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:
// 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> |
9
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.