3599. Partition Array to Minimize XOR

Description

You are given an integer array nums and an integer k.

Your task is to partition nums into k 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 <= 250
1 <= nums[i] <= 10⁹
1 <= k <= n

Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ as the minimum possible value of the maximum XOR among all ways to partition the first $i$ elements into $j$ subarrays. Initially, set $f[0][0] = 0$, and all other $f[i][j] = +\infty$.

To quickly compute the XOR of a subarray, we can use a prefix XOR array $g$, where $g[i]$ represents the XOR of the first $i$ elements. For the subarray $[h + 1…i]$ (with indices starting from $1$), its XOR value is $g[i] \oplus g[h]$.

Next, we iterate $i$ from $1$ to $n$, $j$ from $1$ to $\min(i, k)$, and $h$ from $j - 1$ to $i - 1$, where $h$ represents the end position of the previous subarray (indices starting from $1$). We update $f[i][j]$ using the following state transition equation:

\[f[i][j] = \min_{h \in [j - 1, i - 1]} \max(f[h][j - 1], g[i] \oplus g[h])\]

Finally, we return $f[n][k]$, which is the minimum possible value of the maximum XOR when partitioning the entire array into $k$ subarrays.

The time complexity is $O(n^2 \times k)$, and the space complexity is $O(n \times k)$, where $n$ is the length of the array.

class Solution {
    public int minXor(int[] nums, int k) {
        int n = nums.length;
        int[] g = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            g[i] = g[i - 1] ^ nums[i - 1];
        }

        int[][] f = new int[n + 1][k + 1];
        for (int i = 0; i <= n; ++i) {
            Arrays.fill(f[i], Integer.MAX_VALUE);
        }
        f[0][0] = 0;

        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= Math.min(i, k); ++j) {
                for (int h = j - 1; h < i; ++h) {
                    f[i][j] = Math.min(f[i][j], Math.max(f[h][j - 1], g[i] ^ g[h]));
                }
            }
        }

        return f[n][k];
    }
}

class Solution {
public:
    int minXor(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> g(n + 1);
        for (int i = 1; i <= n; ++i) {
            g[i] = g[i - 1] ^ nums[i - 1];
        }

        const int inf = numeric_limits<int>::max();
        vector f(n + 1, vector(k + 1, inf));
        f[0][0] = 0;

        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= min(i, k); ++j) {
                for (int h = j - 1; h < i; ++h) {
                    f[i][j] = min(f[i][j], max(f[h][j - 1], g[i] ^ g[h]));
                }
            }
        }

        return f[n][k];
    }
};

min = lambda a, b: a if a < b else b
max = lambda a, b: a if a > b else b


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

        f = [[inf] * (k + 1) for _ in range(n + 1)]
        f[0][0] = 0
        for i in range(1, n + 1):
            for j in range(1, min(i, k) + 1):
                for h in range(j - 1, i):
                    f[i][j] = min(f[i][j], max(f[h][j - 1], g[i] ^ g[h]))
        return f[n][k]

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

	const inf = math.MaxInt32
	f := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, k+1)
		for j := range f[i] {
			f[i][j] = inf
		}
	}
	f[0][0] = 0

	for i := 1; i <= n; i++ {
		for j := 1; j <= min(i, k); j++ {
			for h := j - 1; h < i; h++ {
				f[i][j] = min(f[i][j], max(f[h][j-1], g[i]^g[h]))
			}
		}
	}

	return f[n][k]
}

function minXor(nums: number[], k: number): number {
    const n = nums.length;
    const g: number[] = Array(n + 1).fill(0);
    for (let i = 1; i <= n; ++i) {
        g[i] = g[i - 1] ^ nums[i - 1];
    }

    const inf = Number.MAX_SAFE_INTEGER;
    const f: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(inf));
    f[0][0] = 0;

    for (let i = 1; i <= n; ++i) {
        for (let j = 1; j <= Math.min(i, k); ++j) {
            for (let h = j - 1; h < i; ++h) {
                f[i][j] = Math.min(f[i][j], Math.max(f[h][j - 1], g[i] ^ g[h]));
            }
        }
    }

    return f[n][k];
}

3599 - Partition Array to Minimize XOR

3599. Partition Array to Minimize XOR

Description

Solutions

Solution 1: Dynamic Programming

All Problems

All Solutions

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3599. Partition Array to Minimize XOR

Description

Solutions

Solution 1: Dynamic Programming

All Problems

All Solutions