Maximum XOR After Operations

Maximum XOR After Operations - Problem

You are given a 0-indexed integer array nums. In one operation, select any non-negative integer x and an index i, then update nums[i] to be equal to nums[i] AND (nums[i] XOR x).

Note that AND is the bitwise AND operation and XOR is the bitwise XOR operation.

Return the maximum possible bitwise XOR of all elements of nums after applying the operation any number of times.

Input & Output

Example 1 — Basic Case

$ Input: nums = [3,2,4,6]

› Output: 7

💡 Note: Operations can only clear bits. The bitwise OR 3|2|4|6 = 011|010|100|110 = 111 = 7 gives maximum XOR.

Example 2 — Single Element

$ Input: nums = [1]

› Output: 1

💡 Note: With only one element, no XOR operation is needed. The maximum is the element itself: 1.

Example 3 — Powers of 2

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

› Output: 15

💡 Note: OR of powers of 2: 0001|0010|0100|1000 = 1111 = 15. Each bit position has exactly one 1.

Constraints

1 ≤ nums.length ≤ 10⁵
0 ≤ nums[i] ≤ 10⁸

Visualization

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Asked in

G Google 25 f Facebook 18 a Amazon 15

The key insight is that the operation nums[i] AND (nums[i] XOR x) can only clear bits, never set new ones. Therefore, the maximum XOR is achieved by taking the bitwise OR of all elements. Time: O(n), Space: O(1)

Common Approaches

✓ Bit Manipulation - Mathematical Insight

⏱️ Time: O(n) Space: O(1)

The key insight is that the operation nums[i] AND (nums[i] XOR x) can only clear bits from nums[i], never set new ones. Therefore, the maximum XOR is achieved by taking the bitwise OR of all numbers.

Brute Force - Try All Operations

⏱️ Time: O(n × 2^k × n!) Space: O(1)

For each element and each possible x value, apply the operation nums[i] AND (nums[i] XOR x) and check all resulting XOR values. This explores the entire solution space but is extremely inefficient.

Bit Manipulation - Mathematical Insight — Algorithm Steps

Step 1: Understand that AND operation can only turn bits OFF
Step 2: Realize XOR with any x followed by AND preserves or clears bits
Step 3: Maximum is achieved when we keep all possible 1-bits
Step 4: Return bitwise OR of all elements

Visualization

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Step-by-Step Walkthrough

Input

Array [3,2,4,6] in binary

OR Operation

Combine all bits with OR

Result

Maximum XOR = 7

Code -

solution.c — C

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

int solution(int* nums, int n) {
    // The operation nums[i] AND (nums[i] XOR x) can only clear bits
    // Maximum XOR is achieved by keeping all possible 1-bits
    // This equals the bitwise OR of all numbers
    int result = 0;
    for (int i = 0; i < n; i++) {
        result |= nums[i];
    }
    return result;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Simple JSON array parsing
    int nums[100];
    int n = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL && n < 100) {
        nums[n++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array to compute bitwise OR

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for result variable

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Bit Manipulation - Mathematical Insight — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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