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# Count number of trailing zeros in (1^1)*(2^2)*(3^3)*(4^4)*..

Given an integer n, the task is to find the number of trailing zeros in the function i.e. f(n) = 11 * 22 * 33 * … * nn.

Examples:

Input: n = 5
Output:
f(5) = 11 * 22 * 33 * 44 * 55 = 1 * 4 * 27 * 256 * 3125 = 86400000

Input: n = 12
Output: 15

Approach: We know that 5 * 2 = 10 i.e. 1 trailing zero is the result of the multiplication of a single 5 and a single 2. So, if we have x number of 5 and y number of 2 then the number of trailing zeros will be min(x, y)
Now, for every number i in the series, we need to count the number of 2 and 5 in its factors say x and y but the number of 2s and 5s will be x * i and y * i respectively because in the series i is raised to the power itself i.e. ii. Count the number of 2s and 5s in the complete series and print the minimum of them which is the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the number of``// trailing zeros``int` `trailing_zeros(``int` `N)``{` `    ``// To store the number of 2s and 5s``    ``int` `count_of_two = 0, count_of_five = 0;` `    ``for` `(``int` `i = 1; i <= N; i++) {` `        ``int` `val = i;` `        ``while` `(val % 2 == 0 && val > 0) {``            ``val /= 2;` `            ``// If we get a factor 2 then we``            ``// have i number of 2s because``            ``// the power of the number is``            ``// raised to i``            ``count_of_two += i;``        ``}` `        ``while` `(val % 5 == 0 && val > 0) {``            ``val /= 5;` `            ``// If we get a factor 5 then``            ``// we have i number of 5s``            ``// because the power of the``            ``// number is raised to i``            ``count_of_five += i;``        ``}``    ``}` `    ``// Take the minimum of them``    ``int` `ans = min(count_of_two, count_of_five);` `    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `N = 12;` `    ``cout << trailing_zeros(N);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach` `class` `GFG``{``    ` `// Function to return the number of``// trailing zeros``static` `int` `trailing_zeros(``int` `N)``{` `    ``// To store the number of 2s and 5s``    ``int` `count_of_two = ``0``, count_of_five = ``0``;` `    ``for` `(``int` `i = ``1``; i <= N; i++)``    ``{``        ``int` `val = i;``        ``while` `(val % ``2` `== ``0` `&& val > ``0``)``        ``{``            ``val /= ``2``;` `            ``// If we get a factor 2 then we``            ``// have i number of 2s because``            ``// the power of the number is``            ``// raised to i``            ``count_of_two += i;``        ``}` `        ``while` `(val % ``5` `== ``0` `&& val > ``0``)``        ``{``            ``val /= ``5``;` `            ``// If we get a factor 5 then``            ``// we have i number of 5s``            ``// because the power of the``            ``// number is raised to i``            ``count_of_five += i;``        ``}``    ``}` `    ``// Take the minimum of them``    ``int` `ans = Math.min(count_of_two, count_of_five);` `    ``return` `ans;``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `N = ``12``;``    ``System.out.println(trailing_zeros(N));``}``}` `// This code is contributed by chandan_jnu`

## Python3

 `# Python 3 implementation of the approach` `# Function to return the number of``# trailing zeros``def` `trailing_zeros(N):``    ` `    ``# To store the number of 2s and 5s``    ``count_of_two ``=` `0``    ``count_of_five ``=` `0` `    ``for` `i ``in` `range``(``1``, N ``+` `1``, ``1``):``        ``val ``=` `i` `        ``while` `(val ``%` `2` `=``=` `0` `and` `val > ``0``):``            ``val ``/``=` `2` `            ``# If we get a factor 2 then we``            ``# have i number of 2s because``            ``# the power of the number is``            ``# raised to i``            ``count_of_two ``+``=` `i` `        ``while` `(val ``%` `5` `=``=` `0` `and` `val > ``0``):``            ``val ``/``=` `5` `            ``# If we get a factor 5 then we``            ``# have i number of 5s because``            ``# the power of the number is``            ``# raised to i``            ``count_of_five ``+``=` `i``    ` `    ``# Take the minimum of them``    ``ans ``=` `min``(count_of_two, count_of_five)` `    ``return` `ans` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``N ``=` `12` `    ``print``(trailing_zeros(N))` `# This code is contributed by``# Sanjit_Prasad`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG``{``    ` `    ``// Function to return the number of``    ``// trailing zeros``    ``static` `int` `trailing_zeros(``int` `N)``    ``{``    ` `        ``// To store the number of 2s and 5s``        ``int` `count_of_two = 0, count_of_five = 0;``    ` `        ``for` `(``int` `i = 1; i <= N; i++)``        ``{``            ``int` `val = i;``            ``while` `(val % 2 == 0 && val > 0)``            ``{``                ``val /= 2;``    ` `                ``// If we get a factor 2 then we``                ``// have i number of 2s because``                ``// the power of the number is``                ``// raised to i``                ``count_of_two += i;``            ``}``    ` `            ``while` `(val % 5 == 0 && val > 0)``            ``{``                ``val /= 5;``    ` `                ``// If we get a factor 5 then``                ``// we have i number of 5s``                ``// because the power of the``                ``// number is raised to i``                ``count_of_five += i;``            ``}``        ``}``    ` `        ``// Take the minimum of them``        ``int` `ans = Math.Min(count_of_two, count_of_five);``    ` `        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `N = 12;``        ``Console.WriteLine(trailing_zeros(N));``    ``}``}` `// This code is contributed by Ryuga`

## PHP

 ` 0)``        ``{``            ``\$val` `/= 2;` `            ``// If we get a factor 2 then we``            ``// have i number of 2s because``            ``// the power of the number is``            ``// raised to i``            ``\$count_of_two` `+= ``\$i``;``        ``}` `        ``while` `(``\$val` `% 5 == 0 && ``\$val` `> 0)``        ``{``            ``\$val` `/= 5;` `            ``// If we get a factor 5 then``            ``// we have i number of 5s``            ``// because the power of the``            ``// number is raised to i``            ``\$count_of_five` `+= ``\$i``;``        ``}``    ``}` `    ``// Take the minimum of them``    ``\$ans` `= min(``\$count_of_two``, ``\$count_of_five``);` `    ``return` `\$ans``;``}` `// Driver code``\$N` `= 12;``echo` `trailing_zeros(``\$N``);` `// This code is contributed by ita_c``?>`

## Javascript

 ``

Output:

`15`

Time Complexity: O(N * (log2N + log5N))

Auxiliary Space: O(1), since no extra space has been taken.

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