Find the number of consecutive zero at the end after multiplying n numbers

Given an array with n numbers. The task is to print number of consecutive zero’s at the end after multiplying all the n number.

Examples :

Input : arr[] = {100, 10}
Output : 3
Explanation : 100 x 10 = 1000,
3 zero's at the end.

Input : arr[] = {100, 10, 5, 25, 35, 14}
Output : 4
Explanation :
100 x 10 x 5 x 25 x 35 x 14 = 61250000,
4 zero's at the end

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach :
First multiple all the number and save into the string(because if the multiplicative number is big as 2^64 then it give wrong, so store every multiple in string) and then count the number of zero’s.

Efficient approach :
First count the 2’s factor of n numbers and then count the 5’s factor of all n number then print the smallest one.

For Example –

n_number's  | 2's factor | 5's factor 
100         |     2      |    2
10          |     1      |    1
5           |     0      |    1
25          |     0      |    2
35          |     0      |    1 
14          |     1      |    0
Total       |     4      |    7

we can take a minimum so there number of 
zero's is 4

Below is the implementation of above approach :

C++

// CPP program to find the number of consecutive zero 
// at the end after multiplying n numbers 
#include<iostream> 
using namespace std; 
  
// Function to count two's factor 
int two_factor(int n)  
{     
    // Count number of 2s present in n 
    int twocount = 0; 
    while (n % 2 == 0)  
    { 
        twocount++;  
        n = n / 2;  
    } 
    return twocount; 
} 
  
// Function to count five's factor 
int five_factor(int n)  
{ 
    int fivecount = 0; 
    while (n % 5 == 0)  
    { 
        fivecount++;  
        n = n / 5;  
    } 
    return fivecount; 
} 
  
// Function to count number of zeros 
int find_con_zero(int arr[], int n) 
{ 
    int twocount = 0; 
    int fivecount = 0; 
    for (int i = 0; i < n; i++) { 
  
        // Count the two's factor of n number 
        twocount += two_factor(arr[i]); 
  
        // Count the five's factor of n number 
        fivecount += five_factor(arr[i]); 
    } 
  
    // Return the minimum 
    if (twocount < fivecount)  
        return twocount;  
    else 
        return fivecount; 
} 
  
// Driver Code 
int main() 
{ 
    int arr[] = { 100, 10, 5, 25, 35, 14 }; 
    int n = 6; 
    cout << find_con_zero(arr, n); 
} 

Java

// Java program to find the number  
// of consecutive zero at the end 
// after multiplying n numbers 
  
public class GfG{ 
  
    // Function to count two's factor 
    static int two_factor(int n)  
    {      
        // Count number of 2s 
        // present in n 
        int twocount = 0; 
        while (n % 2 == 0)  
        { 
            twocount++;  
            n = n / 2;  
        } 
        return twocount; 
    } 
  
    // Function to count five's 
    // factor 
    static int five_factor(int n)  
    { 
        int fivecount = 0; 
        while (n % 5 == 0)  
        { 
            fivecount++;  
            n = n / 5;  
        } 
        return fivecount; 
    } 
  
    // Function to count number of zeros 
    static int find_con_zero(int arr[], int n) 
    { 
        int twocount = 0; 
        int fivecount = 0;   
          
        for (int i = 0; i < n; i++) {    
  
            // Count the two's factor 
            // of n number 
            twocount += two_factor(arr[i]);  
  
            // Count the five's factor  
            // of n number 
            fivecount += five_factor(arr[i]); 
        } 
  
        // Return the minimum 
        if (twocount < fivecount)  
            return twocount;  
        else
            return fivecount; 
    } 
      
    // driver function 
    public static void main(String s[]) 
    { 
        int arr[] = { 100, 10, 5, 25, 35, 14 }; 
        int n = 6; 
        System.out.println(find_con_zero(arr, n));      
    } 
} 
  
