# Count number of trailing zeros in product of array

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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); ` `    ``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. `

## 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` `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 `

## PHP

 ` `

Output:

``` 9
```

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