Data Structure Lecture Array and Recursion.pdf

DATA STRUCTURES
Lecture 3
Dr. Mili Dhar
Assistant Professor
Galgotias University, Greater Noida

Today’s Agenda
• What is an Array?
• What is an Stack?
• Terminologies related to Stack.
• Diagrammatic view of Stack.
Dr. Mili Dhar

What is an Array?
• Array is a linear Data Structure.
• Array is a Homogeneous collection of multiple data items.
• Open from both the ends.
• Allow random access.
Dr. Mili Dhar

Declaring an Array
#include <conio.h>
#include<stdio.h>
void main()
{
int arr[5];
int i;
for(i=0;i<5;i++)
{
……
……
}
getch();
}
10 20 30 40 50 int i;
0 1 2 3 4
index
arr
Dr. Mili Dhar

Declaration of an Array
type array-name[];
OR
type[] array-name;
Example:
int Arr[];
OR
int [] Arr;
Dr. Mili Dhar

Initialization of an Array
var-name = new type [size];
Example:
int Arr[];
Arr = new int [20];
Dr. Mili Dhar

Insert elements in an Array
int[] myNum = {10, 20, 30, 40};
String[] cars = {"Volvo", "BMW", "Ford", "Mazda"};
Dr. Mili Dhar

Accessing of an Array
for (int i = 0; i < arr.length; i++)
System.out.println("Element at index " +
i + " : "+ arr[i]);
System.out.println(cars[0]);
Dr. Mili Dhar

Array Length
System.out.println(cars.length);
Dr. Mili Dhar

class DS {
public static void main(String[] args)
{
char [] arr; declaration
arr = new char[5]; initialization
arr[0] = ‘10’; inserting elements
arr[1] = ‘20’;
arr[2] = ‘30’;
arr[3] = ‘40’;
arr[4] = ‘50’;
// accessing the elements of the specified array
for (int i = 0; i < arr.length; i++)
System.out.println("Element at index " + i + " : " + arr[i]);
}
}
Dr. Mili Dhar

import java.util.*;
class HelloWorld {
{
int []arr = new int[3];
Scanner sc = new Scanner(System.in);
for (int i =0; i<arr.length; i++) {
System.out.println("Enter the elements "+i);
arr[i]= sc.nextInt(); }
for (int i =0; i<arr.length; i++)
{
System.out.println(arr[i]);
}
}
Dr. Mili Dhar

8/27/2024 12
Address Calculation for 1-D array
Dr. Mili Dhar

• Array of an element of an array say “A[ I ]” is calculated using the
following formula:
Address of A [ I ] = B + W * ( I – LB )
Where,
B = Base address
W = Storage Size of one element stored in the array (in byte)
I = Subscript of element whose address is to be found
LB = Lower limit / Lower Bound of subscript, if not specified assume 0
(zero)
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Dr. Mili Dhar

Example 1:
• Given the base address of an array B[1300…..1900] as
1020 and size of each element is 2 bytes in the memory.
Find the address of B[1700].
• Solution:
The given values are: B = 1020, LB = 1300, W = 2, I = 1700
= 1020 + 2 * (1700 – 1300)
= 1020 + 2 * 400
= 1020 + 800
= 1820 [Ans]
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Dr. Mili Dhar

Example 2:
• Find the base address of an array A[-15:65]. The size of each
element is 2 bytes in the memory and the location of A[37] is 4279.
Solution:
The given values are: LB = -15, UB = 65, W = 2, I = 37, A[37] = 4279
4279= B + 2 * (37 – (-15))
4279= B + 2 * 52
4279= B + 104
B= 4175 [Ans]
8/27/2024 15
Address Calculation
Dr. Mili Dhar

Declaration of 2D-Array
type arr-name[][];
OR
type[][] arr-name;
Example:
int Arr[][];
OR
int [][] Arr;
Dr. Mili Dhar

