Skip to content
Related Articles
Right most non-zero digit in multiplication of array elements
• Difficulty Level : Medium
• Last Updated : 04 May, 2021

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`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live

My Personal Notes arrow_drop_up