Introduction

In this blog post, we will explore how to solve the LeetCode problem "1787" 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 an array nums​​​ and an integer k​​​​​. The XOR of a segment [left, right] where left <= right is the XOR of all the elements with indices between left and right, inclusive: nums[left] XOR nums[left+1] XOR ... XOR nums[right]. Return the minimum number of elements to change in the array such that the XOR of all segments of size k​​​​​​ is equal to zero. Example 1: Input: nums = [1,2,0,3,0], k = 1 Output: 3 Explanation: Modify the array from [1,2,0,3,0] to from [0,0,0,0,0]. Example 2: Input: nums = [3,4,5,2,1,7,3,4,7], k = 3 Output: 3 Explanation: Modify the array from [3,4,5,2,1,7,3,4,7] to [3,4,7,3,4,7,3,4,7]. Example 3: Input: nums = [1,2,4,1,2,5,1,2,6], k = 3 Output: 3 Explanation: Modify the array from [1,2,4,1,2,5,1,2,6] to [1,2,3,1,2,3,1,2,3]. Constraints: 1 <= k <= nums.length <= 2000 ​​​​​​0 <= nums[i] < 210

Explanation

Here's the solution to the problem, including an explanation of the approach, complexity analysis, and the Python code.

• High-Level Approach:

  • The problem requires making the XOR of all segments of length k equal to zero. This is equivalent to ensuring nums[i] ^ nums[i+k] ^ nums[i+2k] ^ ... = 0 for each i from 0 to k-1.
  • For each i from 0 to k-1, consider the subsequence nums[i], nums[i+k], nums[i+2k], .... Within each subsequence, we want to minimize the number of changes to make all elements equal. The optimal choice is to change all elements in the subsequence to the value that appears most frequently in that subsequence.
  • Use dynamic programming to determine the minimum changes to make the XOR of elements at each index i equal to 0, knowing that we can freely modify values in each period k.

• Complexity Analysis:

  • Runtime Complexity: O(n 2¹⁰) = O(n 1024). This comes from iterating over the array n times and potentially looping up to 2^10 times within each such iteration.
  • Storage Complexity: O(k 2¹⁰). This is because the dp array of size `k 2^10` is the primary space cost.

Code

    def minChanges(nums, k):
    n = len(nums)
    max_xor = 1024
    counts = [{} for _ in range(k)]
    for i in range(n):
        idx = i % k
        counts[idx][nums[i]] = counts[idx].get(nums[i], 0) + 1

    dp = [[float('inf')] * max_xor for _ in range(k + 1)]
    dp[0][0] = 0

    for i in range(1, k + 1):
        total_elements = n // k + (1 if i <= n % k else 0)
        for xor_val in range(max_xor):
            # Option 1: Change all elements in the current group to achieve the desired XOR
            dp[i][xor_val] = min(dp[i][xor_val], dp[i - 1][xor_val] + total_elements)

            # Option 2: Keep some element 'num' as is, and change the rest to achieve the xor
            for num in counts[i - 1]:
                required_xor = xor_val ^ num
                dp[i][xor_val] = min(dp[i][xor_val], dp[i - 1][required_xor] + total_elements - counts[i - 1][num])

    return dp[k][0]