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”.
Examples:
Input: pushed = { 1, 2, 3, 4, 5 }, popped = { 4, 5, 3, 2, 1 }
Output: True
Following sequence can be performed:
push(1), push(2), push(3), push(4), pop() -> 4,
push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1
Input: pushed = { 1, 2, 3, 4, 5 }, popped = { 4, 3, 5, 1, 2 }
Output: False
1 can't be popped before 2.
Approach: 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. If empty then print True else print False.
Below is the implementation of above approach:
C++
// C++ implementation of above approach #include <iostream> #include <stack> using namespace std; // Function to check validity of stack sequence bool validateStackSequence(int pushed[], int popped[], int len){ // maintain count of popped elements int j = 0; // an empty stack stack <int> st; for(int i = 0; i < len; i++){ st.push(pushed[i]); // check if appended value is next to be popped out while (!st.empty() && j < len && st.top() == popped[j]){ st.pop(); j++; } } return j == len; } // Driver code int main() { int pushed[] = {1, 2, 3, 4, 5}; int popped[] = {4, 5, 3, 2, 1}; int len = sizeof(pushed)/sizeof(pushed[0]); cout << (validateStackSequence(pushed, popped, len) ? "True" : "False"); return 0; } // This code is contributed by Rituraj Jain |
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Java
// Java program for above implementation import java.util.*; class GfG { // Function to check validity of stack sequence static boolean validateStackSequence(int pushed[], int popped[], int len) { // maintain count of popped elements int j = 0; // an empty stack Stack<Integer> st = new Stack<>(); for (int i = 0; i < len; i++) { st.push(pushed[i]); // check if appended value // is next to be popped out while (!st.empty() && j < len && st.peek() == popped[j]) { st.pop(); j++; } } return j == len; } // Driver code public static void main(String[] args) { int pushed[] = {1, 2, 3, 4, 5}; int popped[] = {4, 5, 3, 2, 1}; int len = pushed.length; System.out.println((validateStackSequence(pushed, popped, len) ? "True" : "False")); } } /* This code contributed by PrinciRaj1992 */ |
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Python3
# Function to check validity of stack sequence def validateStackSequence(pushed, popped): # maintain count of popped elements j = 0 # an empty stack stack = [] for x in pushed: stack.append(x) # check if appended value is next to be popped out while stack and j < len(popped) and stack[-1] == popped[j]: stack.pop() j = j + 1 return j == len(popped) # Driver code pushed = [1, 2, 3, 4, 5] popped = [4, 5, 3, 2, 1] print(validateStackSequence(pushed, popped)) |
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C#
// C# program for above implementation using System; using System.Collections.Generic; class GfG { // Function to check validity of stack sequence static bool validateStackSequence(int []pushed, int []popped, int len) { // maintain count of popped elements int j = 0; // an empty stack Stack<int> st = new Stack<int>(); for (int i = 0; i < len; i++) { st.Push(pushed[i]); // check if appended value // is next to be popped out while (st.Count != 0 && j < len && st.Peek() == popped[j]) { st.Pop(); j++; } } return j == len; } // Driver code public static void Main(String[] args) { int []pushed = {1, 2, 3, 4, 5}; int []popped = {4, 5, 3, 2, 1}; int len = pushed.Length; Console.WriteLine((validateStackSequence(pushed, popped, len) ? "True" : "False")); } } // This code contributed by Rajput-Ji |
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PHP
<?php // PHP implementation of above approach // Function to check validity of stack sequence function validateStackSequence($pushed, $popped, $len) { // maintain count of popped elements $j = 0; // an empty stack $st = array(); for($i = 0; $i < $len; $i++) { array_push($st, $pushed[$i]); // check if appended value is next // to be popped out while (!empty($st) && $j < $len && $st[count($st) - 1] == $popped[$j]) { array_pop($st); $j++; } } return $j == $len; } // Driver code $pushed = array(1, 2, 3, 4, 5); $popped = array(4, 5, 3, 2, 1); $len = count($pushed); echo (validateStackSequence($pushed, $popped, $len) ? "True" : "False"); // This code is contributed by mits ?> |
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Output:
True
Time Complexity: O(N), where N is size of stack.
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