Make the XOR of All Segments Equal to Zero

The first key insight is to understand the condition that the XOR sum of all segments of size k is zero. If nums[i] ^ ... ^ nums[i+k-1] = 0 and nums[i+1] ^ ... ^ nums[i+k] = 0, XORing these two equations gives nums[i] ^ nums[i+k] = 0, which means nums[i] = nums[i+k]. This implies the array must be periodic with period k. All elements at indices j, j+k, j+2k, ... must be identical.

This reduces the problem to partitioning the array into k groups based on index % k. For each group i, we must choose a single value v_i to change all its elements to. The cost for group i is the number of elements not equal to v_i. The chosen values must also satisfy v_0 ^ v_1 ^ ... ^ v_{k-1} = 0.

This structure lends itself to a dynamic programming solution. We can build up the solution one group at a time. Let dp[i][j] be the minimum cost to modify the first i groups such that the XOR sum of their chosen values is j. We can compute dp[i][j] by considering all possible values for the i-th group and using the results from dp[i-1].

• The core condition XOR of all segments of size k is zero implies that nums[i] = nums[i+k] for all 0 <= i < n-k. This means the array must be periodic with period k.
• This property partitions the array indices into k groups based on index % k. All elements within the same group must be changed to an identical value.
• Additionally, the chosen values for each group, let's call them v_0, v_1, ..., v_{k-1}, must satisfy v_0 ^ v_1 ^ ... ^ v_{k-1} = 0.
• The problem is now to select v_0, ..., v_{k-1} to minimize the total number of changes, subject to the XOR sum constraint.
• We can solve this using dynamic programming. Let dp[i][j] be the minimum number of changes for the first i groups (from group 0 to i-1) such that the XOR sum of their chosen values is j.
• State: dp[i][j] = minimum changes for groups 0, ..., i-1 to have an XOR sum of j.
• Initialization: dp[0][0] = 0, and dp[0][j] = infinity for j > 0.
• Transition: To compute dp[i][j], we iterate through all possible values v (from 0 to 1023) for group i-1. If we choose v for group i-1, the XOR sum of the first i-1 groups must have been j ^ v. The cost for changing group i-1 to v is size_{i-1} - freq_{i-1}[v], where size_{i-1} is the number of elements in the group and freq_{i-1}[v] is the frequency of v in that group.
• The recurrence relation is: dp[i][j] = min_{0 <= v < 1024} (dp[i-1][j ^ v] + cost(i-1, v)).
• Final Answer: The minimum changes for all k groups to have an XOR sum of 0 is dp[k][0].

This approach uses a straightforward dynamic programming formulation. We define dp[i][j] as the minimum changes needed for the first i groups (0 to i-1) to have their chosen values' XOR sum equal to j.

We can use a 2D DP table of size (k+1) x 1024. The state transition is as follows: to calculate dp[i][j], we consider every possible value v (from 0 to 1023) that we can assign to group i-1. The cost of assigning v to group i-1 is size_{i-1} - freq_{i-1}[v]. If we choose v, then the XOR sum of the first i-1 groups must have been j ^ v. So, we look up dp[i-1][j ^ v], add the cost for the current group, and take the minimum over all possible v.

To save space, we can notice that dp[i] only depends on dp[i-1]. Thus, we can use only two arrays, one for the previous state and one for the current state, reducing space complexity.

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

class Solution {
    public int minChanges(int[] nums, int k) {
        int n = nums.length;
        int MAXXOR = 1 << 10;
        int INF = (int) 1e9;

        // Group elements and get frequencies
        Map<Integer, Integer>[] freqs = new HashMap[k];
        int[] groupSizes = new int[k];
        for (int i = 0; i < k; i++) {
            freqs[i] = new HashMap<>();
        }
        for (int i = 0; i < n; i++) {
            int groupIdx = i % k;
            freqs[groupIdx].put(nums[i], freqs[groupIdx].getOrDefault(nums[i], 0) + 1);
            groupSizes[groupIdx]++;
        }

        // DP state: dp[j] = min changes for groups processed so far to have XOR sum j
        int[] dp = new int[MAXXOR];
        Arrays.fill(dp, INF);
        dp[0] = 0; // Base case: 0 groups, XOR sum 0, 0 changes

        for (int i = 0; i < k; i++) {
            int[] nextDp = new int[MAXXOR];
            Arrays.fill(nextDp, INF);
            // Iterate over all possible previous XOR sums
            for (int j = 0; j < MAXXOR; j++) {
                if (dp[j] == INF) continue;
                // Iterate over all possible values `v` to set for the current group `i`
                for (int v = 0; v < MAXXOR; v++) {
                    int costToChange = groupSizes[i] - freqs[i].getOrDefault(v, 0);
                    int newXor = j ^ v;
                    nextDp[newXor] = Math.min(nextDp[newXor], dp[j] + costToChange);
                }
            }
            dp = nextDp;
        }

        return dp[0];
    }
}