Maximum XOR for Each Query - Practice Coding Problems

Maximum XOR for Each Query - Problem

You are given a sorted array nums of n non-negative integers and an integer maximumBit. Your task is to perform n queries where each query involves:

Find the optimal value k: Choose a non-negative integer k < 2^maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized.
Remove the last element: After finding k, remove the last element from the current array.

Return an array answer where answer[i] is the answer to the i-th query.

Key Insight: Since we want to maximize the XOR result, we need to find the value k that creates the largest possible binary number when XORed with our current array's XOR.

Input & Output

example_1.py — Basic Example

$ Input: nums = [0,1,1,3], maximumBit = 2

› Output: [0,3,2,3]

💡 Note: Query 1: nums=[0,1,1,3], XOR=3, maxVal=3, optimal k=0 gives 3⊕0=3. Query 2: nums=[0,1,1], XOR=0, optimal k=3 gives 0⊕3=3. Query 3: nums=[0,1], XOR=1, optimal k=2 gives 1⊕2=3. Query 4: nums=[0], XOR=0, optimal k=3 gives 0⊕3=3.

example_2.py — Single Element

$ Input: nums = [2,3,4,7], maximumBit = 3

› Output: [5,2,6,5]

💡 Note: With maximumBit=3, maxVal=7. For each query, we find k that maximizes the XOR result by taking the complement within the 3-bit range.

example_3.py — Edge Case

$ Input: nums = [0], maximumBit = 1

› Output: [1]

💡 Note: Single element array with XOR=0. With maximumBit=1, maxVal=1, so optimal k=0⊕1=1.

Visualization

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Understanding the Visualization

Calculate Array XOR

Find XOR of all elements in current array

Find Bit Complement

To maximize result, flip all bits within maximumBit range

Update for Next Query

Remove last element by XORing it out

Key Takeaway

🎯 Key Insight: To maximize XOR result, use the bitwise complement of the current XOR within the allowed bit range

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array with O(1) operations per element

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space (excluding result array)

✓ Linear Space

Constraints

nums.length == n
1 ≤ n ≤ 10⁵
1 ≤ maximumBit ≤ 20
0 ≤ nums[i] < 2^maximumBit
nums is sorted in ascending order

Asked in

f Meta 25 G Google 20 ⊞ Microsoft 15 a Amazon 10

The optimal solution uses bit manipulation with the key insight that to maximize A XOR k, we should set k to be the bitwise complement of A within the allowed range. By maintaining a running XOR and using k = totalXOR ^ maxVal, we achieve O(n) time complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Optimal Bit Manipulation	O(n)	O(1)	Use prefix XOR and bit complement to find optimal k in O(1) time per query
Brute Force (Try All Possible k)	O(n² × 2^maximumBit)	O(n)	For each query, try every possible k value and find the maximum XOR result

Optimal Bit Manipulation — Algorithm Steps

Calculate total XOR of all elements once
For each query, the optimal k is the complement of current XOR within maximumBit range
k = currentXOR ^ ((1 << maximumBit) - 1)
Update currentXOR by removing the last element (XOR with it)
Add k to result

Visualization

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Step-by-Step Walkthrough

Calculate Total XOR

XOR all elements: 0⊕1⊕1⊕3 = 3

Find Complement

k = totalXOR ⊕ maxVal = 3 ⊕ 7 = 4

Remove Last Element

Update totalXOR by XORing out last element

Code -

solution.c — C

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

int* getMaximumXor(int* nums, int numsSize, int maximumBit, int* returnSize) {
    *returnSize = numsSize;
    int* answer = (int*)malloc(numsSize * sizeof(int));
    
    // Calculate total XOR
    int totalXor = 0;
    for (int i = 0; i < numsSize; i++) {
        totalXor ^= nums[i];
    }
    
    // Maximum possible value with maximumBit bits
    int maxVal = (1 << maximumBit) - 1;
    
    // Process queries from end to start
    for (int i = numsSize - 1; i >= 0; i--) {
        // Optimal k is the complement of current XOR
        int k = totalXor ^ maxVal;
        answer[numsSize - 1 - i] = k;
        
        // Remove last element by XORing it out
        totalXor ^= nums[i];
    }
    
    return answer;
}

int main() {
    // Parse input and call solution
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array with O(1) operations per element

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space (excluding result array)

✓ Linear Space

Constraints

nums.length == n
1 ≤ n ≤ 10⁵
1 ≤ maximumBit ≤ 20
0 ≤ nums[i] < 2^maximumBit
nums is sorted in ascending order

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Optimal Bit Manipulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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