Minimum Flips to Make a OR b Equal to c

Minimum Flips to Make a OR b Equal to c - Problem

Bit Manipulation Medium

Given three positive numbers a, b, and c, return the minimum number of bit flips required to make (a OR b) == c.

A flip operation consists of changing any single bit from 1 to 0 or from 0 to 1 in the binary representation of a number.

The bitwise OR operation returns 1 if at least one of the corresponding bits is 1, otherwise 0.

Input & Output

Example 1 — Basic Case

$ Input: a = 2, b = 7, c = 5

› Output: 2

💡 Note: Binary: a=010, b=111, c=101. Current a|b=111, target=101. Need to flip bits 1 in both a and b to make them 0.

Example 2 — No Flips Needed

$ Input: a = 4, b = 2, c = 6

› Output: 0

💡 Note: Binary: a=100, b=010, c=110. Current a|b=110 equals target c=110, so no flips needed.

Example 3 — Need More 1s

$ Input: a = 1, b = 2, c = 7

› Output: 1

💡 Note: Binary: a=001, b=010, c=111. Current a|b=011, need 111. Must flip bit 2 in either a or b to 1.

Constraints

1 ≤ a, b, c ≤ 10⁹

Visualization

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Asked in

f Facebook 15 M Microsoft 12 a Amazon 8

The key insight is to process each bit position and apply OR logic rules: when target bit is 1, ensure at least one input is 1; when target bit is 0, both inputs must be 0. Best approach uses bit manipulation to count flips efficiently. Time: O(log n), Space: O(1)

Common Approaches

✓ Bit-by-Bit Comparison

⏱️ Time: O(log n) Space: O(1)

Iterate through each bit position, check the current OR result against the target bit, and count how many flips are needed to make them match.

Optimized Bit Manipulation

⏱️ Time: O(log n) Space: O(1)

Calculate the XOR between (a|b) and c to find all mismatched bit positions, then count the required flips based on the bit patterns.

Bit-by-Bit Comparison — Algorithm Steps

Step 1: Extract bits at each position for a, b, and c
Step 2: Calculate current OR result
Step 3: Count flips needed based on target bit

Visualization

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Step-by-Step Walkthrough

Extract Bits

Get the least significant bit of each number

Check OR Result

Calculate a|b and compare with target c bit

Count Flips

Add required flips based on the difference

Code -

solution.c — C

#include <stdio.h>

int solution(int a, int b, int c) {
    int flips = 0;
    
    // Check each bit position
    while (a > 0 || b > 0 || c > 0) {
        int bitA = a & 1;
        int bitB = b & 1;
        int bitC = c & 1;
        
        int currentOr = bitA | bitB;
        
        if (currentOr != bitC) {
            if (bitC == 1) {
                // Need at least one of a or b to be 1
                flips += 1;
            } else {
                // Need both a and b to be 0
                flips += bitA + bitB;
            }
        }
        
        a >>= 1;
        b >>= 1;
        c >>= 1;
    }
    
    return flips;
}

int main() {
    int a, b, c;
    scanf("%d", &a);
    scanf("%d", &b);
    scanf("%d", &c);
    
    int result = solution(a, b, c);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

We iterate through at most 32 bits (log of maximum integer value)

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra variables for counting

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Bit-by-Bit Comparison — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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