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# 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 ``using` `namespace` `std;` `// Function to calculate minimum``// time to exchange note``void` `minTimeToExchange(``int` `k[], ``int` `m[],``                                     ``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[] = {{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

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## Javascript

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Output:

`480`

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