LeetCode-in-Java
3599. Partition Array to Minimize XOR
Medium
You are given an integer array nums and an integer k.
Your task is to partition nums into k non-empty **non-empty subarrays. For each subarray, compute the bitwise **XOR of all its elements.
Return the minimum possible value of the maximum XOR among these k subarrays.
Example 1:
Input: nums = [1,2,3], k = 2
Output: 1
Explanation:
The optimal partition is [1] and [2, 3].
- XOR of the first subarray is
1. - XOR of the second subarray is
2 XOR 3 = 1.
The maximum XOR among the subarrays is 1, which is the minimum possible.
Example 2:
Input: nums = [2,3,3,2], k = 3
Output: 2
Explanation:
The optimal partition is [2], [3, 3], and [2].
- XOR of the first subarray is
2. - XOR of the second subarray is
3 XOR 3 = 0. - XOR of the third subarray is
2.
The maximum XOR among the subarrays is 2, which is the minimum possible.
Example 3:
Input: nums = [1,1,2,3,1], k = 2
Output: 0
Explanation:
The optimal partition is [1, 1] and [2, 3, 1].
- XOR of the first subarray is
1 XOR 1 = 0. - XOR of the second subarray is
2 XOR 3 XOR 1 = 0.
The maximum XOR among the subarrays is 0, which is the minimum possible.
Constraints:
1 <= nums.length <= 2501 <= nums[i] <= 1091 <= k <= n
Solution
import java.util.Arrays;
public class Solution {
public int minXor(int[] nums, int k) {
int n = nums.length;
// Step 1: Prefix XOR array
int[] pfix = new int[n + 1];
for (int i = 1; i <= n; i++) {
pfix[i] = pfix[i - 1] ^ nums[i - 1];
}
// Step 2: DP table
int[][] dp = new int[n + 1][k + 1];
for (int[] row : dp) {
Arrays.fill(row, Integer.MAX_VALUE);
}
for (int i = 0; i <= n; i++) {
// Base case: 1 partition
dp[i][1] = pfix[i];
}
// Step 3: Fill DP for partitions 2 to k
for (int parts = 2; parts <= k; parts++) {
for (int end = parts; end <= n; end++) {
for (int split = parts - 1; split < end; split++) {
int segmentXOR = pfix[end] ^ pfix[split];
int maxXOR = Math.max(dp[split][parts - 1], segmentXOR);
dp[end][parts] = Math.min(dp[end][parts], maxXOR);
}
}
}
return dp[n][k];
}
}