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 notevoid 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 functionint 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// notesimport 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 notesfrom sys import maxsize# Function to calculate minimum# time to exchange notedef 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 Codeif __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 notesusing 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 notefunction 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
