Minimize XOR

Minimize XOR - Problem

Given two positive integers num1 and num2, find the positive integer x such that:

x has the same number of set bits as num2, and
The value x XOR num1 is minimal.

Note that XOR is the bitwise XOR operation.

Return the integer x. The test cases are generated such that x is uniquely determined.

The number of set bits of an integer is the number of 1's in its binary representation.

Input & Output

Example 1 — Perfect Match

$ Input: num1 = 3, num2 = 5

› Output: 3

💡 Note: num1 = 3 (binary: 011) has 2 set bits. num2 = 5 (binary: 101) has 2 set bits. Since 3 already has the required number of bits, x = 3 gives XOR = 0, which is minimal.

Example 2 — Need More Bits

$ Input: num1 = 1, num2 = 12

› Output: 3

💡 Note: num1 = 1 (binary: 001) has 1 set bit. num2 = 12 (binary: 1100) has 2 set bits. We need x with 2 set bits. Starting from 1, we add the lowest bit to get 3 (binary: 011). XOR: 3 ⊕ 1 = 2.

Example 3 — Need Fewer Bits

$ Input: num1 = 25, num2 = 6

› Output: 17

💡 Note: num1 = 25 (binary: 11001) has 3 set bits. num2 = 6 (binary: 110) has 2 set bits. We need x with 2 set bits. Remove the highest bit from 25 to get 17 (binary: 10001) with 2 set bits.

Constraints

1 ≤ num1, num2 ≤ 10⁹
num1 and num2 are positive integers

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is to minimize XOR by making x as similar to num1 as possible while maintaining the exact number of set bits from num2. The greedy approach starts with num1 and optimally adjusts the bit count. Time: O(log n), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(2^n) Space: O(1)

Generate all possible numbers with the same number of set bits as num2, then find the one that minimizes XOR with num1. This involves checking all combinations which becomes computationally expensive for large numbers.

Greedy Bit Manipulation

⏱️ Time: O(log n) Space: O(1)

To minimize XOR, we want x to be as similar to num1 as possible. First, copy as many 1-bits from num1 as we can. Then, if we need more bits, set the lowest possible bits. If we have too many bits, clear the highest bits that don't contribute to minimizing XOR.

Brute Force — Algorithm Steps

Count set bits in num2
Generate all numbers with that many set bits
Calculate XOR with num1 for each
Return the number with minimum XOR

Visualization

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Step-by-Step Walkthrough

Count Bits

Count set bits in num2

Generate Numbers

Try all numbers with same bit count

Find Minimum

Calculate XOR and keep minimum

Code -

solution.c — C

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

int countBits(int n) {
    int count = 0;
    while (n) {
        count += n & 1;
        n >>= 1;
    }
    return count;
}

int solution(int num1, int num2) {
    int targetBits = countBits(num2);
    int minXor = INT_MAX;
    int result = 0;
    
    // Try numbers up to a reasonable limit
    for (int x = 1; x < (1 << 20); x++) {
        if (countBits(x) == targetBits) {
            int xorVal = x ^ num1;
            if (xorVal < minXor) {
                minXor = xorVal;
                result = x;
            }
        }
    }
    
    return result;
}

int main() {
    int num1, num2;
    scanf("%d %d", &num1, &num2);
    
    printf("%d\n", solution(num1, num2));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Must check exponential combinations of bit positions

⚠ Quadratic Growth

Space Complexity

O(1)

Only storing current minimum and result

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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