Count number of trailing zeros in product of array

2.6

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.

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 << countZeroso(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 countZeroso(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 

Python 3

# 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.


Output:

 9

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