Count number of trailing zeros in product of array

Last Updated : 01 Aug, 2022

Given a array size of n, we need to find the total number of zeros in the product of array.

Examples:

Input  : a[] = {100, 20, 40, 25, 4} 
Output : 6
Product is 100 * 20 * 40 * 25 * 4
which is 8000000 and has 6 trailing 0s.

Input  : a[] = {10, 100, 20, 30, 25, 4, 
                43, 25, 50, 90, 12, 80}
Output : 13

A simple solution is simply multiply and count trailing 0s in product. This solution may cause integer overflow. A better solution is based on the fact that zeros are formed by a combination of 2 and 5. Hence the number of zeros will depend on the number of pairs of 2’s and 5’s that can be formed.
Ex.: 8 * 3 * 5 * 23 * 17 * 25 * 4 * 11
2³ * 3¹ * 5¹ * 23¹ * 17¹ * 5² * 2² * 11¹

In this example there are 5 twos and 3 fives. Hence, we shall be able to form only 3 pairs of (2*5). Hence will be 3 Zeros in the product.

Implementation:

C++

// CPP program for count total zero in product of array
#include <iostream>
using namespace std;
 
// Returns count of zeros in product of array
int countZeros(int a[], int n)
{
    int count2 = 0, count5 = 0;
    for (int i = 0; i < n; i++) {
 
        // count number of 2s in each element
        while (a[i] % 2 == 0) {
            a[i] = a[i] / 2;
            count2++;
        }
 
        // count number of 5s in each element
        while (a[i] % 5 == 0) {
            a[i] = a[i] / 5;
            count5++;
        }
    }
    // return the minimum
    return (count2 < count5) ? count2 : count5;
}
 
// Driven Program
int main()
{
    int a[] = { 10, 100, 20, 30, 50, 90, 12, 80 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << countZeros(a, n);
    return 0;
}

Java

// Java program for count total
// zero in product of array
import java.util.*;
import java.lang.*;
 
public class GfG
{
    // Returns count of zeros in product of array
    public static int countZeros(int[] a, int n)
    {
        int count2 = 0, count5 = 0;
        for (int i = 0; i < n; i++) 
        {
 
            // count number of 2s 
            // in each element
            while (a[i] % 2 == 0) 
            {
                a[i] = a[i] / 2;
                count2++;
            }
 
            // count number of 5s 
            // in each element
            while (a[i] % 5 == 0) 
            {
                a[i] = a[i] / 5;
                count5++;
            }
        }
        // return the minimum
        return (count2 < count5) ? count2 : count5;
    }
     
    // Driver function 
    public static void main(String argc[])
    {
        int[] a = new int[]{ 10, 100, 20, 30, 
                            50, 91, 12, 80 };
        int n = 8;
        System.out.println(countZeroso(a, n));
    }
     
}
 
// This code is contributed 
// by Sagar Shukla 

Python3

# Python 3 program for count 
# total zero in product of array
 
# Returns count of zeros 
# in product of array
def countZeros(a, n) :
    count2 = 0
    count5 = 0
    for i in range(0, n) :
         
        # count number of 2s 
        # in each element
        while (a[i] % 2 == 0) :
            a[i] = a[i] // 2
            count2 = count2 + 1
         
         
        # count number of 5s 
        # in each element
        while (a[i] % 5 == 0) :
            a[i] = a[i] // 5
            count5 = count5 + 1
         
         
    # return the minimum
    if(count2 < count5) :
        return count2
    else : 
        return count5
         
 
# Driven Program
a = [ 10, 100, 20, 30, 50, 90, 12, 80 ]
n = len(a)
print(countZeros(a, n))
 
# This code is contributed 
# by Nikita Tiwari.

C#

// C# program for count total
// zero in product of array
using System;
 
public class GfG
{
    // Returns count of zeros in product of array
    public static int countZeros(int[] a, int n)
    {
        int count2 = 0, count5 = 0;
        for (int i = 0; i < n; i++) 
        {
 
            // count number of 2s 
            // in each element
            while (a[i] % 2 == 0) 
            {
                a[i] = a[i] / 2;
                count2++;
            }
 
            // count number of 5s 
            // in each element
            while (a[i] % 5 == 0) 
            {
                a[i] = a[i] / 5;
                count5++;
            }
        }
        // return the minimum
        return (count2 < count5) ? count2 : count5;
    }
     
    // Driver function 
    public static void Main()
    {
        int[] a = new int[]{ 10, 100, 20, 30, 
                            50, 91, 12, 80 };
        int n = 8;
        Console.WriteLine(countZeroso(a, n));
    }
     
}
 
// This code is contributed 
// by vt_m

PHP

<?php
// PHP program for count total
// zero in product of array
 
function countZeros($a, $n)
{
     
    $count2 = 0; $count5 = 0;
    for ($i = 0; $i < $n; $i++) 
    {
 
        // count number of 2s 
        // in each element
        while ($a[$i] % 2 == 0)
        {
            $a[$i] = $a[$i] / 2;
            $count2++;
        }
 
        // count number of 5s
        // in each element
        while ($a[$i] % 5 == 0)
        {
            $a[$i] = $a[$i] / 5;
            $count5++;
        }
    }
     
    // return the minimum
    return ($count2 < $count5) ? $count2 : $count5;
}
 
// Driver Code
$a = array(10, 100, 20, 30, 50, 90, 12, 80);
$n = sizeof($a);
echo(countZeros($a, $n));
 
// This code is contributed by Ajit.
?>

Javascript

<script>
 
// Javascript program for count total
// zero in product of array
 
// Returns count of zeros in product of array
function countZeros(a, n)
{
    let count2 = 0, count5 = 0;
    for(let i = 0; i < n; i++)
    {
         
        // Count number of 2s in each element
        while (a[i] % 2 == 0) 
        {
            a[i] = parseInt(a[i] / 2);
            count2++;
        }
 
        // Count number of 5s in each element
        while (a[i] % 5 == 0) 
        {
            a[i] = parseInt(a[i] / 5);
            count5++;
        }
    }
     
    // Return the minimum
    return (count2 < count5) ? count2 : count5;
}
 
// Driver code
let a = [ 10, 100, 20, 30, 50, 90, 12, 80 ];
let n = a.length;
 
document.write(countZeros(a, n));
 
// This code is contributed by souravmahato348
 
</script>

Output

Time Complexity: O(n * (log₂m + log₅m)), where n is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Python program for removing i-th character from a string

jaingyayak

Improve

Article Tags :

Practice Tags :

Count number of trailing zeros in product of array

C++

Java

Python3

C#

PHP

Javascript

Similar Reads

Thank You!

What kind of Experience do you want to share?