Initialization of an 2D-Array
var-name = new type [size][size];
Example:
int Arr[] [];
Arr = new int [20][20];
Dr. Mili Dhar

int[][] myNum = {{10, 20}, {30, 40}};
String[] []cars = {{"Volvo", "BMW“}, {"Ford", "Mazda"}};
Dr. Mili Dhar

import java.util.*;
class HelloWorld {public static void main(String[] args)
{
int [][]arr = new int[3][3];
Scanner sc = new Scanner(System.in);
for (int i =0; i<arr.length; i++) {
for (int j =0; j<arr.length; j++) {
System.out.println("Enter the elements "+i,j);
arr[i][j]= sc.nextInt(); } }
for (int i =0; i<arr.length; i++)
{
for (int j =0; j<arr.length; j++) {
System.out.println(arr[i][j]);
}}
}
Dr. Mili Dhar

Display elements of 2D Array
public class Main
{
{
int[][] myNumbers = { {1, 2, 3, 4}, {5, 6, 7} };
for (int i = 0; i < myNumbers.length; ++i)
{
for(int j = 0; j < myNumbers[i].length; ++j)
{
System.out.println(myNumbers[i][j]);
}
}
}
}
Dr. Mili Dhar

Row Major Array
In row major order, the elements of a particular row are
stored at adjacent memory locations. The first element
of the array (arr[0][0]) is stored at the first location
followed by the arr[0][1] and so on. After the first row,
elements of the next row are stored next.
Dr. Mili Dhar

Column Major Array
In column major order, the elements of a column are
stored adjacent to each other in the memory.The first
element of the array (arr[0][0]) is stored at the first
location followed by the arr[1][0] and so on. After the
first column, elements of the next column are stored
stating from the top.
Dr. Mili Dhar

address[i][j] = I + W * [(i - l_row) * N + (j - l_col)]
•I : Base address
l_row : lower bound for row
l_col : lower bound for column
W : sizeof (data type)
N : Number of columns
8/27/2024 23
Address Calculation row major array
Dr. Mili Dhar

• Example:
Consider an integer array of size 3X3. The address of the first
element is 1048. Calculate the address of the element at index i =
2, j = 1. (0 based index)
• I = 1048, l_row = 0 = l_col, i = 2, j = 1, W = 2, N = 3
• address[2][1] = I + W * [(i-l_row) * N + (j - l_col)]
• address[2][1] = 1048 + 2 *[ 2 * 3 + 1 ]= 1048 + 12 + 2 = 1062
8/27/2024 24
Dr. Mili Dhar

• Example:
Example: Given an array, arr[1………10][1………15] with base
value 100 and the size of each element is 1 Byte in memory. Find
the address of arr[8][6] with the help of row-major order.
8/27/2024 25
Base address B = 100
W = 1 Bytes, I = 8, J = 6
LR = 1, LC = 1
N = Upper Bound – Lower Bound + 1
= 15 – 1 + 1 = 15
Solution:
Address of A[8][6] = 100 + 1 * ((8 – 1) * 15 + (6 – 1))
= 100 + 1 * ((7) * 15 + (5))
= 100 + 1 * (110)
Address of A[I][J] = 210
Dr. Mili Dhar

address[i][j] = I + W * ((j – l_col) * M + (i – l_row))
I : Base address
l_row : lower bound for row
l_col : lower bound for column
W : sizeof (data type)
M : Number of rows
8/27/2024 26
Address Calculation column major array
Dr. Mili Dhar

• Example:
• Consider an integer array of size 3X3. The address of the first
element is 1048. Calculate the address of the element at index i = 2, j
= 1. (0 based index)
• I = 1048, l_row = 0 = l_col, i = 2, j = 1, W = 2, M = 3
• address[2][1] = I + W * [(j - l_col) * M + (i - l_row)]
• address[2][1] = 1048 + 2 * [1 * 3 + 2 ]= 1048 + 6 + 4 = 1058
8/27/2024 27
Dr. Mili Dhar

• Example:
Example: Given an array, arr[1………10][1………15] with base
value 100 and the size of each element is 1 Byte in memory. Find
the address of arr[8][6] with the help of column-major order.
8/27/2024 28
Base address B = 100
W = 1 Bytes, I = 8, J = 6
LR = 1, LC = 1
N = Upper Bound – Lower Bound + 1
= 10 – 1 + 1 = 10
Solution:
Address of A[8][6] = 100 + 1 * ((6 – 1) * 10 + (8 – 1))
= 100 + 1 * ((5) * 10 + (7))
= 100 + 1 * (57)
Address of A[I][J] = 157
Dr. Mili Dhar

