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vitfil [10]
3 years ago
5

Consider the following Stack operations:

Computers and Technology
1 answer:
maxonik [38]3 years ago
6 0

Answer:

Sequence of popped values: h,s,f.

State of stack (from top to bottom): m, d

Explanation:

Assuming that stack is  initially empty. Suppose that p contains the popped values. The state of the stack is where the top and bottom are pointing to in the stack. The top of the stack is that end of the stack where the new value is entered and existing values is removed. The sequence works as following:

push(d) -> enters d to the Stack

Stack:  

d ->top

push(h) -> enters h to the Stack

Stack:

h ->top

d ->bottom

pop() -> removes h from the Stack:

Stack:

d ->top

p: Suppose p contains popped values so first popped value entered to p is h

p = h

push(f) -> enters f to the Stack

Stack:

f ->top

d ->bottom

push(s) -> enters s to the Stack

Stack:

s ->top

f

d ->bottom

pop() -> removes s from the Stack:

Stack:

f ->top

d -> bottom

p = h, s

pop() -> removes f from the Stack:

Stack:

d ->top

p = h, s, f

push(m) -> enters m to the Stack:

Stack:

m ->top

d ->bottom

So looking at p the sequence of popped values is:

h, s, f

the final state of the stack:

m, d

end that is the top of the stack:

m

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30 points) Suppose you are given a string containing only the characters ( and ). In this problem, you will write a function to
navik [9.2K]

Answer:

The screenshot is attached below

Explanation:

The below program check balence of paranthesis using stack.

if the input charecter is ( then push to stack

if the input charecter is ) then pop top element and compair both elements for match.

Program:

public class Main

{

static class stack // create class for stack

{

int top=-1; // initialy stack is empty so top = -1

char items[] = new char[50]; // Create an array for stack to store stack elements

 

void push(char a) // Function push element to stack

{

if (top == 49) // chack stack is full ?

{

System.out.println("Stack full");

}

else

{

items[++top] = a; // increment the array index and store the element

}

}

 

char pop() // function to return the elements from stack

{

if (top == -1) // check the stack is empty

{

System.out.println("Empty stack");

return '\0';

}

else

{

char element = items[top]; // return the top element

top--; // Decrement the array index by one

return element;

}

}

 

boolean isEmpty() // Check the stack is empty or not

{

return (top == -1) ? true : false;

}

}

static boolean isMatch(char character1, char character2) // Check the input charecters are matching pairs or not

{

if (character1 == '(' && character2 == ')') // If they match return true

return true;

else

return false;

}

 

static boolean isBalanced(String parantheses) // check the input string is balenced or not

{

char[] exp = new char[parantheses.length()]; // Create a charecter array

for (int i = 0; i < parantheses.length(); i++) { // Convert the string parantheses to charecter array

exp[i] = parantheses.charAt(i);

}

stack st=new stack(); // Declare an empty character stack    

for(int i=0;i<exp.length;i++)

{  

if (exp[i] == '('))   // if input charecter is '(' push to stack

st.push(exp[i]);

if (exp[i] == ')') // if input charecter is ')' pop top element from stack

{

if (st.isEmpty())

{

return false;

}

 

else if ( !isMatch(st.pop(), exp[i]) ) // Call isMatch function

{

return false;

}

}

 

}

 

if (st.isEmpty())

return true; //balanced

else

{

return false; //not balanced

}

}

 

 

public static void main(String[] args)

{

 

System.out.println(isBalanced("()()()")); // Output true if string is balenced

 

}

 

}

Screenshot:

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Answer:

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import java.text.SimpleDateFormat;

import java.time.LocalDateTime;

import java.time.format.DateTimeFormatter;

import java.util.Date;

import java.util.Scanner;  

public class Main {

   private static final DateFormat dateformat = new SimpleDateFormat("yyyy/MM/dd HH:mm:ss");

   private static final DateTimeFormatter datetimeformat = DateTimeFormatter.ofPattern("yyyy/MM/dd HH:mm:ss");

   public static void main(String[] args) {

       

       System.out.println("Enter your ID and Name");

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       Scanner sc1 = new Scanner(System.in);  

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       int ID = sc1.nextInt();

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}

Explanation:

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Answer:

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{

public static void main(String[] args) {

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}

public static int min(int n1, int n2, int n3){

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}

}

Explanation:

*The code is in Java.

Create a method named min that takes three parameters, n1, n2, and n3

Inside the method:

Call the method Math.min() to find the smallest among n1 and n2. Then, pass the result of this method to Math.min() again with n3 to find the min among three of them and return it. Note that Math.min() returns the smallest number among two parameters.

In the main:

Call the method with parameters given in the example and print the result

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3 years ago
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