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++ 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 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 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# 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 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.
?>

<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>

45

Time Complexity: O(log₂Y * log₅Y)

Space Complexity: O(log₅Y)

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