unit-3 test-2

Give the correct option form the question

• a

• b

• c

• d

Answer explanation

Give the correct option from the question

• a

• b

• c

• d

Select the correct option from the question

• a

• b

• c

• d

select the correct option from the question

• a

• b

• c

• d

select the option form the question

• a

• b

• c

• d

Answer explanation

select the correct option from the question

• a

• b

• c

• d

Answer explanation

Algorithm for converting infix to postfix notation 1. Initialize an empty stack and an empty output string. 2. Scan the infix expression from left to right. 3. If the current element is an operand, append it to the output string. 4. If the current element is an operator, then: a. Pop operators from the stack until an operator with lower precedence than the current operator is found, and append them to the output string. b. Push the current operator onto the stack. 5. If the current element is a left parenthesis, push it onto the stack. 6. If the current element is a right parenthesis, then: a. Pop operators from the stack and append them to the output string until a left parenthesis is found. b. Discard the left parenthesis from the stack. 7. Repeat steps 3 to 6 until all elements have been scanned. 8. Pop any remaining operators from the stack and append them to the output string. Maximum number of symbols on the stack In the given infix expression, the maximum number of symbols on the stack at one time can be calculated as follows: 1. Initialize an empty stack. 2. Scan the infix expression from left to right. 3. Keep track of the maximum size of the stack during the scan. 4. If the current element is an operand, do not push anything onto the stack. 5. If the current element is an operator or a parenthesis, push it onto the stack. 6. If the current element is a right parenthesis, pop all elements from the stack until a left parenthesis is found and count the number of symbols popped. 7. Update the maximum size of the stack if it is greater than the previous maximum. 8. After scanning the entire expression, return the maximum size of the stack. For the given infix expression, the maximum size of the stack can be calculated as follows: 1. Initialize an empty stack and a variable to keep track of the maximum size. 2. Scan the infix expression from left to right. 3. The first element is 4, an operand. Do not push anything onto the stack. 4. The next element is 3, an operand. Do not push anything onto the stack. 5. The next element is , an operator. Push it onto the stack. 6. The next element is (, a left parenthesis. Push it onto the stack. 7. The next element is 6, an operand. Do not push anything onto the stack. 8. The next element is , an operator. Push it onto the stack. 9. The next element is 3, an operand. Do not push anything onto the stack. 10. The next element is -, an operator. Push it onto the stack. 11. The next element is 12, an operand. Do not push anything onto the stack. 12. The next element is ), a right parenthesis. Pop all elements from the stack until a left parenthesis is found. The symbols popped are - * 6 3. The size of the stack is 1 after the left parenthesis is discarded. 13. Update the maximum size of the stack to 4. 14. The last element is . Do not push anything onto the stack. 15. Return the maximum size of the stack, which is 4.

Therefore, the correct answer is option D, 4.

select the option from the given question

• a

• b

• c

• d

Answer explanation

And top2) are used to indicate the top elements of both stacks. Explain how this array can be used to implement two stacks and discuss the time complexity of the operations. To implement two stacks using a single array, we can allocate the first half of the array to one stack and the second half to the other stack. For example, if the array size is MAXSIZE, we can allocate the first MAXSIZE/2 elements to the first stack and the remaining MAXSIZE/2 elements to the second stack. To keep track of the top elements of both stacks, we can use two variables top1 and top2. top1 will initially point to the first element of the first stack, and top2 will initially point to the last element of the second stack. As elements are pushed onto or popped from either stack, these variables will be updated accordingly. To push an element onto the first stack, we can increment top1 and store the element at A[top1]. Similarly, to push an element onto the second stack, we can decrement top2 and store the element at A[top2]. To pop an element from the first stack, we can read the element at A[top1] and then decrement top1. Similarly, to pop an element from the second stack, we can read the element at A[top2] and then increment top2. The time complexity of these operations depends on the implementation. If we use a simple array with no additional data structures, the time complexity of push and pop operations will be O(1) as they involve simple array access and update operations. However, if we want to ensure that both stacks have equal size and avoid overflow or underflow, we can use additional data structures such as a counter or a flag to keep track of the number of elements in each stack. This will increase the time complexity of push and pop operations to O(2) or O(3) as we may need to update the counter or flag in addition to the array access and update operations.

select the answer from the question

• a

• c

• c

• d

Answer explanation

The given sequence of operations involves pushing and popping elements in a stack, as well as enqueuing and dequeuing elements in a queue. We need to determine the value of variables 's' and 'q' after performing these operations. 1. Stack Operations: - Push(54): This operation pushes the element 54 onto the stack. - Push(52): This operation pushes the element 52 onto the stack. - Pop(): This operation removes and returns the top element of the stack, which is 52 in this case. - Push(55): This operation pushes the element 55 onto the stack. - Push(62): This operation pushes the element 62 onto the stack. - s = Pop(): This operation removes and returns the top element of the stack, which is 62 in this case. Thus, the value of 's' is 62. 2. Queue Operations: - Enqueue(21): This operation adds the element 21 to the queue. - Enqueue(24): This operation adds the element 24 to the queue. - Dequeue(): This operation removes and returns the front element of the queue, which is 21 in this case. - Enqueue(28): This operation adds the element 28 to the queue. - Enqueue(32): This operation adds the element 32 to the queue. - q = Dequeue(): This operation removes and returns the front element of the queue, which is 24 in this case. Thus, the value of 'q' is 24.

Therefore, the value of 's' is 62 and the value of 'q' is 24. The correct answer is option 'A' (86)

select the correct option from the question

• a

• b

• c

• d

Answer explanation

select the correct option from the question

• a

• b

• c

• d

Answer explanation

To DEQUEUE an item, simply POP. To ENQUEUE an item, we can do following 3 operations 1) REVERSE 2) PUSH 3) REVERSE