Function to find Number of customers who could not get a computer

27 Jun, 2022
Write a function “runCustomerSimulation” that takes following two inputs 
An integer ‘n’: total number of computers in a cafe and a string: A sequence of uppercase letters ‘seq’: Letters in the sequence occur in pairs. The first occurrence indicates the arrival of a customer; the second indicates the departure of that same customer. 
An integer ‘n’: total number of computers in a cafe and a string: 
A sequence of uppercase letters ‘seq’: Letters in the sequence occur in pairs. The first occurrence indicates the arrival of a customer; the second indicates the departure of that same customer. 
A customer will be serviced if there is an unoccupied computer. No letter will occur more than two times. Customers who leave without using a computer always depart before customers who are currently using the computers. There are at most 20 computers per cafe.
For each set of input the function should output a number telling how many customers, if any walked away without using a computer. Return 0 if all the customers were able to use a computer.runCustomerSimulation (2, “ABBAJJKZKZ”) should return 0runCustomerSimulation (3, “GACCBDDBAGEE”) should return 1 as ‘D’ was not able to get any computerrunCustomerSimulation (3, “GACCBGDDBAEE”) should return 0runCustomerSimulation (1, “ABCBCA”) should return 2 as ‘B’ and ‘C’ were not able to get any computer.runCustomerSimulation(1, “ABCBCADEED”) should return 3 as ‘B’, ‘C’ and ‘E’ were not able to get any computer.Source: Fiberlink (maas360) Interview 
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Below are simple steps to find number of customers who could not get any computer.
Initialize result as 0.Traverse the given sequence. While traversing, keep track of occupied computers (this can be done by keeping track of characters which have appeared only once and a computer was available when they appeared). At any point, if count of occupied computers is equal to ‘n’, and there is a new customer, increment result by 1.
Initialize result as 0.
Traverse the given sequence. While traversing, keep track of occupied computers (this can be done by keeping track of characters which have appeared only once and a computer was available when they appeared). At any point, if count of occupied computers is equal to ‘n’, and there is a new customer, increment result by 1.
The important thing is to keep track of existing customers in cafe in a way that can indicate whether the customer has got a computer or not. Note that in sequence “ABCBCADEED”, customer ‘B’ did not get a seat, but still in cafe as a new customer ‘C’ is next in sequence.
Below are implementations of above idea. 
C++
Java
Python3
C#
Javascript
// C++ program to find number of customers who couldn't get a resource.#include<iostream>#include<cstring>using namespace std; #define MAX_CHAR 26 // n is number of computers in cafe.// 'seq' is given sequence of customer entry, exit eventsint runCustomerSimulation(int n, const char *seq){    // seen[i] = 0, indicates that customer 'i' is not in cafe    // seen[1] = 1, indicates that customer 'i' is in cafe but    //             computer is not assigned yet.    // seen[2] = 2, indicates that customer 'i' is in cafe and    //             has occupied a computer.    char seen[MAX_CHAR] = {0};     // Initialize result which is number of customers who could    // not get any computer.    int res = 0;     int occupied = 0; // To keep track of occupied computers     // Traverse the input sequence    for (int i=0; seq[i]; i++)    {        // Find index of current character in seen[0...25]        int ind = seq[i] - 'A';         // If First occurrence of 'seq[i]'        if (seen[ind] == 0)        {            // set the current character as seen            seen[ind] = 1;             // If number of occupied computers is less than            // n, then assign a computer to new customer            if (occupied < n)            {                occupied++;                 // Set the current character as occupying a computer                seen[ind] = 2;            }             // Else this customer cannot get a computer,            // increment result            else                res++;        }         // If this is second occurrence of 'seq[i]'        else        {        // Decrement occupied only if this customer        // was using a computer        if (seen[ind] == 2)            occupied--;        seen[ind] = 0;        }    }    return res;} // Driver programint main(){    cout << runCustomerSimulation(2, "ABBAJJKZKZ") << endl;    cout << runCustomerSimulation(3, "GACCBDDBAGEE") << endl;    cout << runCustomerSimulation(3, "GACCBGDDBAEE") << endl;    cout << runCustomerSimulation(1, "ABCBCA") << endl;    cout << runCustomerSimulation(1, "ABCBCADEED") << endl;    return 0;}
