Find the minimum time after which one can exchange notes

Given n number of cashiers exchanging the money. At the moment, cashier had number of people in front of him. The person in the line to cashier had notes. 
Find, how early can one exchange his notes.
Time taken by the cashiers: 

• The cashier took 5 seconds to scan a single note.
• After the cashier scanned every note for the customer, he took 15 seconds to exchange the notes.

Examples: 

Input : n = 5 
k[] = 10 10 10 10 10 
m1[] = 6 7 8 6 8 5 9 8 10 5 
m2[] = 9 6 9 8 7 8 8 10 8 5 
m3[] = 8 7 7 8 7 5 6 8 9 5 
m4[] = 6 5 10 5 5 10 7 8 5 5 
m5[] = 10 9 8 7 6 9 7 9 6 5 
Output : 480 
Explanation: The cashier takes 5 secs for every note of each customer, therefore add 5*m[i][j]. Each cashier spends 15 seconds for every customer, therefore add 15*k[] to the answer. The minimum time obtained after calculating the time taken by each cashier is our answer. Cashier m4 takes the minimum time i.e. 480. 
Input : n = 1 
k[] = 1 
m1[] = 100 
Output : 515 

Approach: Calculate the total time for every cashier and minimum time obtained among all the cashier’s time is the desired answer.
Below is the implementation of above approach: 

C++

// CPP code to find minimum
// time to exchange notes
#include <bits/stdc++.h>
using namespace std;

// Function to calculate minimum
// time to exchange note
void minTimeToExchange(int k[], int m[][10],
                                     int n)
{
    int min = INT_MAX;

    // Checking for every cashier
    for (int i = 0; i < n; i++)
    {
        // Time for changing the notes
        int temp = k[i] * 15;

        // Calculating scanning time
        // for every note
        for (int j = 0; j < k[i]; j++)
        {
            temp += m[i][j] * 5;
        }

        // If value in temp is minimum
        if (temp < min)
            min = temp;
    }

    cout << min;
}

// Driver function
int main()
{
    // number of cashiers
    int n = 5;

    // number of customers with
    // each cashier
    int k[] = {10, 10, 10, 10, 10};

    // number of notes with each customer
    int m[][10] = {{6, 7, 8, 6, 8, 5, 9, 8, 10, 5},
                    {9, 6, 9, 8, 7, 8, 8, 10, 8, 5},
                    {8, 7, 7, 8, 7, 5, 6, 8, 9, 5},
                    {6, 5, 10, 5, 5, 10, 7, 8, 5, 5},
                    {10, 9, 8, 7, 6, 9, 7, 9, 6, 5}};

    // Calling function
    minTimeToExchange(k, m, n);

    return 0;
}

Java

// Java code to find minimum time to exchange
// notes
import java.io.*;

public class GFG {

    // Function to calculate minimum
    // time to exchange note
    static void minTimeToExchange(int []k,
                           int [][]m, int n)
    {

        int min = Integer.MAX_VALUE;

        // Checking for every cashier
        for (int i = 0; i < n; i++)
        {
            // Time for changing the notes
            int temp = k[i] * 15;

            // Calculating scanning time
            // for every note
            for (int j = 0; j < k[i]; j++)
            {
                temp += m[i][j] * 5;
            }

            // If value in temp is minimum
            if (temp < min)
                min = temp;
        }

        System.out.println(min);
    }

    // Driver function
    static public void main (String[] args)
    {

        // number of cashiers
        int n = 5;

        // number of customers with
        // each cashier
        int []k = {10, 10, 10, 10, 10};

        // number of notes with each customer
        int [][]m = {
                {6, 7, 8, 6, 8, 5, 9, 8, 10, 5},
                {9, 6, 9, 8, 7, 8, 8, 10, 8, 5},
                {8, 7, 7, 8, 7, 5, 6, 8, 9, 5},
                {6, 5, 10, 5, 5, 10, 7, 8, 5, 5},
                {10, 9, 8, 7, 6, 9, 7, 9, 6, 5}};

        // Calling function
        minTimeToExchange(k, m, n);
    }
}

// This code is contributed by vt_m.

