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Make the XOR of All Segments Equal to Zero

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

Solution Explanation:

The problem asks for the minimum number of elements to change in the array nums such that the XOR of all segments of size k is equal to zero. This means that for every consecutive subarray of length k, the XOR of its elements must be 0.

The key observation is that if the XOR of all segments of size k is 0, then the array must exhibit a pattern with a period of k. In other words, nums[i] == nums[i + k] for all valid indices i. This is because:

  • The XOR of the segment [i, i + k - 1] is 0.
  • The XOR of the segment [i + 1, i + k] is 0.
  • XORing these two segments eliminates all elements except nums[i] and nums[i + k], resulting in nums[i] ^ nums[i + k] == 0, which implies nums[i] == nums[i + k].

This cyclic nature allows us to solve the problem efficiently using dynamic programming.

Approach: Dynamic Programming

  1. Preprocessing:

    • We create a 2D array cnt of size k x 1024. cnt[i][j] stores the count of elements with value j in the i-th group (modulo k).
    • We also create an array size of size k. size[i] stores the number of elements in the i-th group.

  2. Dynamic Programming:

    • We use dynamic programming to find the minimum number of changes needed. We define f[i] as the minimum number of changes required such that the XOR sum of the first i groups is 0. The initialization f[0] = 0.
    • We iterate through each group. For each group, we have two options for each element:
      • Option 1 (Modify all to the same value): We can change all elements in the current group to the same value. The minimum cost for this is the minimum value in the f array (representing the previous groups' XOR sum) plus the size of the current group.
      • Option 2 (Use existing elements): We can enumerate all values in current group and use each element to minimize number of changes. Let's say we decide to set the XOR sum to j. We then look for the best cost from a previous state using f[j ^ v], which is the cost from the previous state where the XOR sum is j ^ v. Add the changes needed to make the current group have XOR sum j.

  3. Result: The final result is f[0], which represents the minimum number of changes required to make the XOR of all segments of size k equal to 0.

Time Complexity Analysis

The time complexity is dominated by the nested loops in the dynamic programming step. The outer loop iterates k times (number of groups). The inner loops iterates through all possible XOR sums (1024). For each XOR sum, another loop iterates through the elements of the current group (maximum size of a group is n/k).

Therefore, the overall time complexity is approximately O(k * 210 * (n/k)) = O(n * 210), where n is the length of nums. The 2<sup>10</sup> comes from considering all possible values (0-1023) of each element. In practice, because of the dynamic programming optimization, this would be a significant underestimation of the runtime. The 2^10 stems from the fact that a number's value can be any value from 0 to 1023. Therefore, we need to check a large number of states when finding the minimum changes needed in the dynamic programming step.

Space Complexity Analysis

The space complexity is determined by the size of the cnt array and the f array. The cnt array has size O(k * 210), and the f array has size O(210). Therefore, the overall space complexity is O(k * 210).

The provided code in Python, Java, C++, and Go implements this dynamic programming solution. Remember that the constant factor hidden within the Big O notation is significant due to the 210 term. For large values of k, the runtime might become substantial.