Right most non-zero digit in multiplication of array elements

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

Approach:

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.
Now what we can do is divide each array element into its shortest divisible form by 5 and increase count of such occurrences.
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.
Set the multiplier value as either 1 or 5 in case count of 5 is not 0 after above two loops.
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 <bits/stdc++.h>
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

Javascript

<script>
 
// Javascript implementation of the approach
 
    // Function to return the rightmost non-zero
    // digit in the multiplication
    // of the array elements
    function rightmostNonZero(a , n) 
    {
        // To store the count of times 5 can
        // divide the array elements
        var c5 = 0;
 
        // Divide the array elements by 5
        // as much as possible
        for (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 (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--;
            }
        }
 
        var ans = 1;
        for (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
     
        var a = [ 7, 42, 11, 64 ];
        var n = a.length;
 
        document.write(rightmostNonZero(a, n));
 
// This code contributed by aashish1995 
 
</script>

Output:

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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