What is the value of x after the following code? stack S; S.push(3); S.push(2); S.pop0; intx=S.top) S.push(5); S.pop0; x=x+S.top0;
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A: The original queue(myData) is 12, 24, 36 where 12 is front. Thus 36 is rear.
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A: Here we have given complete code for the asked program. you can find the solution in step 2.
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Q: What is the value of x after the following code? stack S; S.push(17); S.push(5); S.push(3); S.pop0;…
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A: Lets see the solution.
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Q: Given a stack myData: 34, 56, 78, 12, 66 (top is 34) what is the output after the following…
A: Given:-
Q: What is the sum of all the removed values in the following sequence of operations using generic push…
A: The answer is:24. Explanation below:
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A: This is a basic stack question from Java.
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A:
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A:
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A: Pushing 4: Pushing 5: Popping:
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- public interface StackInterface void push (T element) throws StackoverfloWException; void pop () throws StackUnderflowException; T top () throws StackUnderflowException; boolean isFul1 (); boolean isEmpty(); Show what is written by the following segments of code (see printın on the code), given that iteml, item2, and item3 are int variables, and ali is an object that fits the abstract description of a stack as shown above StackInterface. Assume that you can store and retrieve variables of type int on ali. iteml = 2; item2 = 0; item3 = 6; ali.push (item2); ali.push (item1); ali.push ( iteml+item3); item2 = ali.top ( ); ali.push (item3*item3); ali.push (item2); ali.push (3); iteml = ali.top ( ); ali.pop( ); System.out.println (iteml + " " + item2 + while (!ali.isEmpty( )) + item3); iteml = ali.top( ) ; ali.pop () ; System.out.println (item1) ;Homework* - Prime factors and prime numbers: Write a program that prompts the user to enter a positive integer which is greater than one and displays all its smallest factors in decreasing order. For example, if the integer is 140, the smallest factors are displayed as 7, 5, 2, 2. Use the StackOflntegers class to store the factors (e.g., 2, 2, 5, 7) and retrieve and display them in reverse order. After that, write a program that displays all the prime numbers less than the number – which is used in the prime factors (e.g: 140)- in decreasing order. Use the StackOflntegers class to store the prime numbers (e.g., 2, 3, 5, ...) and retrieve and display them in reverse order. Here is the sample runs: To use the prime factors you must enter a positive integer which is greater than one: 140 7 5 2 2 Printing prime numbers under 140: 139 137 131 127 113 109 107 103 101 97 89 83 79 73 71 67 61 59 53 47 43 41 37 31 29 23 19 17 13 11 7 5 3 2 * This homework has a part of Challenge Question.Create an application that reads a statement from the user and outputs it with the characters of each word reversed. To reverse the letters in each phrase, use a stack.
- Create an application that reads a statement from the user and outputs it with the characters of each word reversed. To reverse the characters in each word, use a stack.Create an application that reads a statement from the user and outputs it with the characters of each word reversed. To reverse the characters in each word, use a stack.What is the value of x after the following code? stack S; S.push(17); S.push(5); S.push(3); S.pop0; int x = S.top() S.push(4); S.pop0; S.pop0; x =x+S.top(0;You are given a stack of 10 matches. Players take turns taking matches from the stack. A player can take 1, 3, 5 or 7 matches at a time. The player who takes the last match wins. Which of the following is a losing state? Select one or more: a. 9 b. 4 c. 8 d. 2
- public void OnSend(View v){ String phoneNumber = number.getText().toString(); String smsmessage = message.getText().toString(); if(phoneNumber == null || phoneNumber.length() == 0 || smsmessage == null || smsmessage.length() == 0){ return; } if(checkPermission(Manifest.permission.SEND_SMS)){ SmsManager smsManager = SmsManager.getDefault(); smsManager.sendTextMessage(phoneNumber, null, smsmessage, null, null); Toast.makeText(this, "Message sent", Toast.LENGTH_SHORT).show(); } Question 2: Explain the following code: button.setOnClickListener(new View.OnClickListener() { @Override public void onClick(View v) { FragmentManager fm = getSupportFragmentManager(); FragmentTransaction ft = fm.beginTransaction(); XFragment xfragment = new XFragment(); ft.add(R.id.Main_fragment, xfragment);…7. What is output by the following code section? QueueInterface aQueue = new QueueReference Based(); int numl, num2; for (int i = 1; i <= 5; i++) { aQueue.enqueue(i); } // end for for (int i = 1; i <= 5; i++) { numl = (Integer) aQueue.dequeue (); num2 (Integer) aQueue.dequeue (); aQueue.enqueue (numl + num2); aQueue.enqueue (num2 - numl); } // end for while (!aQueue.isEmpty()) { System.out.print (aQueue.dequeue () + } // end forget_points("safe", "loft") // => List(A,A,C,A)get_points("safe", "gate") // => List(A,C,A,C)get_points("safe", "star") // => List(C,A,P,A)get_points("safe", "sums") // => List(C,A,A,P) Create a function called get_points which calculates the returns ,as shown above in the tests, by comparing the second word with the first word. The get_points function should output a list of 4 elements of type Letter.if a letter is present in the first word and in the correct place it should return with a "C". If the letter is present but not in the right place it should be a "P".And if the letter is not in the hidden_word then it should an "A" To do this you need to use the finder function to calculate if all the letters that are present in the first wordex. finder("safe", "loft") -> List(f) using the fix_spot function recursively analyse whether if the letter should be A,C or P. The finder function is a wrapper for the fix_spot function calling the fix_spot function with appropriate…
- Stacks can be used to evaluate a post-fix expressions. What is the result if the following postfix expression is evaluated? ´ 7 7 2 3 + 8 * + 3 + *3. Card Flipper: You walk into a room, and see a row of n cards. Each one has a number x; written on it, where i ranges from 1 to n. However, initially all the cards are face down. Your goal is to find a local minimum: that is, a card i whose number is less than or equal to those of its neighbors, xj-1 = X; <= Xj+1. The first and last cards can also be local minima, and they only have one neighbor to compare to. There can be many local minima, but you are only responsible for finding one of them. Obviously you can solve this problem by turning over all n cards, and scanning through them. However, show that you can find such a minimum by turning over only O(log n) cards.What does the following code do? In [1]: def mystery (a, b) : if b == 1: return a else: return a + mystery (a, b - 1) In [2]: mystery (2, 10) Out [2]: ?????