achieved efficiently using the XOR operation itself.

• Runtime Complexity: O(n), where n is the number of elements in nums. Storage Complexity: O(1).

Code

    def minOperations(nums, k):
    """
    Calculates the minimum number of bit flips required to make the XOR of
    the elements in nums equal to k.

    Args:
        nums: A list of integers.
        k: The target XOR value.

    Returns:
        The minimum number of operations (bit flips).
    """
    xor_sum = 0
    for num in nums:
        xor_sum ^= num

    xor_diff = xor_sum ^ k
    operations = 0
    while xor_diff > 0:
        operations += xor_diff & 1
        xor_diff >>= 1

    return operations