# 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 rightmost 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:
Result of multiplication is 206976
So the rightmost digit is 6

Approach:

1. The question is too 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);`   `    ``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`

## Javascript

 ``

Output:

`6`

Time Complexity: O(N)
Here, N is the number of elements in the array. The rightmostNonZero() function iterates over the array once and takes constant time for each iteration.

Space Complexity: O(1)
No extra space is required.

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