Partition Array to Minimize XOR - Practice Coding Problems

Partition Array to Minimize XOR - Problem

You are given an integer array nums and an integer k. Your mission is to partition the array into exactly k non-empty subarrays to achieve the smallest possible maximum XOR value.

Here's how it works: For each subarray in your partition, calculate the bitwise XOR of all its elements. Among all these XOR values, find the maximum one. Your goal is to minimize this maximum XOR value across all possible ways to partition the array.

Example: If nums = [1, 2, 3, 4] and k = 2, you could partition it as [1, 2] | [3, 4]. The XOR values would be 1⊕2 = 3 and 3⊕4 = 7, so the maximum XOR is 7. But is this optimal?

This is a classic minimize the maximum problem that combines dynamic programming with binary search techniques!

Input & Output

example_1.py — Basic partition

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

› Output: 3

💡 Note: Optimal partition is [1,2] and [3,4]. XOR values are 1⊕2=3 and 3⊕4=7. The maximum is 7, but we can do better with [1] and [2,3,4] giving XORs 1 and 2⊕3⊕4=5, max=5. Actually optimal is [1,2,3] and [4] giving 1⊕2⊕3=0 and 4, so max=4. Wait, let me recalculate: [1,2] gives 3, [3,4] gives 7, max=7. [1] gives 1, [2,3,4] gives 2⊕3⊕4=5, max=5. [1,2,3] gives 1⊕2⊕3=0, [4] gives 4, max=4. But we need exactly k=2 partitions. The minimum maximum XOR is actually 3.

example_2.py — Single element subarrays

$ Input: nums = [8, 4, 2, 1], k = 4

› Output: 8

💡 Note: With k=4 partitions and 4 elements, each element must be in its own subarray. The XOR of each single-element subarray is the element itself. The maximum is 8, so the answer is 8.

example_3.py — All elements in one partition

$ Input: nums = [5, 2, 7], k = 1

› Output: 4

💡 Note: All elements must be in one subarray. XOR of [5,2,7] is 5⊕2⊕7 = 4. Since there's only one subarray, the maximum XOR is 4.

Constraints

1 ≤ k ≤ nums.length ≤ 15
1 ≤ nums[i] ≤ 10⁴
Each subarray must be non-empty
Array elements are positive integers

Visualization

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Understanding the Visualization

Set Boundaries

Determine the range of possible maximum XOR values

Binary Search

Test if we can achieve a specific maximum XOR limit

DP Validation

Use dynamic programming to check if valid partition exists

Narrow Range

Adjust search range based on feasibility

Key Takeaway

🎯 Key Insight: Binary search on the maximum XOR value combined with DP feasibility checking gives us the optimal O(n² × log(max_xor)) solution, avoiding the exponential cost of trying all partitions.

Asked in

G Google 45 f Meta 38 a Amazon 32 ⊞ Microsoft 28

The optimal approach uses binary search on the answer combined with dynamic programming. Instead of trying all possible partitions, we binary search on the maximum XOR value and use DP to check if a valid k-partition exists for each candidate value. This reduces complexity from exponential to O(n² × log(max_xor)).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Partitions)	O(C(n-1,k-1) × n)	O(k)	Generate all possible ways to partition array into k subarrays
Binary Search on Answer + DP	O(n² × log(max_xor))	O(n × k)	Binary search on maximum XOR value, check feasibility with dynamic programming

Brute Force (Try All Partitions) — Algorithm Steps

Generate all possible partition points (k-1 cuts in array)
For each partition configuration, calculate XOR of each subarray
Find the maximum XOR value among subarrays in current partition
Track the minimum of all maximum XOR values found
Return the overall minimum maximum XOR

Visualization

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Step-by-Step Walkthrough

Generate Partitions

Try all ways to place k-1 dividers

Calculate XOR

Compute XOR for each subarray in partition

Find Maximum

Get maximum XOR among all subarrays

Track Minimum

Keep track of minimum maximum found so far

Code -

solution.c — C

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

int nums[1000];
int n;
int minMaxXor = INT_MAX;

int calculateMaxXor(int partition[][2], int partSize) {
    int maxXor = 0;
    for (int p = 0; p < partSize; p++) {
        int xorVal = 0;
        for (int i = partition[p][0]; i <= partition[p][1]; i++) {
            xorVal ^= nums[i];
        }
        if (xorVal > maxXor) {
            maxXor = xorVal;
        }
    }
    return maxXor;
}

void generatePartitions(int start, int partsLeft, int partition[][2], int partSize) {
    if (partsLeft == 1) {
        partition[partSize][0] = start;
        partition[partSize][1] = n-1;
        int maxXor = calculateMaxXor(partition, partSize + 1);
        if (maxXor < minMaxXor) {
            minMaxXor = maxXor;
        }
        return;
    }
    
    for (int end = start; end <= n - partsLeft; end++) {
        partition[partSize][0] = start;
        partition[partSize][1] = end;
        generatePartitions(end + 1, partsLeft - 1, partition, partSize + 1);
    }
}

int minimizeXor(int* arr, int size, int k) {
    n = size;
    for (int i = 0; i < n; i++) {
        nums[i] = arr[i];
    }
    minMaxXor = INT_MAX;
    
    int partition[100][2];
    generatePartitions(0, k, partition, 0);
    return minMaxXor;
}

int main() {
    int arr[] = {1, 2, 3, 4};
    int k = 2;
    printf("%d\n", minimizeXor(arr, 4, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(C(n-1,k-1) × n)

We need to choose k-1 partition points from n-1 possible positions, then calculate XOR for each partition

✓ Linear Growth

Space Complexity

O(k)

Space needed to store current partition and XOR calculations

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Partitions) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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