Factorial Trailing Zeroes

Factorial Trailing Zeroes - Problem

Math Medium

Given an integer n, return the number of trailing zeroes in n!.

Note that n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1.

Follow up: Could you write a solution that works in logarithmic time complexity?

Input & Output

Example 1 — Small Factorial

$ Input: n = 3

› Output: 0

💡 Note: 3! = 6, which has no trailing zeros

Example 2 — One Trailing Zero

$ Input: n = 5

› Output: 1

💡 Note: 5! = 120, which has 1 trailing zero (from 2×5=10)

Example 3 — Multiple Zeros

$ Input: n = 25

› Output: 6

💡 Note: 25! has 6 trailing zeros from factors: 5 multiples of 5 plus 1 extra from 25=5×5

Constraints

0 ≤ n ≤ 10⁴

Visualization

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Asked in

M Microsoft 35 a Amazon 28 🍎 Apple 22 G Google 18

The key insight is that trailing zeros come from factors of 10, which are created by pairs of 2 and 5. Since there are always more factors of 2 than 5 in n!, we only need to count factors of 5. Best approach is the mathematical method counting multiples of 5, 25, 125, etc. Time: O(log n), Space: O(1)

Common Approaches

✓ Brute Force - Calculate Factorial

⏱️ Time: O(n) Space: O(1)

Compute the factorial of n and then count how many zeros appear at the end of the result. This approach works for small values but fails for large n due to integer overflow.

Mathematical Approach - Count Factors of 5

⏱️ Time: O(log n) Space: O(1)

Trailing zeros are created by factors of 10, which come from pairs of 2 and 5. Since there are always more factors of 2 than 5 in n!, we only need to count factors of 5. We count multiples of 5, 25, 125, etc.

Brute Force - Calculate Factorial — Algorithm Steps

Calculate n! = n × (n-1) × ... × 2 × 1
Convert result to string
Count zeros from the end until non-zero digit

Visualization

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Step-by-Step Walkthrough

Calculate

Multiply all numbers from 1 to n

Count

Count trailing zeros in result

Result

Return zero count

Code -

solution.c — C

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

int solution(int n) {
    if (n < 0) return 0;
    
    // Use the mathematical approach to avoid overflow
    // Brute force factorial calculation overflows for large n
    int count = 0;
    long long powerOf5 = 5;
    
    while (powerOf5 <= n) {
        count += n / powerOf5;
        powerOf5 *= 5;
    }
    
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n)

Need to multiply n numbers to calculate factorial

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Related Problems

Common Approaches

Brute Force - Calculate Factorial — Algorithm Steps

Visualization

Code -

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