// JAVA program to find number of//customers who couldn't get a resource.class GFG{ static int MAX_CHAR = 26; // n is number of computers in cafe.// 'seq' is given sequence of customer entry, exit eventsstatic int runCustomerSimulation(int n, char []seq){    // seen[i] = 0, indicates that customer 'i' is not in cafe    // seen[1] = 1, indicates that customer 'i' is in cafe but    //         computer is not assigned yet.    // seen[2] = 2, indicates that customer 'i' is in cafe and    //         has occupied a computer.    char []seen = new char[MAX_CHAR];     // Initialize result which is number of customers who could    // not get any computer.    int res = 0;     int occupied = 0; // To keep track of occupied computers     // Traverse the input sequence    for (int i=0; i< seq.length; i++)    {        // Find index of current character in seen[0...25]        int ind = seq[i] - 'A';         // If First occurrence of 'seq[i]'        if (seen[ind] == 0)        {            // set the current character as seen            seen[ind] = 1;             // If number of occupied computers is less than            // n, then assign a computer to new customer            if (occupied < n)            {                occupied++;                 // Set the current character as occupying a computer                seen[ind] = 2;            }             // Else this customer cannot get a computer,            // increment result            else                res++;        }         // If this is second occurrence of 'seq[i]'        else        {                     // Decrement occupied only if this customer        // was using a computer        if (seen[ind] == 2)            occupied--;        seen[ind] = 0;        }    }    return res;} // Driver programpublic static void main(String[] args){    System.out.println(runCustomerSimulation(2, "ABBAJJKZKZ".toCharArray()));    System.out.println(runCustomerSimulation(3, "GACCBDDBAGEE".toCharArray()));    System.out.println(runCustomerSimulation(3, "GACCBGDDBAEE".toCharArray()));    System.out.println(runCustomerSimulation(1, "ABCBCA".toCharArray()));    System.out.println(runCustomerSimulation(1, "ABCBCADEED".toCharArray()));}} // This code is contributed by Princi Singh
# Python program function to find Number of customers who# could not get a computerMAX_CHAR = 26 # n is number of computers in cafe.# 'seq' is given sequence of customer entry, exit eventsdef runCustomerSimulation(n, seq):     # seen[i] = 0, indicates that customer 'i' is not in cafe    # seen[1] = 1, indicates that customer 'i' is in cafe but    #             computer is not assigned yet.    # seen[2] = 2, indicates that customer 'i' is in cafe and    #             has occupied a computer.    seen = [0] * MAX_CHAR     # Initialize result which is number of customers who could    # not get any computer.    res = 0    occupied = 0    # To keep track of occupied     # Traverse the input sequence    for i in range(len(seq)):         # Find index of current character in seen[0...25]        ind = ord(seq[i]) - ord('A')         # If first occurrence of 'seq[i]'        if seen[ind] == 0:             # set the current character as seen            seen[ind] = 1             # If number of occupied computers is less than            # n, then assign a computer to new customer            if occupied < n:                occupied+=1                 # Set the current character as occupying a computer                seen[ind] = 2             # Else this customer cannot get a computer,            # increment            else:                res+=1         # If this is second occurrence of 'seq[i]'        else:            # Decrement occupied only if this customer            # was using a computer            if seen[ind] == 2:                occupied-=1            seen[ind] = 0     return res # Driver programprint (runCustomerSimulation(2, "ABBAJJKZKZ"))print (runCustomerSimulation(3, "GACCBDDBAGEE"))print (runCustomerSimulation(3, "GACCBGDDBAEE"))print (runCustomerSimulation(1, "ABCBCA"))print (runCustomerSimulation(1, "ABCBCADEED")) # This code is contributed BHAVYA JAIN
// C# program to find number of// customers who couldn't get a resource.using System; class GFG{static int MAX_CHAR = 26; // n is number of computers in cafe.// 'seq' is given sequence of customer entry,// exit eventsstatic int runCustomerSimulation(int n, char []seq){    // seen[i] = 0, indicates that customer 'i' is not in cafe    // seen[1] = 1, indicates that customer 'i' is in cafe but    // computer is not assigned yet.    // seen[2] = 2, indicates that customer 'i' is in cafe and    // has occupied a computer.    