• A matrix is a two-dimensional data object made of m rows and
n columns, therefore having total m x n values. If most of the
elements of the matrix have 0 value, then it is called a sparse
matrix.
Why to use Sparse Matrix instead of simple matrix ?
• Storage: There are lesser non-zero elements than zeros and
thus lesser memory can be used to store only those elements.
• Computing time: Computing time can be saved by logically
designing a data structure traversing only non-zero elements..
8/27/2024 29
Sparse Matrix and its representations
Dr. Mili Dhar

8/27/2024 Dr. Pooja Unit -1 30
• Representing a sparse matrix by a 2D array leads to wastage of lots of
memory as zeroes in the matrix are of no use in most of the cases.
So, instead of storing zeroes with non-zero elements, we only store
non-zero elements. This means storing non-zero elements with
triples- (Row, Column, value).
Dr. Mili Dhar

8/27/2024 31
• Sparse Matrix Representations can be done in many ways following
are two common representations:
• Array representation
• Linked list representation
Dr. Mili Dhar

8/27/2024 32
int size = 0;
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < 5; j++)
{
if (sparseMatrix[i][j] != 0)
{
size++;
}
}
}
class DS
{
public static void main(String[]
args)
{
int sparseMatrix[][]
= {
{0, 0, 3, 0, 4},
{0, 0, 5, 7, 0},
{0, 0, 0, 0, 0},
{0, 2, 6, 0, 0}
};
Dr. Mili Dhar

8/27/2024 33
// number of columns in compactMatrix (size) must be
// equal to number of non - zero elements in sparseMatrix
int compactMatrix[][] = new int[3][size];
// Making of new matrix
int k = 0;
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < 5; j++)
{
Dr. Mili Dhar

8/27/2024 34
if (sparseMatrix[i][j] != 0)
{
compactMatrix[0][k] = i;
compactMatrix[1][k] = j;
compactMatrix[2][k] = sparseMatrix[i][j];
k++;
}
}
}
Dr. Mili Dhar

8/27/2024 35
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < size; j++)
{
System.out.printf("%d ", compactMatrix[i][j]);
}
System.out.printf("n");
}
}
}
Dr. Mili Dhar

8/27/2024 36
Arithmetic operations on matrices
Dr. Mili Dhar

8/27/2024 37
public class MatrixAdditionExample{
public static void main(String args[]){
//creating two matrices
int a[][]={{1,3,4},{2,4,3},{3,4,5}};
int b[][]={{1,3,4},{2,4,3},{1,2,4}};
//creating another matrix to store the sum of two matrices
int c[][]=new int[3][3]; //3 rows and 3 columns
//adding and printing addition of 2 matrices
for(int i=0;i<3;i++){
for(int j=0;j<3;j++){
c[i][j]=a[i][j]+b[i][j]; //use - for subtraction
System.out.print(c[i][j]+" ");
}
System.out.println(); //new line
}
}}
Dr. Mili Dhar

8/27/2024 38
Dr. Mili Dhar

8/27/2024 39
public class MM{
public static void main(String
args[]){
//creating two matrices
int a[][]={{1,1,1},{2,2,2},{3,3,3}};
int b[][]={{1,1,1},{2,2,2},{3,3,3}};
//creating another matrix to store
the multiplication of two matrices
int c[][]=new int[3][3];
//multiplying and printing
multiplication of 2 matrices
for(int i=0;i<3;i++){
for(int j=0;j<3;j++){
c[i][j]=0;
for(int k=0;k<3;k++)
{
c[i][j]+=a[i][k]*b[k][j];
}
System.out.print(c[i][j]+" ");}
System.out.println();//new line
}
}}
Dr. Mili Dhar

