Partition Array to Minimize XOR

Partition Array to Minimize XOR - Problem

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.

A subarray is a contiguous part of an array.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3,4], k = 2

› Output: 4

💡 Note: Optimal partition: [1,2,3] with XOR = 0, and [4] with XOR = 4. Maximum XOR = max(0,4) = 4.

Example 2 — Single Elements

$ Input: nums = [5,6,7], k = 3

› Output: 7

💡 Note: Each element forms its own partition: [5], [6], [7]. Maximum XOR = max(5,6,7) = 7.

Example 3 — All Together

$ Input: nums = [1,1,1,1], k = 1

› Output: 0

💡 Note: All elements in one partition: [1,1,1,1]. XOR = 1⊕1⊕1⊕1 = 0.

Constraints

1 ≤ nums.length ≤ 10³
1 ≤ nums[i] ≤ 10⁶
1 ≤ k ≤ nums.length

Visualization

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Asked in

G Google 15 M Microsoft 12 A Amazon 8

The key insight is to use binary search on the maximum XOR value. We search between 0 and the maximum possible XOR, then for each candidate maximum, greedily check if we can form k partitions where no partition's XOR exceeds the candidate. Best approach is Binary Search on Answer with greedy validation. Time: O(n × log(maxXOR)), Space: O(1).

Common Approaches

✓ Binary Search on Answer

⏱️ Time: O(n × log(maxXOR)) Space: O(1)

Binary search on the possible maximum XOR values. For each candidate maximum, use a greedy approach to check if we can partition the array into k subarrays where each subarray's XOR is at most the candidate value.

Brute Force Recursion

⏱️ Time: O(2^n) Space: O(n)

Recursively try every possible partition point and compute XOR for each subarray. Keep track of the minimum possible maximum XOR across all valid partitions.

Dynamic Programming with Memoization

⏱️ Time: O(n² × k) Space: O(n × k)

Use memoization to store results of solve(index, parts_remaining) to avoid recalculating the same subproblems multiple times.

Binary Search on Answer — Algorithm Steps

Binary search on maximum XOR value
For each candidate, greedily check feasibility
Extend current partition until XOR exceeds candidate
Count total partitions needed

Visualization

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Step-by-Step Walkthrough

Search Space

Binary search between 0 and maximum possible XOR

Check Feasibility

For each candidate, greedily check if k partitions possible

Narrow Range

If feasible, try smaller maximum; else try larger

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int prefix_xor[1001];
int dp[1001][1001];

int get_xor(int l, int r) {
    return prefix_xor[r + 1] ^ prefix_xor[l];
}

int canPartition(int numsSize, int k, int maxXOR) {
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (int i = 1; i <= numsSize; i++) {
        for (int p = 1; p <= k && p <= i; p++) {
            for (int j = p - 1; j < i; j++) {
                if (dp[j][p - 1] && get_xor(j, i - 1) <= maxXOR) {
                    dp[i][p] = 1;
                    break;
                }
            }
        }
    }
    return dp[numsSize][k];
}

int solution(int* nums, int numsSize, int k) {
    prefix_xor[0] = 0;
    for (int i = 0; i < numsSize; i++) {
        prefix_xor[i + 1] = prefix_xor[i] ^ nums[i];
    }

    int left = 0;
    int right = 0;
    for (int i = 0; i < numsSize; i++) {
        int xorVal = 0;
        for (int j = i; j < numsSize; j++) {
            xorVal ^= nums[j];
            if (xorVal > right) right = xorVal;
        }
    }

    int result = right;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (canPartition(numsSize, k, mid)) {
            result = mid;
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);

    static int nums[1000];
    int numsSize = 0;

    char* p = line;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;

    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        char* endptr;
        long val = strtol(p, &endptr, 10);
        if (endptr != p) {
            nums[numsSize++] = (int)val;
            p = endptr;
        } else {
            break;
        }
    }

    int k;
    scanf("%d", &k);

    int result = solution(nums, numsSize, k);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × log(maxXOR))

Binary search has log(maxXOR) iterations, each validation takes O(n) time

⚡ Linearithmic

Space Complexity

O(1)

Only uses constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Binary Search on Answer — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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