Reverse arrays using Stacks in java

Problem:

Create a java program that reverses an array in java using stacks.

Output:

90 78 40 9 6 3 1
10.5 6.9 3.2 2.5

Solution:

import java.util.Stack;

public class ReverseArrayApp {

 public static void main(String[] args){
  Integer[] arr1 = {1, 3, 6, 9, 40, 78, 90};
  Double[] arr2 = {2.5, 3.2, 6.9, 10.5};
  
 
  Number[] arr1_rev = reverse(arr1);
  Number[] arr2_rev = reverse(arr2);
  
  
  for(Number eltInt : arr1_rev) 
   System.out.print(eltInt + " ");
  System.out.println();
  for(Number eltDouble : arr2_rev)
   System.out.print(eltDouble + " ");
  }

 
 private static Number[] reverse(Number[] arr) {
  Stack<Number> temp = new Stack<>();
  Number [] output = new Number[arr.length];
  
  for(Number elt: arr) 
   temp.push(elt);
  
  int index = 0;
  
  while(!temp.empty()){
   output[index] = temp.pop();
   index++;
  }
  
  return output;
  
 }
 
 
 }
 

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Check if a character is balanced using Stacks in Java

Problem:

Write a java method that checks if a character is balanced using stacks.

Output:

Not applicable.

Solution:

import java.util.EmptyStackException;
import java.util.Scanner;
import java.util.Stack;

public class ParanChecker
{
  public static void main(String[] args)
  {
    //We assume that the Character is balanced from the beginning
    boolean balanced = true;
    
    String OPEN = "([{";
    String CLOSE = ")]}";
    
    Scanner scan = new Scanner(System.in);
    Stack <Character> s = new Stack<Character>();
    
    System.out.println("Enter your expression:");
    String expression = scan.nextLine();
    
    try 
    {
      int index = 0;
      
      while (balanced && index < expression.length())
      {
        char nextCh = expression.charAt(index);
        if ( OPEN.indexOf(nextCh) != -1 )
          s.push(nextCh);
        else if ( CLOSE.indexOf(nextCh) != -1)
        {
          char topCh = s.pop();
          balanced = (OPEN.indexOf(topCh) == CLOSE.indexOf(nextCh));
        }
      index++;    
      }
    }
    
    catch(EmptyStackException emptyStackException)
    {
      balanced = false;
    }
    
    if ( balanced && s.empty())
      System.out.println("The Character is balanced");
    else
      System.out.println("The Character is not balanced");
  }
}

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Convert from Decimal to Binary using Stacks in Java

Problem:

To convert a number from decimal to binary, you simply divide by two until a quotient of zero is reached, then use the successive remainders in reverse order as the binary representation. Use a stack to store the remainder of the division and finally print the binary representation. The first line of input n is the number test cases. Each test case begins with an integer t representing the number of integers in the list, then t integers follow.

Output:

3
5
7
10

101
111
1010

Solution:

import java.util.Scanner;
import java.util.Stack;

public class ProblemD {

 public static void main(String[] args) {
  
  Stack<Integer> stack = new Stack<Integer>();
  
  Scanner scan = new Scanner(System.in);
  
  int numberOfExpressions = scan.nextInt();
  int[] num = new int[numberOfExpressions];
  
  for (int i = 0;i<numberOfExpressions;i++)
  {
   num[i] = scan.nextInt();
  }
  
  for (int i = 0;i<numberOfExpressions;i++)
  {
   if (num[i] == 0) {
                System.out.println("0000");
   } else if (num[i] == 1) {
                System.out.println("0001");
   } else {
                while (num[i] != 0) {
                        int a = num[i] / 2;
                        int b = num[i] % 2;
                        stack.push(b);
                        num[i] = a;
                }
        }
   
   while (!stack.empty()) {
                System.out.print(stack.pop());
        }

   System.out.println(" ");
  }

 }
}

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Calculate Postfix Expressions using Queues in Java

Problem:

You are given a list of postfix expressions. Calculate the result of the expressions using a stack. The first line of input n is the number of test cases followed by n expressions. Each expression begins with an integer t representing the number of characters in the expression, and then t characters follow which are either a number or an operand.

Output:

3
3 2 3 *
5 5 4 2 + -
9 5 6 2 + * 12 4 / -

6
1
37

Solution:

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
public class ProblemC {

 public static void main(String[] args) {
  
  Queue<Integer> q = new LinkedList<Integer>();
  String output = "";
  Scanner scan = new Scanner(System.in);
  
  int testCases = scan.nextInt();
  
  for (int cases = 0; cases < testCases; cases++) {
   
   int t = scan.nextInt();
   
   for (int i = 0; i < t; i++)
    q.offer(scan.nextInt());
   
   output +=( "Test case "+(cases+1)+":" + "\n");
   
   while (!q.isEmpty()) {
    if (q.peek()%2 == 0)
     output += (q.poll() + "\n");
    else {
     output+= q.poll()+" ";
     output += q.poll() + "\n";
    }
   }
  }
  
  System.out.println(output);
 }
}

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Check if a String is a Palindrome using Stacks in Java

Problem:

You are given a list of Strings as input. Check whether each string is a palindrome or not using a stack data structure. The first line of input n is the number of test cases followed by n strings.

Output:

3
test
baab
uigui

No
Yes
Yes

Solution:

import java.util.Scanner;
import java.util.Stack;

public class ProblemA {

 public static void main(String[] args) {
  
  Stack<Character> stack = new Stack<Character>();
  
  Scanner scan = new Scanner(System.in);
  
  int numberOfExpressions = scan.nextInt();
  String[] expression = new String[numberOfExpressions];
  
  for (int i = 0;i<numberOfExpressions;i++)
  {
   expression[i] = scan.next();
  }
  
  for (int i = 0;i<numberOfExpressions;i++)
  {      
   String checker = "";
   
   for (int j = 0; j < expression[i].length(); j++)
    stack.push(expression[i].charAt(j));
   
   while (!stack.isEmpty()) 
    checker += stack.pop();
   
   if (checker.equals(expression[i])) 
    System.out.println("Yes");
   else
    System.out.println("No");
  
  }
 }
}

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