Write a program to insert an element in ith
position of the array.
import java.io.*;
import java.lang.*;
import java.util.*;
class InsertElement
{
// Function to insert x in arr at position pos
public static int[] insertX(int n, int arr[], int x, int pos)
{
int i;
// create a new array of size n+1
int newarr[] = new int[n + 1];
Dr. Mili Dhar

// insert the elements from the old array into the new
// array. insert all elements till pos then insert x at
pos then insert rest of the elements
for (i = 0; i < n + 1; i++) {
if (i < pos - 1)
newarr[i] = arr[i];
else if (i == pos - 1)
newarr[i] = x;
else
newarr[i] = arr[i - 1];
}
return newarr;
}
Dr. Mili Dhar

// Main function
{
int n = 10;
int i;
// initial array of size 10
int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
// print the original array
System.out.println("Initial Array:n" + Arrays.toString(arr));
// element to be inserted
int x = 50;
Dr. Mili Dhar

// position at which element is to be inserted
int pos = 5;
// call the method to insert x in arr at position pos
arr = insertX(n, arr, x, pos);
// print the updated array
System.out.println("nArray with " + x + " inserted at
position " + pos + ":n" + Arrays.toString(arr));
}
}
Dr. Mili Dhar

Write a program to delete ith element
of the array.
import java.util.Arrays;
class DeleteElement {
// Function to remove the element
public static int[] removeTheElement(int[] arr, int index)
{
// If the array is empty
// or the index is not in array range
// return the original array
if (arr == null || index < 0 || index >= arr.length) {
return arr;
}
Dr. Mili Dhar

of the array.
// Create another array of size one less
int[] anotherArray = new int[arr.length - 1];
// Copy the elements except the index
// from original array to the other array
for (int i = 0, k = 0; i < arr.length; i++) {
// if the index is the removal element index
if (i == index) {
continue;
}
Dr. Mili Dhar

of the array.
// if the index is not the removal element index
anotherArray[k++] = arr[i];
}
// return the resultant array
return anotherArray;
}
// Driver Code
{
// Get the array
int[] arr = { 1, 2, 3, 4, 5 };
// Print the resultant array
System.out.println("Original Array: " + Arrays.toString(arr));
Dr. Mili Dhar

of the array.
// Get the specific index
int index = 2;
// Print the index
System.out.println("Index to be removed: " + index);
// Remove the element
arr = removeTheElement(arr, index);
// Print the resultant array
System.out.println("Resultant Array: " +
Arrays.toString(arr));
}
}
Dr. Mili Dhar

Write a program to reverse the element
of the array.
import java.util.Scanner;
public class ArrayReverser {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter the size of the array:");
int size = scanner.nextInt();
int[] array = new int[size];
System.out.println("Enter " + size + " elements:");
for (int i = 0; i < size; i++) {
array[i] = scanner.nextInt();
}
Dr. Mili Dhar

Write a program to reverse the element
of the array.
reverseArray(array);
System.out.println("Reversed array:");
for (int num : array) {
System.out.print(num + " ");
}
}
Dr. Mili Dhar

of the array.
public static void reverseArray(int[] array) {
int startIndex = 0;
int endIndex = array.length - 1;
while (startIndex < endIndex) {
int temp = array[startIndex];
array[startIndex] = array[endIndex];
array[endIndex] = temp;
startIndex++;
endIndex--;
}
}
}
Dr. Mili Dhar

Tail Recursion
Tail recursion is defined as a recursive function in which the
recursive call is the last statement that is executed by the
function. So basically nothing is left to execute after the
recursion call.
static void print(int n)
{
if (n < 0)
return;
System.out.print(" " + n);
print(n - 1);
}
Dr. Mili Dhar

Head Recursion
Head recursion is defined as a recursive function in which the
recursive call is the first statement that is executed by the
function. So basically nothing is left to execute before the
recursion call.
static void print(int n)
{
if (n < 0)
return;
else
print(n - 1);
System.out.print(" " + n);
}
Dr. Mili Dhar

Tree Recursion
Tree Recursion: If a recursive function calling itself for one time then
it’s known as Linear Recursion. Otherwise if a recursive function
calling itself for more than one time then it’s known as Tree
Recursion.
Dr. Mili Dhar

