Given n number of cashiers exchanging the money. At the moment,
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
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