// This code is contributed by Gitanjali 

Python3

# Python3 code to find the number of consecutive zero 
# at the end after multiplying n numbers 
  
# Function to count two's factor 
def two_factor( n ): 
  
    # Count number of 2s present in n 
    twocount = 0
    while n % 2 == 0: 
        twocount+=1
        n =int( n / 2) 
    return twocount 
      
# Function to count five's factor 
def five_factor( n ): 
    fivecount = 0
    while n % 5 == 0: 
        fivecount+=1
        n = int(n / 5) 
    return fivecount 
  
# Function to count number of zeros 
def find_con_zero( arr, n ): 
    twocount = 0
    fivecount = 0
    for i in range(n): 
  
        # Count the two's factor of n number 
        twocount += two_factor(arr[i]) 
          
        # Count the five's factor of n number 
        fivecount += five_factor(arr[i]) 
      
    # Return the minimum 
    if twocount < fivecount: 
        return twocount 
    else: 
        return fivecount 
  
# Driver Code 
arr = [ 100, 10, 5, 25, 35, 14 ] 
n = 6
print(find_con_zero(arr, n)) 
  
# This code is contributed by "Sharad_Bhardwaj". 

C#

// C# program to find the number  
// of consecutive zero at the end 
// after multiplying n numbers 
using System; 
  
public class GfG { 
  
    // Function to count two's factor 
    static int two_factor(int n)  
    {  
          
        // Count number of 2s 
        // present in n 
        int twocount = 0; 
          
        while (n % 2 == 0)  
        { 
            twocount++;  
            n = n / 2;  
        } 
          
        return twocount; 
    } 
  
    // Function to count five's 
    // factor 
    static int five_factor(int n)  
    { 
        int fivecount = 0; 
          
        while (n % 5 == 0)  
        { 
            fivecount++;  
            n = n / 5;  
        } 
          
        return fivecount; 
    } 
  
    // Function to count number of zeros 
    static int find_con_zero(int []arr, int n) 
    { 
        int twocount = 0; 
        int fivecount = 0;  
          
        for (int i = 0; i < n; i++) {  
  
            // Count the two's factor 
            // of n number 
            twocount += two_factor(arr[i]);  
  
            // Count the five's factor  
            // of n number 
            fivecount += five_factor(arr[i]); 
        } 
  
        // Return the minimum 
        if (twocount < fivecount)  
            return twocount;  
        else
            return fivecount; 
    } 
      
    // driver function 
    public static void Main() 
    { 
          
        int []arr = { 100, 10, 5, 25, 35, 14 }; 
        int n = 6; 
          
        Console.WriteLine(find_con_zero(arr, n));  
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to find the number  
// of consecutive zero at the end  
// after multiplying n numbers 
  
// Function to count two's factor 
function two_factor($n)  
{  
    // Count number of  
    // 2s present in n 
    $twocount = 0; 
    while ($n % 2 == 0)  
    { 
        $twocount++;  
        $n = (int)($n / 2);  
    } 
    return $twocount; 
} 
  
// Function to count  
// five's factor 
function five_factor($n)  
{ 
    $fivecount = 0; 
    while ($n % 5 == 0)  
    { 
        $fivecount++;  
        $n = (int)($n / 5);  
    } 
    return $fivecount; 
} 
  
// Function to count 
// number of zeros 
function find_con_zero($arr, $n) 
{ 
    $twocount = 0; 
    $fivecount = 0; 
    for ($i = 0; $i < $n; $i++)  
    { 
  
        // Count the two's  
        // factor of n number 
        $twocount += two_factor($arr[$i]); 
  
        // Count the five's  
        // factor of n number 
        $fivecount += five_factor($arr[$i]); 
    } 
  
    // Return the minimum 
    if ($twocount < $fivecount)  
        return $twocount;  
    else
        return $fivecount; 
} 
  
// Driver Code 
$arr= array(100, 10, 5, 
            25, 35, 14); 
$n = 6; 
echo find_con_zero($arr, $n); 
  
// This code is contributed by mits 
?> 

Output :

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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