Python3

# Python3 code to find minimum
# time to exchange notes
from sys import maxsize

# Function to calculate minimum
# time to exchange note
def minTimeToExchange(k, m, n):
    minn = maxsize

    # Checking for every cashier
    for i in range(n):

        # Time for changing the notes
        temp = k[i] * 15

        # Calculating scanning time
        # for every note
        for j in range(k[i]):
            temp += m[i][j] * 5

        # If value in temp is minimum
        if temp < minn:
            minn = temp

    print(minn)

# Driver Code
if __name__ == "__main__":

    # number of cashiers
    n = 5

    # number of customers with
    # each cashier
    k = [10, 10, 10, 10, 10]

    # number of notes with each customer
    m = [[6, 7, 8, 6, 8, 5, 9, 8, 10, 5],
         [9, 6, 9, 8, 7, 8, 8, 10, 8, 5],
         [8, 7, 7, 8, 7, 5, 6, 8, 9, 5],
         [6, 5, 10, 5, 5, 10, 7, 8, 5, 5],
         [10, 9, 8, 7, 6, 9, 7, 9, 6, 5]]

    # Calling function
    minTimeToExchange(k, m, n)

# This code is contributed by
# sanjeev2552

C#

// C# code to find minimum
// time to exchange notes
using System;

public class GFG {

    // Function to calculate minimum
    // time to exchange note
    static void minTimeToExchange(int []k,
                          int [,]m, int n)
    {

        int min = int.MaxValue;

        // Checking for every cashier
        for (int i = 0; i < n; i++)
        {
            // Time for changing the notes
            int temp = k[i] * 15;

            // Calculating scanning time
            // for every note
            for (int j = 0; j < k[i]; j++)
            {
                temp += m[i,j] * 5;
            }

            // If value in temp is minimum
            if (temp < min)
                min = temp;
        }

        Console.WriteLine(min);
    }

    // Driver function
    static public void Main (){
        // number of cashiers
        int n = 5;

        // number of customers with
        // each cashier
        int []k = {10, 10, 10, 10, 10};

        // number of notes with each customer
        int [,]m = {{6, 7, 8, 6, 8, 5, 9, 8, 10, 5},
                    {9, 6, 9, 8, 7, 8, 8, 10, 8, 5},
                    {8, 7, 7, 8, 7, 5, 6, 8, 9, 5},
                    {6, 5, 10, 5, 5, 10, 7, 8, 5, 5},
                    {10, 9, 8, 7, 6, 9, 7, 9, 6, 5}};

        // Calling function
        minTimeToExchange(k, m, n);
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP code to find minimum
// time to exchange notes

// Function to calculate minimum
// time to exchange note
function minTimeToExchange($k, $m,
                               $n)
{
    $min = PHP_INT_MAX;

    // Checking for every cashier
    for ( $i = 0; $i < $n; $i++)
    {

        // Time for changing the notes
        $temp = $k[$i] * 15;

        // Calculating scanning time
        // for every note
        for ($j = 0; $j < $k[$i]; $j++)
        {
            $temp += $m[$i][$j] * 5;
        }

        // If value in temp is minimum
        if ($temp < $min)
            $min = $temp;
    }

    echo $min;
}

    // Driver Code
    // number of cashiers
    $n = 5;

    // number of customers with
    // each cashier
    $k = array(10, 10, 10, 10, 10);

    // number of notes with
    // each customer
    $m = array(array(6, 7, 8, 6, 8, 5, 9, 8, 10, 5),
               array(9, 6, 9, 8, 7, 8, 8, 10, 8, 5),
               array(8, 7, 7, 8, 7, 5, 6, 8, 9, 5),
               array(6, 5, 10, 5, 5, 10, 7, 8, 5, 5),
               array(10, 9, 8, 7, 6, 9, 7, 9, 6, 5));

    // Calling function
    minTimeToExchange($k, $m, $n);

// This code is contributed by anuj_67.
?>

Javascript

<script>

    // JavaScript code to find minimum
    // time to exchange notes

    // Function to calculate minimum
    // time to exchange note
    function minTimeToExchange(k, m, n)
    {

        let min = Number.MAX_VALUE;

        // Checking for every cashier
        for (let i = 0; i < n; i++)
        {
            // Time for changing the notes
            let temp = k[i] * 15;

            // Calculating scanning time
            // for every note
            for (let j = 0; j < k[i]; j++)
            {
                temp += m[i][j] * 5;
            }

            // If value in temp is minimum
            if (temp < min)
                min = temp;
        }

        document.write(min + "</br>");
    }

    // number of cashiers
    let n = 5;

    // number of customers with
    // each cashier
    let k = [10, 10, 10, 10, 10];

    // number of notes with each customer
    let m = [
      [6, 7, 8, 6, 8, 5, 9, 8, 10, 5],
      [9, 6, 9, 8, 7, 8, 8, 10, 8, 5],
      [8, 7, 7, 8, 7, 5, 6, 8, 9, 5],
      [6, 5, 10, 5, 5, 10, 7, 8, 5, 5],
      [10, 9, 8, 7, 6, 9, 7, 9, 6, 5]];

    // Calling function
    minTimeToExchange(k, m, n);

</script>

Output:  

480