Capgemini ADAPT Sprint 1 Solution Java | LSAByCustomerName Problem Solution

Capgemini ADAPT Sprint 1 Solution Java LSAByCustomerName Problem Solution
Capgemini ADAPT Sprint 1 Solution Java LSAByCustomerName Problem Solution

In this post, you will get the solution to the first problem of Capgemini ADAPT Sprint 1 – JBR-CS5A-L2-2-LSAByCustomerName.

Note: The solutions are provided for learning purposes only. 

Sprint 1 – Problem 4: JBR-CS5A-L2-2-LSAByCustomerName:


Problem Description:


Create a program to search the customer based on the customer’s Name using Linear Search Algorithm.

This program should print “No Record Found” if the name is not available in the given array.

Note: Use Linear Search Algorithm

Sample Input:


Simrath

Sample Output:


1002

Simrath

Amristar

Sample Input:


King

Sample Output:


No Record Found

Execution Time Limit:


10 seconds

The solution to JBR-CS5A-L2-2-LSAByCustomerName:


 

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Source{
      static String customerDetails[][] = new String [5][3];
      static{
         customerDetails[0][0] = "1001";
         customerDetails[0][1] = "Raj";
         customerDetails[0][2] = "Chennai";

         customerDetails[1][0] = "1008";
         customerDetails[1][1] = "Akshay";
         customerDetails[1][2] = "Pune";
  
         customerDetails[2][0] = "1002";
         customerDetails[2][1] = "Simrath";
         customerDetails[2][2] = "Amristar";

         customerDetails[3][0] = "1204";
         customerDetails[3][1] = "Gaurav";
         customerDetails[3][2] = "Delhi";

         customerDetails[4][0] = "1005";
         customerDetails[4][1] = "Ganesh";
         customerDetails[4][2] = "Chennai";
  }
  public static void main(String args[]) throws Exception{
     Scanner sc = new Scanner(System.in);
     String key = sc.nextLine();
     int index = -1;
     Boolean found = false;
     int n = customerDetails.length;
     for(int i = 0; i < n; i ++)
     {
         if(key.equalsIgnoreCase(customerDetails[i][1]))
          {
             found = true;
             index = i;
             break;
          }
      }
      if(found)
      {
          for(int i = 0; i < 3; i++)
           {
               System.out.println(customerDetails[index[i]]))
           }
       }
       else
       {
            System.out.println("No Record Found");
        }
    }
}