Number of Common Factors

Number of Common Factors - Problem

Given two positive integers a and b, return the number of common factors of a and b.

An integer x is a common factor of a and b if x divides both a and b.

Input & Output

Example 1 — Basic Case

$ Input: a = 12, b = 18

› Output: 4

💡 Note: Common factors are 1, 2, 3, and 6. Both 12 and 18 are divisible by each of these numbers.

Example 2 — Small Numbers

$ Input: a = 25, b = 30

› Output: 2

💡 Note: Common factors are 1 and 5. GCD(25,30) = 5, and 5 has factors 1 and 5.

Example 3 — Coprime Numbers

$ Input: a = 7, b = 13

› Output: 1

💡 Note: 7 and 13 are prime numbers with no common factors except 1.

Constraints

1 ≤ a, b ≤ 1000

Visualization

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Asked in

G Google 15 M Microsoft 12

The key insight is that common factors of two numbers are exactly the factors of their GCD. Best approach uses GCD + square root optimization to count factors efficiently. Time: O(log(min(a,b)) + √gcd), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(min(a, b)) Space: O(1)

Iterate through all possible divisors from 1 to the minimum of the two numbers. For each number, check if it divides both a and b evenly (remainder is 0). Count all such common divisors.

GCD + Square Root Optimization

⏱️ Time: O(log(min(a,b)) + sqrt(gcd)) Space: O(1)

First compute the GCD of both numbers, then count factors of the GCD. Since common factors of two numbers are exactly the factors of their GCD, we only need to check up to the square root of GCD and count pairs of factors.

Brute Force — Algorithm Steps

Iterate i from 1 to min(a, b)
Check if i divides both a and b
If yes, increment counter

Visualization

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Step-by-Step Walkthrough

Start

Begin checking from i = 1

Check

Test if i divides both numbers

Count

Increment counter if it's a common factor

Code -

solution.c — C

#include <stdio.h>

int solution(int a, int b) {
    int count = 0;
    int minVal = (a < b) ? a : b;
    for (int i = 1; i <= minVal; i++) {
        if (a % i == 0 && b % i == 0) {
            count++;
        }
    }
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(min(a, b))

Single loop iterates through all numbers from 1 to min(a,b)

✓ Linear Growth

Space Complexity

O(1)

Only uses a counter variable, no extra data structures

✓ Linear Space

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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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