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Find the minimum time after which one can exchange notes

  • Difficulty Level : Basic
  • Last Updated : 25 May, 2021

Given n number of cashiers exchanging the money. At the moment, i^{th}  cashier had k_{i}  number of people in front of him. The j^{th}  person in the line to i^{th}  cashier had m_{i,j}  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

 




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