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 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
# 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# 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 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.
?>
<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>
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.