## Problem Statement

Given a number **n, **find the number of trailing zeroes in **n!**.

**Sample Test Cases**

**Input 1:** n = 5**Output 1:** 1**Explanation 1:** 5! = 120 which has only 1 trailing zero.

### Confused about your next job?

**Input 2:** n = 100**Output 2:** 24**Explanation 2:** The number of trailing zeroes of 100! can be found to have 24 trailing zeroes.

## Naive Approach

The naive approach to solve this problem is to calculate the value of **n!** and then to find the number of trailing zeroes in it.

We can find the number of trailing zeroes in a number by repeatedly dividing it by 10 until its last digit becomes non-zero.

### C++ Implementation

int getTrailingZeroes(int n) { int factorial = 1; for (int i = 1; i <= n; i++) { factorial *= i; } int zeroes = 0; while (factorial % 10 == 0) { zeroes++; factorial /= 10; } return zeroes; }

### Java Implementation

public static int getTrailingZeroes(int n) { int factorial = 1; for (int i = 1; i <= n; i++) { factorial *= i; } int zeroes = 0; while (factorial % 10 == 0) { zeroes++; factorial /= 10; } return zeroes; }

### Python Implementation

def getTrailingZeroes(n): factorial = 1 zeroes = 0 for i in range(1, n + 1): factorial *= i while factorial % 10 == 0: factorial //= 10 zeroes += 1 return zeroes

### Complexity Analysis

**Time Complexity:**O(n)**Space Complexity:**O(1)

## Optimal Approach

Observe that number of trailing zeroes in a number will be the highest power of 10 present in the number.

Now we know that **10**^{n}** = 2**^{n}** * 5**** ^{n}**. So, the highest power of 10 will be equal to the minimum of the highest power of 2 and the highest power of 5 present in

**n!**. We can observe that the highest power of 2 is always going to be less than or equal to the highest power of 5 in any value of

**n!**. So, our problem boils down to finding the highest power of 5 in

**n!**.

The highest power of a prime number **p**, present in any factorial is given by a formula known as **Legendre’s Formula**,

Where **V**_{p}**(n!)** represents the highest exponent in **p**, such that **p**** ^{n}** divides

**n!**where

**p**is a prime number.

For the number of trailing zeroes, the above formula transforms to,

Where k = **floor of log _{5}n**.

### C++ Code

int getTrailingZeroes(int n) { int zeroes = 0; int power = 5; for (int i = 1;; i++) { zeroes += (n / power); if (n / power == 0) { break; } power *= 5; } return zeroes; }

### Java Code

public static int getTrailingZeroes(int n) { int zeroes = 0; int power = 5; for (int i = 1;; i++) { zeroes += (n / power); if (n / power == 0) { break; } power *= 5; } return zeroes; }

### Python Code

def getTrailingZeroes(n): zeroes = 0 power = 5 while (n // power) != 0: zeroes += n // power power *= 5 return zeroes

### Complexity Analysis

**Time Complexity:**O(log_{5}n)**Space Complexity:**O(1)

**Practice Problem**

## FAQs

**Q. What is the main drawback of the naive approach?**

A. The main drawback of the naive approach is that it calculates the factorial of **n **before computing the result, whose value can grow very large extremely quickly and can cause an overflow issue.

**Q. How is the time complexity of the optimal approach calculated?**

A. In the optimal approach, n is divided by 5 at each iteration till it reaches 0. So the number of iterations till which the algorithm will continue is equal to **floor(log _{5}n)**, which is our required time complexity.