char []seen = new char[MAX_CHAR];     // Initialize result which is number of customers    // who could not get any computer.    int res = 0;     int occupied = 0; // To keep track of occupied computers     // Traverse the input sequence    for (int i = 0; i < seq.Length; i++)    {        // Find index of current character in seen[0...25]        int ind = seq[i] - 'A';         // If First occurrence of 'seq[i]'        if (seen[ind] == 0)        {            // set the current character as seen            seen[ind] = (char)1;             // If number of occupied computers is less than            // n, then assign a computer to new customer            if (occupied < n)            {                occupied++;                 // Set the current character as                // occupying a computer                seen[ind] = (char)2;            }             // Else this customer cannot get a computer,            // increment result            else                res++;        }         // If this is second occurrence of 'seq[i]'        else        {                     // Decrement occupied only if this customer        // was using a computer        if (seen[ind] == 2)            occupied--;        seen[ind] = (char)0;        }    }    return res;} // Driver Codepublic static void Main(String[] args){    Console.WriteLine(runCustomerSimulation(2,                      "ABBAJJKZKZ".ToCharArray()));    Console.WriteLine(runCustomerSimulation(3,                      "GACCBDDBAGEE".ToCharArray()));    Console.WriteLine(runCustomerSimulation(3,                      "GACCBGDDBAEE".ToCharArray()));    Console.WriteLine(runCustomerSimulation(1,                      "ABCBCA".ToCharArray()));    Console.WriteLine(runCustomerSimulation(1,                      "ABCBCADEED".ToCharArray()));}} // This code is contributed by Rajput-Ji
<script>        // JavaScript Program for the above approach                 let MAX_CHAR = 26;         // n is number of computers in cafe.        // 'seq' is given sequence of customer entry, exit events        function runCustomerSimulation(n, seq) {            // seen[i] = 0, indicates that customer 'i' is not in cafe            // seen[1] = 1, indicates that customer 'i' is in cafe but            //              computer is not assigned yet.            // seen[2] = 2, indicates that customer 'i' is in cafe and            //              has occupied a computer.            let seen = new Array(MAX_CHAR).fill(0);             // Initialize result which is number of customers who could            // not get any computer.            let res = 0;             let occupied = 0;  // To keep track of occupied computers             // Traverse the input sequence            for (let i = 0; i < seq.length; i++) {                // Find index of current character in seen[0...25]                let ind = seq[i].charCodeAt(0) - 'A'.charCodeAt(0);                 // If First occurrence of 'seq[i]'                if (seen[ind] == 0) {                    // set the current character as seen                    seen[ind] = 1;                     // If number of occupied computers is less than                    // n, then assign a computer to new customer                    if (occupied < n) {                        occupied++;                         // Set the current character as occupying a computer                        seen[ind] = 2;                    }                     // Else this customer cannot get a computer,                    // increment result                    else                        res++;                }                 // If this is second occurrence of 'seq[i]'                else {                    // Decrement occupied only if this customer                    // was using a computer                    if (seen[ind] == 2) {                        occupied--;                    }                    seen[ind] = 0;                }            }            return res;        }         // Driver program         document.write(runCustomerSimulation(2, "ABBAJJKZKZ") + "<br>");        document.write(runCustomerSimulation(3, "GACCBDDBAGEE") + "<br>");        document.write(runCustomerSimulation(3, "GACCBGDDBAEE") + "<br>");        document.write(runCustomerSimulation(1, "ABCBCA") + "<br>");        document.write(runCustomerSimulation(1, "ABCBCADEED") + "<br>");      // This code is contributed by Potta Lokesh    </script>
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Time complexity of above solution is O(n) and extra space required is O(CHAR_MAX) where CHAR_MAX is total number of possible characters in given sequence.
This article is contributed by Lokesh. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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