LeetCode 3599 - Partition Array to Minimize XOR

This problem asks us to partition an array of integers, nums, into exactly k non-empty contiguous subarrays, and then compute the bitwise XOR of each subarray. The goal is to minimize the maximum XOR among these subarrays.

leetcodemediumarraydynamic-programmingbit-manipulationprefix-sum

LeetCode Problem 3599

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming, Bit Manipulation, Prefix Sum

Solution

Problem Understanding

This problem asks us to partition an array of integers, nums, into exactly k non-empty contiguous subarrays, and then compute the bitwise XOR of each subarray. The goal is to minimize the maximum XOR among these subarrays. Essentially, we are trying to distribute the elements into k groups such that the "largest XOR" we get is as small as possible.

The input consists of an integer array nums with length n (1 ≤ n ≤ 250) and an integer k (1 ≤ k ≤ n). Each element nums[i] ranges from 1 to 10^9. The output is a single integer, the minimum achievable maximum XOR across all valid partitions.

Important points from the constraints:

  • The array is relatively small (up to 250 elements), which allows for dynamic programming approaches that are polynomial in n.
  • XOR operations can be large (up to 10^9), but the values themselves do not need additional handling beyond standard integer types.
  • Each subarray must be contiguous; arbitrary grouping is not allowed.

Key edge cases include:

  1. k = 1, which means the whole array is a single subarray.
  2. k = n, where each element forms its own subarray.
  3. Arrays with repeated numbers, which can result in zero XOR for some subarrays.
  4. Arrays where optimal partitioning requires non-obvious grouping to minimize the maximum XOR.

Approaches

Brute Force

A brute-force solution would consider all possible ways to partition nums into k contiguous subarrays. For each partition, we compute the XOR of each subarray and record the maximum XOR. The final answer would be the minimum of these maxima.

While correct, this approach is computationally infeasible because the number of ways to partition an array of length n into k subarrays is combinatorial, specifically C(n-1, k-1) possibilities. This grows extremely fast with n and k.

Optimal Approach

The key observation is that XOR is associative and can be computed efficiently using a prefix XOR array. This allows us to compute the XOR of any subarray in constant time. We can then use dynamic programming to track the minimum possible maximum XOR when partitioning the first i elements into j subarrays.

We define dp[i][j] as the minimum possible maximum XOR for partitioning the first i elements into j subarrays. The recurrence relation is:

dp[i][j] = min(max(dp[p][j-1], XOR(p+1, i))) for all p < i

Here, XOR(p+1, i) is the XOR of elements from index p+1 to i. Using the prefix XOR array, this is computed as prefixXOR[i] ^ prefixXOR[p]. This reduces the problem from exponential to polynomial complexity.

Approach Time Complexity Space Complexity Notes
Brute Force O(C(n-1, k-1) * n) O(n) Enumerates all partitions and computes XORs
Optimal O(n^2 * k) O(n*k) Uses DP with prefix XOR to efficiently compute subarray XORs

Algorithm Walkthrough

  1. Compute a prefix XOR array, prefixXOR, where prefixXOR[i] is the XOR of nums[0..i-1]. This allows O(1) XOR computation for any subarray.
  2. Initialize a DP table dp[i][j] where i ranges from 0 to n and j ranges from 0 to k. Set all values to infinity except dp[0][0] = 0.
  3. Iterate over the number of partitions j from 1 to k.
  4. For each subarray end index i from 1 to n, consider all possible previous cut points p from 0 to i-1.
  5. For each cut point p, compute the XOR of the subarray nums[p..i-1] using prefixXOR[i] ^ prefixXOR[p].
  6. Update dp[i][j] as the minimum of its current value and max(dp[p][j-1], subarrayXOR).
  7. After filling the table, dp[n][k] will hold the minimum possible maximum XOR for k partitions.

Why it works: The DP table keeps track of the optimal solution for all prefixes and all possible numbers of partitions. The recurrence ensures that every possible last cut is considered, so the minimum maximum XOR is guaranteed.

Python Solution

from typing import List

class Solution:
    def minXor(self, nums: List[int], k: int) -> int:
        n = len(nums)
        prefixXOR = [0] * (n + 1)
        for i in range(n):
            prefixXOR[i + 1] = prefixXOR[i] ^ nums[i]

        dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
        dp[0][0] = 0

        for j in range(1, k + 1):
            for i in range(1, n + 1):
                for p in range(i):
                    subXOR = prefixXOR[i] ^ prefixXOR[p]
                    dp[i][j] = min(dp[i][j], max(dp[p][j - 1], subXOR))

        return dp[n][k]

This implementation first computes the prefix XOR to enable constant-time subarray XOR computations. The DP table dp[i][j] stores the minimum possible maximum XOR for the first i elements split into j partitions. Nested loops fill this table, considering every possible last cut, and the result is dp[n][k].

Go Solution

func minXor(nums []int, k int) int {
    n := len(nums)
    prefixXOR := make([]int, n+1)
    for i := 0; i < n; i++ {
        prefixXOR[i+1] = prefixXOR[i] ^ nums[i]
    }

    dp := make([][]int, n+1)
    for i := range dp {
        dp[i] = make([]int, k+1)
        for j := range dp[i] {
            dp[i][j] = 1<<31 - 1
        }
    }
    dp[0][0] = 0

    for j := 1; j <= k; j++ {
        for i := 1; i <= n; i++ {
            for p := 0; p < i; p++ {
                subXOR := prefixXOR[i] ^ prefixXOR[p]
                maxVal := dp[p][j-1]
                if subXOR > maxVal {
                    maxVal = subXOR
                }
                if maxVal < dp[i][j] {
                    dp[i][j] = maxVal
                }
            }
        }
    }
    return dp[n][k]
}

In Go, we handle integer limits using 1<<31 - 1 to represent infinity. Slice allocation and nested loops mirror the Python version. Subarray XOR computation is identical using the prefix XOR array.

Worked Examples

Example 1: nums = [1,2,3], k = 2

Step-by-step table:

i j Subarray XORs considered dp[i][j]
1 1 [1] 1
2 1 [1,2] 3
2 2 [1],[2] 2
3 2 [1,2],[3], [1],[2,3] 1

Result: dp[3][2] = 1.

Example 2: nums = [2,3,3,2], k = 3

Optimal partition: [2],[3,3],[2] → XORs: 2,0,2 → max XOR = 2.

Example 3: nums = [1,1,2,3,1], k = 2

Optimal partition: [1,1],[2,3,1] → XORs: 0,0 → max XOR = 0.

Complexity Analysis

Measure Complexity Explanation
Time O(n^2 * k) Each dp[i][j] considers i previous cuts, nested loops over n and k
Space O(n*k) DP table size is n+1 by k+1, prefix XOR array is O(n)

The quadratic dependence on n is feasible because n ≤ 250. Using prefix XOR reduces subarray XOR computation to O(1).

Test Cases

# Provided examples
assert Solution().minXor([1,2,3], 2) == 1  # minimal max XOR is 1
assert Solution().minXor([2,3,3,2], 3) == 2  # minimal max XOR is 2
assert Solution().minXor([1,1,2,3,1], 2) == 0  # minimal max XOR is 0

# Edge cases
assert