Count number of trailing zeros in product of array

• Difficulty Level : Easy
• Last Updated : 28 Apr, 2021

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 using namespace std; // Returns count of zeros in product of arrayint 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 Programint main(){    int a[] = { 10, 100, 20, 30, 50, 90, 12, 80 };    int n = sizeof(a) / sizeof(a);    cout << countZeros(a, n);    return 0;}

Java

 // Java program for count total// zero in product of arrayimport 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

Python3

 # Python 3 program for count# total zero in product of array # Returns count of zeros# in product of arraydef 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 Programa = [ 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 arrayusing System; 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()    {        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



Javascript



Output:

9

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