Count trailing zeroes in factorial of a number
Last Updated : 22 Mar, 2025
Given an integer n, write a function that returns count of trailing zeroes in n!.
Examples :
Input: n = 5
Output: 1
Explanation: Factorial of 5 is 120 which has one trailing 0.
Input: n = 20
Output: 4
Explanation: Factorial of 20 is 2432902008176640000 which has 4 trailing zeroes.
Input: n = 100
Output: 24
[Naive Approach] By calculating the factorial - O(n) Time and O(1) Space
The idea is to calculate the factorial of the number and then count the number of trailing zeros by repeatedly dividing the factorial by 10 till the last digit is 0. The number of times the factorial is divided by 10 is the number of trailing zeros. This approach is not useful for large numbers as calculating their factorial will cause overflow.
C++ #include <iostream> using namespace std; // Function to calculate factorial of a number int factorial(int n) { int fact = 1; for (int i = 1; i <= n; ++i) { fact *= i; } return fact; } // Function to count trailing zeros in factorial int countTrailingZeros(int n) { int fact = factorial(n); int count = 0; // Count trailing zeros by dividing the factorial by 10 while (fact % 10 == 0) { count++; fact /= 10; } return count; } int main() { int n = 5; cout << countTrailingZeros(n) << endl; // Output only the count return 0; } public class Main { // Function to calculate factorial of a number public static int factorial(int n) { int fact = 1; for (int i = 1; i <= n; ++i) { fact *= i; } return fact; } // Function to count trailing zeros in factorial public static int countTrailingZeros(int n) { int fact = factorial(n); int count = 0; // Count trailing zeros by dividing the factorial by 10 while (fact % 10 == 0) { count++; fact /= 10; } return count; } public static void main(String[] args) { int n = 5; System.out.println(countTrailingZeros(n)); // Output only the count } } # Function to calculate factorial of a number def factorial(n): fact = 1 for i in range(1, n + 1): fact *= i return fact # Function to count trailing zeros in factorial def count_trailing_zeros(n): fact = factorial(n) count = 0 # Count trailing zeros by dividing the factorial by 10 while fact % 10 == 0: count += 1 fact //= 10 return count # Driver code n = 5 print(count_trailing_zeros(n)) # Output only the count
using System; class Program { // Function to calculate factorial of a number public static int Factorial(int n) { int fact = 1; for (int i = 1; i <= n; ++i) { fact *= i; } return fact; } // Function to count trailing zeros in factorial public static int CountTrailingZeros(int n) { int fact = Factorial(n); int count = 0; // Count trailing zeros by dividing the factorial by 10 while (fact % 10 == 0) { count++; fact /= 10; } return count; } static void Main() { int n = 5; Console.WriteLine(CountTrailingZeros(n)); // Output only the count } } // Function to calculate factorial of a number function factorial(n) { let fact = 1; for (let i = 1; i <= n; ++i) { fact *= i; } return fact; } // Function to count trailing zeros in factorial function countTrailingZeros(n) { let fact = factorial(n); let count = 0; // Count trailing zeros by dividing the factorial by 10 while (fact % 10 === 0) { count++; fact /= 10; } return count; } // Driver code const n = 5; console.log(countTrailingZeros(n)); // Output only the count [Expected Approach] Repeated division by 5 - O(log n base 5) Time and O(1) Space
The idea is to consider all prime factors of factorial n. A trailing zero is always produced by prime factors 2 and 5. If we can count the number of 5s and 2s in n!, our task is done.
How does this work? A trailing 0 is formed by multiplication of 5 and 2. So if we consider all prime factors of all numbers from 1 to n, there would be more 2s than 5s. So the number of 0s is limited by number of 5s. If we count number of 5s in prime factors, we get the result. Consider the following examples:
n = 5: There is one 5 and three 2s in prime factors of 5! (2 * 2 * 2 * 3 * 5). So a count of trailing 0s is min(1, 3) = 1.
n = 11: There are two 5s and eight 2s in prime factors of 11! (2 8 * 34 * 52 * 7). So the count of trailing 0s is min(2, 8) = 2.
We can observe that the number of 2s in prime factors is always more than or equal to the number of 5s. So, if we count 5s in prime factors, we are done.
How to count the total number of 5s in prime factors of n! ?
A simple way is to calculate floor(n/5). For example, 7! has one 5, 10! has two 5s. But, numbers like 25, 125, etc have more than 5 instead of floor (n / 5). For example, if we consider 28! we get one extra 5 and the number of 0s becomes 6. Handling this is simple, first, divide n by 5 and remove all single 5s, then divide by 25 to remove extra 5s, and so on.
Following is the summarized formula for counting trailing 0s:
Trailing 0s in n! = Count of 5s in prime factors of n! = floor(n/5) + floor(n/25) + floor(n/125) + ..
C++ // C++ program to count trailing 0s in n! #include <iostream> using namespace std; // Function to return trailing 0s in factorial of n int countTrailingZeros(int n) { // Negative Number Edge Case if (n < 0) return -1; // Initialize result int count = 0; // Keep dividing n by powers of // 5 and update count for (int i = 5; n / i >= 1; i *= 5) count += n / i; return count; } // Driver Code int main() { int n = 100; cout << countTrailingZeros(n); return 0; } // C program to count trailing 0s in n! #include <stdio.h> // Function to return trailing // 0s in factorial of n int countTrailingZeros(int n) { // Negative Number Edge Case if (n < 0) return -1; // Initialize result int count = 0; // Keep dividing n by powers of // 5 and update count for (int i = 5; n / i >= 1; i *= 5) count += n / i; return count; } // Driver Code int main() { int n = 100; printf("%d\n", countTrailingZeros(n)); return 0; } // Java program to count // trailing 0s in n! import java.io.*; class GFG { // Function to return trailing // 0s in factorial of n static int countTrailingZeros(int n) { if (n < 0) // Negative Number Edge Case return -1; // Initialize result int count = 0; // Keep dividing n by powers // of 5 and update count for (int i = 5; n / i >= 1; i *= 5) count += n / i; return count; } // Driver Code public static void main(String[] args) { int n = 100; System.out.println(countTrailingZeros(n)); } } # Python3 program to count trailing 0s in n! # Function to return trailing 0s in factorial of n def countTrailingZeros(n): # Negative Number Edge Case if (n < 0): return -1 # Initialize result count = 0 # Keep dividing n by # 5 & update Count while (n >= 5): n //= 5 count += n return count # Driver program n = 100 print(countTrailingZeros(n))
// C# program to count trailing 0s in n! using System; class GFG { // Function to return trailing // 0s in factorial of n static int countTrailingZeros(int n) { if (n < 0) // Negative Number Edge Case return -1; // Initialize result int count = 0; // Keep dividing n by powers // of 5 and update count for (int i = 5; n / i >= 1; i *= 5) count += n / i; return count; } // Driver Code public static void Main() { int n = 100; Console.WriteLine(countTrailingZeros(n)); } } // JavaScript program to count trailing 0s in n! // Function to return trailing // 0s in factorial of n function countTrailingZeros(n) { if (n < 0) // Negative Number Edge Case return -1; // Initialize result let count = 0; // Keep dividing n by powers of // 5 and update count for (let i = 5; Math.floor(n / i) >= 1; i *= 5) count += Math.floor(n / i); return count; } // Driver Code let n = 100; console.log(countTrailingZeros(n)); Similar Reads
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