Notice, that the problem is equivalent to the following problem: we need to make a_0 = a_k = a_2k, a_1 = a_(k+1) = and so on. We can reformulate it: we have k groups of numbers and we need to change minimum possible number of numbers, such that in each group we have equal numbers and XOR of all groups is equal to 0.

Define as dp[i][mask] the maximum number of values we can not change, such that we have mask is XOR of first i groups (it is easier to work with maximums, not minimums, because groups can have different size).

Then, let us calculate i-th row of matrix dp, if we know i-1-th row. Then we can have two options:

  1. On the last step we take mask, which is not in numbers of dp[i-1] row, it this case we need to choose maximum from dp[i-1]
  2. Or we can take mask, which is inside group i-1 of our numbers, in this case we need to check dp[i-1][j^mask] + freq[i-1][mask].

Complexity: denote by T number of bits each number can be restricted and by n length of nums. We have 3 loops here, but loop for i and for mask will give you in total O(n) complexity, and loop for j will give you O(2^T) complexity, so totally we have O(2^T n) time complexity and O(2^T k) space complexity.

class Solution:
    def minChanges(self, nums, k):
        N = 1<<10
        
        freq = defaultdict(Counter)
        for i, num in enumerate(nums):
            freq[i%k][num] += 1
            
        dp = [[0] * N for _ in range(k+1)] 
        for i in range(1, N): dp[0][i] = -float("inf")
            
        for i in range(1, k+1):
            max_row = max(dp[i-1])
            
            for j in range(N):
                for mask in freq[i-1]:
                    dp[i][j] = max(dp[i][j], max_row, dp[i-1][j^mask] + freq[i-1][mask])
            
        return len(nums) - dp[-1][0]

Shorter version, using greedy idea

This solution is inspired by delphih solution https://leetcode.com/problems/make-the-xor-of-all-segments-equal-to-zero/discuss/1097266/Python-Solution-with-some-explanation, please go and upvote this post!

First of all, it can be proven, that one of the two cases holds:

  1. All numbers except one in each group are numbers from this groups.
  2. All numbers in each group are numbers from this group.

It can be proven, that it can not happen, that we have 2 or more numbers from outside groups, I will write it a later if I have time.

For the first one, we need to check groups with the maximum frequencies of elements and choose the k-1 most frequent.

For the second case, we can use almost the same logic we have previously, but now we do not need to think about numbers outside groups! So we just have case 1 from previous algorithm and we can forget about case 2.

Complexity: what is important, complexity is the same as it was previously: O(2^T n) time complexity and O(2^T k) space complexity.

class Solution:
    def minChanges(self, A, k):
        freq = [Counter(A[j] for j in range(i, len(A), k)) for i in range(k)]
        mxs = [freq[i].most_common(1)[0][1] for i in range(k)]

        @lru_cache(None)
        def dp(i, j):
            if i == 0: return -10000*j  
            return max(dp(i-1,j^m) + freq[i-1][m] for m in freq[i-1])

        return len(A) - max(sum(mxs) - min(mxs), dp(k, 0))