1787. Make the XOR of All Segments Equal to Zero

Explanation:

To solve this problem, we can iterate through all possible values of XOR for segments of size k, and keep track of the minimum number of elements that need to be changed in the array to make the XOR of all segments of size k equal to zero. We can use dynamic programming to store the minimum number of changes needed for each possible XOR value.

We will create a map to store the count of each XOR value encountered at a given position i in the array.
Initialize a 2D array dp to store the minimum number of changes needed for each XOR value at each position.
Iterate through each segment of size k starting from index i to i+k-1.
For each segment, update the count of XOR value for that segment in the map.
Update the dp array with the minimum number of changes needed for each XOR value.
Finally, return the minimum number of changes needed to make all XOR values of segments of size k equal to zero.

Time Complexity: O(n * 2^k), where n is the length of the input array and k is the segment size. Space Complexity: O(n * 2^k) for the dp array and the map.

:

View Leetcode Problem

class Solution {
    public int minChanges(int[] nums, int k) {
        int n = nums.length;
        int[][] dp = new int[n][1024];
        int[] count = new int[1024];
        
        for (int i = 0; i < n; i++) {
            count[nums[i]]++;
        }
        
        for (int i = 0; i < n; i++) {
            int[] next = new int[1024];
            for (int j = 0; j < 1024; j++) {
                next[j] = Integer.MAX_VALUE;
            }
            
            int size = Math.min(k, n - i);
            int xor = 0;
            int best = 0;
            
            for (int j = 0; j < size; j++) {
                next[xor] = Math.min(next[xor], best + count[j]);
                best = Math.min(best, dp[i][j]);
                xor ^= nums[i + j];
            }
            
            for (int j = 0; j < 1024; j++) {
                dp[i + 1][j] = Math.min(dp[i + 1][j], next[j]);
            }
        }
        
        return dp[n][0];
    }
}

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