Find the smallest number X such that X! contains at least Y trailing zeros.

Given an integer Y, find the smallest number X such that X! contains at least Y trailing zeros. 
Prerequisites – Count trailing zeroes in factorial of a number

Examples:

Input : Y = 2 
Output : 10 
10! = 3628800, which has 2 trailing zeros. 9! = 362880, which has 1 trailing zero. Hence, 10 is the correct answer.

Input : Y = 6 
Output : 25 
25! = 15511210043330985984000000, which has 6 trailing zeros. 24! = 620448401733239439360000, which has 4 trailing zeros. Hence, 25 is the correct answer. 

Approach: The problem can be easily solved by using Binary Search. The number of trailing zeros in N! is given by the count of the factors 5 in N!.Read this article for prerequisites. The countFactor(5, N) function returns the count of factor 5 in N! which is equal to count of trailing zeros in N!. The smallest number X such that X! contains at least Y trailing zeros can be computed quickly by using binary search on a range [0, 5 * Y] using this function.

Below is the implementation of above approach.  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the number
// of factors P in X!
int countFactor(int P, int X)
{
    if (X < P)
        return 0;

    return (X / P + countFactor(P, X / P));
}

// Function to find the smallest X such
// that X! contains Y trailing zeros
int findSmallestX(int Y)
{
    int low = 0, high = 5 * Y;
    int N = 0;
    while (low <= high) {
        int mid = (high + low) / 2;

        if (countFactor(5, mid) < Y) {
            low = mid + 1;
        }

        else {
            N = mid;
            high = mid - 1;
        }
    }

    return N;
}

// Driver code
int main()
{
    int Y = 10;

    cout << findSmallestX(Y);

    return 0;
}

Java

// Java implementation of the above approach
class GFG
{

    // Function to count the number
    // of factors P in X!
    static int countFactor(int P, int X)
    {
        if (X < P)
            return 0;

        return (X / P + countFactor(P, X / P));
    }

    // Function to find the smallest X such
    // that X! contains Y trailing zeros
    static int findSmallestX(int Y)
    {
        int low = 0, high = 5 * Y;
        int N = 0;
        while (low <= high)
        {
            int mid = (high + low) / 2;

            if (countFactor(5, mid) < Y)
            {
                low = mid + 1;
            }

            else
            {
                N = mid;
                high = mid - 1;
            }
        }

        return N;
    }

    // Driver code
    public static void main(String args[])
    {
        int Y = 10;

        System.out.println(findSmallestX(Y));
    }
}

// This code is contributed by Ryuga

Python3

# Python3 implementation of the approach

# Function to count the number
# of factors P in X!
def countFactor(P, X):
    if (X < P):
        return 0;

    return (X // P + countFactor(P, X // P));

# Function to find the smallest X such
# that X! contains Y trailing zeros
def findSmallestX(Y):

    low = 0;
    high = 5 * Y;
    N = 0;
    while (low <= high):
        mid = (high + low) // 2;

        if (countFactor(5, mid) < Y):
            low = mid + 1;

        else:
            N = mid;
            high = mid - 1;

    return N;

# Driver code
Y = 10;

print(findSmallestX(Y));

# This code is contributed by mits

C#

// C# implementation of the approach
class GFG
{

// Function to count the number
// of factors P in X!
static int countFactor(int P, int X)
{
    if (X < P)
        return 0;

    return (X / P + countFactor(P, X / P));
}

// Function to find the smallest X such
// that X! contains Y trailing zeros
static int findSmallestX(int Y)
{
    int low = 0, high = 5 * Y;
    int N = 0;
    while (low <= high)
    {
        int mid = (high + low) / 2;

        if (countFactor(5, mid) < Y)
        {
            low = mid + 1;
        }

        else
        {
            N = mid;
            high = mid - 1;
        }
    }

    return N;
}

// Driver code
static void Main()
{
    int Y = 10;

    System.Console.WriteLine(findSmallestX(Y));
}
}

// This code is contributed by mits

PHP

<?php
// PHP implementation of the above approach

// Function to count the number
// of factors P in X!
function countFactor($P, $X)
{
    if ($X < $P)
        return 0;

    return ((int)($X / $P) +
             countFactor($P, ($X / $P)));
}

// Function to find the smallest X such
// that X! contains Y trailing zeros
function findSmallestX($Y)
{
    $low = 0; $high = 5 * $Y;
    $N = 0;
    while ($low <= $high)
    {
        $mid = (int)(($high + $low) / 2);

        if (countFactor(5, $mid) < $Y)
        {
            $low = $mid + 1;
        }

        else
        {
            $N = $mid;
            $high = $mid - 1;
        }
    }

    return $N;
}

// Driver code
$Y = 10;

echo(findSmallestX($Y));

// This code is contributed by Code_Mech.
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to count the number
// of factors P in X!
function countFactor(P, X)
{
    if (X < P)
        return 0;

    return (parseInt(X / P) +
    countFactor(P, parseInt(X / P)));
}

// Function to find the smallest X such
// that X! contains Y trailing zeros
function findSmallestX(Y)
{
    let low = 0, high = 5 * Y;
    let N = 0;

    while (low <= high)
    {
        let mid = parseInt((high + low) / 2);

        if (countFactor(5, mid) < Y)
        {
            low = mid + 1;
        }
        else
        {
            N = mid;
            high = mid - 1;
        }
    }
    return N;
}

// Driver code
let Y = 10;

document.write(findSmallestX(Y));

// This code is contributed by subhammahato348

</script>

Output:

45

Time Complexity: O(log₂Y * log₅Y)

Space Complexity: O(log₅Y)

The extra space is used due to recursion call stack of countFactor() function.