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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 zerosint 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 codeint 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 zerosstatic 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 codepublic 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 zerosdef 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 codeif __name__ == '__main__':    N = 12     print(trailing_zeros(N)) # This code is contributed by# Sanjit_Prasad

## C#

 // C# implementation of the above approachusing 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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