Given push and pop sequences with distinct values. The task is to check if this could have been the result of a sequence of push and pop operations on an initially empty stack. Return “True” if it otherwise returns “False”.
If the element X has been pushed to the stack then check if the top element in the pop sequence is X or not.
If it is X then pop it right now else top of the push sequence will be changed and make the sequences invalid. So, similarly, do the same for all the elements and check if the stack is empty or not in the last.
This approach simulates the stack operations using the pushed array and checks if the final stack state matches the popped array. If the sequence is valid, the program returns True, else it returns False.
Two integer variables i and j are initialized to 0, representing the indices of the pushed and popped arrays, respectively.
For each element val in the pushed array, the program first pushes it onto the simulated stack by incrementing the i index and setting the value of pushed[i] to val.
The program then checks if the top element of the simulated stack (pushed[i-1]) is equal to the next element to be popped (popped[j]). If they are equal, it means that the top element of the simulated stack matches the next element to be popped, so the program pops the element from the simulated stack by decrementing the i index and incrementing the j index.
Steps 2-3 are repeated for each element in the pushed array.
If the pushed array has been fully processed and the simulated stack is empty (i == 0), then the sequence is valid and the function returns True. Otherwise, the sequence is invalid and the function returns False.
Note: this approach does not use an actual stack data structure, but instead simulates the stack using the pushed array itself, which acts as a stack. The program uses two pointers i and j to keep track of the current state of the simulated stack and the next element to be popped, respectively.
Below is the implementation of the above approach:
// Function to check validity of stack sequence using
Please Login to comment...