TCS Pre-ILP Assignments[Basics of Programming]


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To who ever is going to copy paste and submit the same copy as below, Answers can vary depending on thoughts. There is more than one way for a solution, please do NOT blindly Copy Paste.


Thanks to the Community - TCS ILP 2011 and Ayan Jana for providing the solutions.

Assignment 1 - Please zip and upload the files.
Question 1
a. Search for a name
 Write a program to accept an array of names and a name and check whether the name is present in the array. Return the count of occurrence. Use the following array as input
{“Dave”, “Ann”, “George”, “Sam”, “Ted”, “Gag”, “Saj”, “Agati”, “Mary”, “Sam”, “Ayan”, “Dev”, “Kity”, “Meery”, “Smith”, “Johnson”, “Bill”, “Williams”, “Jones”, “Brown”, “Davis”, “Miller”, “Wilson”, “Moore”, “Taylor, “Anderson”, “Thomas”, “Jackson”}

Program : 

import java.io.*;
import java.util.Scanner;
class program1
  {
  public static void main(String args[])
  {
   String[] names={“Dave”, “Ann”, “George”, “Sam”, “Ted”, “Gag”, “Saj”, “Agati”, “Mary”, “Sam”,   “Ayan”, “Dev”, “Kity”, “Meery”, “Smith”, “Johnson”, “Bill”, “Williams”, “Jones”, “Brown”, “Davis”, “Miller”, “Wilson”, “Moore”, “Taylor, “Anderson”, “Thomas”, “Jackson”};
 Scanner sr= new Scanner(System.in);
 System.out.println("Enter a name to check");
 String name=sr.nextLine();
 int count=0;
 for(int i=0;i<names.length;i++)
  {
  if(names[i].equals(name))
  count++;
  }
 if(count!=0)
  System.out.println("The Number of occurance of the Name is " + count );
 else
  System.out.println("Word not Found");
  }
  }


b.  Improve the understandability of the below given code:


import java.util.*;
class problem3
 {
int[] numArray = new int[10];
public static void incrementElements (int[] integerArray)
 {
  int arraylen = integerArray.length;
  for (int i = 0; i < arraylen; i ++)
{
 System.out.println(integerArray[i]);
}
  for (int i = 0; i < arraylen; i ++)
{
 integerArray[i] = integerArray[i] + 10;
}
  for (int i=0; i < arraylen; i ++)
{
 System.out.println(integerArray[i]);
}
 }



CODE DESCRIPTION:

The code defines that when any element of the integer array is occured then it is printed first and after that it's increamented by 10 ..
and then the resultant array is printed, the same code is being defined in the component classes of java.util.*-The Component class is
the abstract superclass of the nonmenu-related Abstract Window Toolkit components. Class Component can also be extended directly to create
a lightweight component. A lightweight component is a component that is not associated with a native opaque window.the code is predefined
in the package and when ever we want to execute any application
which wants the code...den without wrting the whole code we can directly import the package and execute thAt
application,...\.Here if we explain the code step by step, we can see that an array named numarray is assigned
containing 10 elements, which defines the size of the array;

int num[]array=new int[10];

next the arraylength is calculated using integerArray.length,

and next using for loop we set a loop from 0 to the obtained length of the array

for (int i = 0; i < arraylen; i ++)

next the present integer value is incremented, by a value of 10 and then
the control passes to the next element in the array.

next we have the display the resulted array using the final code:

for (int i=0; i < arraylen; i ++)

{

System.out.println(integerArray[i]);

}

Question 2
Greatest common divisor
Calculate the greatest common divisor of two positive numbers a and b.
gcd(a,b) is recursively defined as
gcd(a,b) = a if a =b
gcd(a,b) = gcd(a-b, b) if a >b
gcd(a,b) = gcd(a, b-a) if b > a

Program:


import java.io.*;
import java.util.Scanner;
class program2
 {
public static void main(String args[])
 {
Scanner sr= new Scanner(System.in);
System.out.println("Enter The number a");
int a=sr.nextInt();
System.out.println("Enter The number b");
int b=sr.nextInt();
int gcd;
if(a==b)
   gcd=a;
else if(a>b)
   gcd=findgcd(a-b,b);
else
   gcd=findgcd(b,b-a);
System.out.println("The greatest common divisor of numbers " + a + " and " + b + " is " + gcd);
 }
public static int findgcd(int c,int d)
 {
if (d == 0)
   return c;
return findgcd(d, c % d);
 }
 }


(b) Improve the understandability of the below given code:

class Problem1
 {
  int[] a;
  int nElems;
  public ArrayBub(int max)
   {
a = new int[max];
   }
  public void insert(int value)
   {
a[nElems] = value;
nElems++;
   }
  public void Sort()
   {
int out, in;
for(out=nElems-1; out>1; out--)
 for(in=0; in<out; in++)
if( a[in] > a[in+1] )
swap(in, in+1);
   }
  public void swap(int one, int two)
   {
long temp = a[one];
a[one] = a[two];
a[two] = temp;
   }
 }


The above code explains bubble sorting implementation.

Bubble sort is also known as exchange sort. Bubble sort is a simplest sorting algorithm.
In bubble sort algorithm array is traversed from 0 to the length-1 index of the array and
compared one element to the next element and swap values in between if the next element is
less than the previous element. In other words, bubble sorting algorithm compare two values
and put the largest value at largest index. The algorithm follow the same steps repeatedly
until the values of array is sorted. In worst-case the complexity of bubble sort is O(n2) and
in best-case the complexity of bubble sort is O(n).


Say we have an array unsorted A[0],A[1],A[2]................ A[n-1] and A[n] as input. Then the following steps are followed by bubble sort algorithm
to sort the values of an array.
1.Compare A[0] and A[1] .
2.If A[0]>A[1] then Swap A[0] and A[1].
3.Take next A[1] and A[2].
4.Comapre these values.
5.If A[1]>A[2] then swap A[1] and A[2]
...............................................................
................................................................
at last compare A[n-1] and A[n]. If A[n-1]>A[n] then swap A[n-1] and A[n]. As we see the highest value
is reached at nth position. At next iteration leave nth value. Then apply the same steps repeatedly on A[0],A[1],A[2]................ A[n-1] elements
repeatedly until the values of array is sorted.


So in the above code we obtain:

step 1: we initialise number of elements and an array.

step 2:we initialise max for a value and then set a varialble say a= maximum.

step3: we accept the value for the elements.

step4: we set two variables out and in and we compare a[0] with a[1] and if a[0] > a[1] then swap them. As we see the highest value is reached at nth position.
At next iteration leave nth value. Then apply the same steps repeatedly on A[0],A[1],A[2]................ A[n-1] elements repeatedly until the values of array
is sorted.