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 much 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

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



Output:

480

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