# Right most non-zero digit in multiplication of array elements

Given an array arr[] of N non-negative integers. The task is to find the right most non zero digit in the product of array elements.

Examples:

Input: arr[] = {3, 5, 6, 90909009}
Output: 7

Input: arr[] = {7, 42, 11, 64}
Output: 6
Result of multiplication is 206976
So the rightmost digit is 6

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

Approach:

1. The question is too much simple if you know basic maths. It is given that you have to find the rightmost positive digit. Now a digit is made multiple of 10, if there are 2 and 5. They produce a number with last digit 0.
2. Now what we can do is divide each array element into its shortest divisible form by 5 and increase count of such occurrences.
3. Now divide each array element into its shortest divisible form by 2 and decrease count of such occurrences. This way we are not considering the multiplication of 2 and a 5 in our multiplication.
4. Set the multiplier value as either 1 or 5 in case count of 5 is not 0 after above two loops.
5. Multiply each array variable now and store just last digit by taking remainder by 10

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the rightmost non-zero ` `// digit in the multiplication ` `// of the array elements ` `int` `rightmostNonZero(``int` `a[], ``int` `n) ` `{ ` `    ``// To store the count of times 5 can ` `    ``// divide the array elements ` `    ``int` `c5 = 0; ` ` `  `    ``// Divide the array elements by 5 ` `    ``// as much as possible ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``while` `(a[i] > 0 && a[i] % 5 == 0) { ` `            ``a[i] /= 5; ` `            ``// increase count of 5 ` `            ``c5++; ` `        ``} ` `    ``} ` ` `  `    ``// Divide the array elements by ` `    ``// 2 as much as possible ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``while` `(c5 && a[i] > 0 && !(a[i] & 1)) { ` `            ``a[i] >>= 1; ` ` `  `            ``// Decrease count of 5, because a '2' and ` `            ``// a '5' makes a number with last digit '0' ` `            ``c5--; ` `        ``} ` `    ``} ` `    ``long` `long` `ans = 1; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``ans = (ans * a[i] % 10) % 10; ` `    ``} ` ` `  `    ``// If c5 is more than the multiplier ` `    ``// should be taken as 5 ` `    ``if` `(c5) ` `        ``ans = (ans * 5) % 10; ` ` `  `    ``if` `(ans) ` `        ``return` `ans; ` ` `  `    ``return` `-1; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 7, 42, 11, 64 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` ` `  `    ``cout << rightmostNonZero(a, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the rightmost non-zero ` `// digit in the multiplication ` `// of the array elements ` `static` `int` `rightmostNonZero(``int` `a[], ``int` `n) ` `{ ` `    ``// To store the count of times 5 can ` `    ``// divide the array elements ` `    ``int` `c5 = ``0``; ` ` `  `    ``// Divide the array elements by 5 ` `    ``// as much as possible ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``while` `(a[i] > ``0` `&& a[i] % ``5` `== ``0``) ` `        ``{ ` `            ``a[i] /= ``5``; ` `             `  `            ``// increase count of 5 ` `            ``c5++; ` `        ``} ` `    ``} ` ` `  `    ``// Divide the array elements by ` `    ``// 2 as much as possible ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``while` `(c5 != ``0` `&& a[i] > ``0` `&&  ` `                         ``(a[i] & ``1``) == ``0``)  ` `        ``{ ` `            ``a[i] >>= ``1``; ` ` `  `            ``// Decrease count of 5, because a '2' and ` `            ``// a '5' makes a number with last digit '0' ` `            ``c5--; ` `        ``} ` `    ``} ` `     `  `    ``int` `ans = ``1``; ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``ans = (ans * a[i] % ``10``) % ``10``; ` `    ``} ` ` `  `    ``// If c5 is more than the multiplier ` `    ``// should be taken as 5 ` `    ``if` `(c5 != ``0``) ` `        ``ans = (ans * ``5``) % ``10``; ` ` `  `    ``if` `(ans != ``0``) ` `        ``return` `ans; ` ` `  `    ``return` `-``1``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `a[] = { ``7``, ``42``, ``11``, ``64` `}; ` `    ``int` `n = a.length; ` ` `  `    ``System.out.println(rightmostNonZero(a, n)); ` `} ` `} ` ` `  `// This code is contributed by ` `// Surendra_Gangwar `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the rightmost non-zero ` `# digit in the multiplication ` `# of the array elements ` `def` `rightmostNonZero(a, n): ` `     `  `    ``# To store the count of times 5 can ` `    ``# divide the array elements ` `    ``c5 ``=` `0` ` `  `    ``# Divide the array elements by 5 ` `    ``# as much as possible ` `    ``for` `i ``in` `range``(n): ` `        ``while` `(a[i] > ``0` `and` `a[i] ``%` `5` `=``=` `0``): ` `            ``a[i] ``/``/``=` `5` `             `  `            ``# increase count of 5 ` `            ``c5 ``+``=` `1` ` `  `    ``# Divide the array elements by ` `    ``# 2 as much as possible ` `    ``for` `i ``in` `range``(n): ` `        ``while` `(c5 ``and` `a[i] > ``0` `and` `(a[i] & ``1``) ``=``=` `0``): ` `            ``a[i] >>``=` `1` ` `  `            ``# Decrease count of 5, because a '2' and ` `            ``# a '5' makes a number with last digit '0' ` `            ``c5 ``-``=` `1` ` `  `    ``ans ``=` `1` `    ``for` `i ``in` `range``(n): ` `        ``ans ``=` `(ans ``*` `a[i] ``%` `10``) ``%` `10` ` `  `    ``# If c5 is more than the multiplier ` `    ``# should be taken as 5 ` `    ``if` `(c5): ` `        ``ans ``=` `(ans ``*` `5``) ``%` `10` ` `  `    ``if` `(ans): ` `        ``return` `ans ` ` `  `    ``return` `-``1` ` `  `# Driver code ` `a ``=` `[``7``, ``42``, ``11``, ``64``] ` `n ``=` `len``(a) ` ` `  `print``(rightmostNonZero(a, n)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the rightmost non-zero ` `// digit in the multiplication ` `// of the array elements ` `static` `int` `rightmostNonZero(``int``[] a, ``int` `n) ` `{ ` `     `  `    ``// To store the count of times 5 can ` `    ``// divide the array elements ` `    ``int` `c5 = 0; ` ` `  `    ``// Divide the array elements by 5 ` `    ``// as much as possible ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``while` `(a[i] > 0 && a[i] % 5 == 0) ` `        ``{ ` `            ``a[i] /= 5; ` `             `  `            ``// increase count of 5 ` `            ``c5++; ` `        ``} ` `    ``} ` ` `  `    ``// Divide the array elements by ` `    ``// 2 as much as possible ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``while` `(c5 != 0 && a[i] > 0 &&  ` `                         ``(a[i] & 1) == 0)  ` `        ``{ ` `            ``a[i] >>= 1; ` ` `  `            ``// Decrease count of 5, because a '2' and ` `            ``// a '5' makes a number with last digit '0' ` `            ``c5--; ` `        ``} ` `    ``} ` `     `  `    ``int` `ans = 1; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``ans = (ans * a[i] % 10) % 10; ` `    ``} ` ` `  `    ``// If c5 is more than the multiplier ` `    ``// should be taken as 5 ` `    ``if` `(c5 != 0) ` `        ``ans = (ans * 5) % 10; ` ` `  `    ``if` `(ans != 0) ` `        ``return` `ans; ` ` `  `    ``return` `-1; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int``[] a = { 7, 42, 11, 64 }; ` `    ``int` `n = a.Length; ` ` `  `    ``Console.WriteLine(rightmostNonZero(a, n)); ` `} ` `} ` ` `  `// This code is contributed by ` `// Code_@Mech `

Output:

```6
```

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