Tree Recursion
class tree
{
static void fun(int n)
{
if (n > 0) {
System.out.print(" "+ n);
fun(n - 1);
fun(n - 1);
}
}
{
fun(3);
}
}
Dr. Mili Dhar

Tree Recursion
import java.util.*;
class Tree
{
static int fib(int n)
{
// Stop condition
if (n == 0)
return 0;
if (n == 1 || n == 2)
return 1;
else
return (fib(n - 1) + fib(n - 2));
}
Dr. Mili Dhar

Tree Recursion
public static void main(String []args)
{
// Initialize variable n.
int n = 5;
System.out.print("Fibonacci series of 5 numbers is: ");
// for loop to print the fibonacci series.
for (int i = 0; i < n; i++)
{
System.out.print(fib(i)+" ");
}
}
}
Dr. Mili Dhar

Nested Recursion
Nested Recursion: In this recursion, a recursive function
will pass the parameter as a recursive call. That means
“recursion inside recursion”.
import java.util.*;
class NestedRecursion {
static int fun(int n)
{
if (n > 100)
return n - 10;
return fun(fun(n + 11));
}
public static void main(String
args[])
{
int r;
r = fun(95);
System.out.print(" "+ r);
} }
Dr. Mili Dhar

Write a recursive program to print an
array in reverse order.
import java.util.Scanner;
public class PrintArrayReverseOrder {
static void ReverseArray(int arr[], int n)
{
int i;
if(n>0)
{
i=n-1;
System.out.print(arr[i]+" ");
ReverseArray(arr,i);
}
}
Dr. Mili Dhar

Write a recursive program to print an
array in reverse order.
Scanner cs=new Scanner(System.in);
int n,i;
System.out.println("Enter your Array Size:");
n=cs.nextInt();
int arr[]=new int[n];
System.out.println("Enter the Array Element:");
for(i=0;i<n;i++)
{
arr[i]=cs.nextInt();
}
System.out.print("After reversing Array Element Are:");
ReverseArray(arr,n);
cs.close();
}
}
Dr. Mili Dhar

Write a recursion Program to find a^b
class Power {
int base = 3, powerRaised = 4;
int result = power(base, powerRaised);
System.out.println(base + "^" + powerRaised + "=" + result);
}
public static int power(int base, int powerRaised)
{
if (powerRaised != 0) {
return (base * power(base, powerRaised - 1));
}
else {
return 1;
}
}
}
Dr. Mili Dhar

What is a Stack?
• A Stack is a linear Data Structure just like an array.
• Stack is a restricted form of Array.
• Stack is Open from one end only and closed from the other
end.
• Stack doesn’t allow random access of data, rather the data is
accessed in the sequential manner.
• Like array it is also a homogeneous collection of elements.
• Like array it is also index based.
Dr. Mili Dhar

Terminologies related to Stack
1. PUSH
1. POP
1. LIFO
1. PEEP
1. TOS
1. OVERFLOW
1. UNDERFLOW
Dr. Mili Dhar

PUSH, POP & LIFO
• PUSH - Push is a process of inserting elements in the
Stack, one at a time.
• POP - Pop is the process of removing elements from the
Stack, one at a time.
• LIFO - Since stack is open from one end only, the element
inserted(PUSHED) last will be the one to be
removed(POPPED) first. Thus we say a stack is based on
LAST IN FIRST OUT order.
Dr. Mili Dhar

PEEP, TOS, OVERFLOW &
UNDERFLOW
PEEP: The process of accessing (not removing) the top elements
of the stack.
OVERFLOW & UNDERFLOW: These are conditions which arise
when we try to PUSH data in a FULL STACK, or we try to POP data
from an EMPTY stack, Respectively.
Dr. Mili Dhar

Diagrammatic View Of a Stack
20
10
10
arr
tos
-1
4
3
2
1
0
4
3
2
1
0
arr
0
tos
1
tos
arr
push(10); push(20);
4
3
2
1
0
Dr. Mili Dhar

Data Structure Lecture Array and Recursion.pdf

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