Check if the given push and pop sequences of Stack is valid or not - GeeksforGeeks

14 May, 2021
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++
Java
Python3
C#
PHP
Javascript
// C++ implementation of above approach#include <iostream>#include <stack> using namespace std; // Function to check validity of stack sequencebool 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 codeint 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
// Java program for above implementationimport 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 */
# Function to check validity of stack sequencedef 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 codepushed = [1, 2, 3, 4, 5]popped = [4, 5, 3, 2, 1]print(validateStackSequence(pushed, popped))
// C# program for above implementationusing 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
<?php// PHP implementation of above approach // Function to check validity of stack sequencefunction 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?>
<script> // Javascript implementation of above approach function validateStackSequence( pushed, popped, len){         // maintain count of popped elements    var j = 0;         // an empty stack    var st=[];    for(var i = 0; i < len; i++){        st.push(pushed[i]);                 // check if appended value is next        // to be popped out        while (!st.length==0 && j < len &&        st[st.length-1] == popped[j])        {            st.pop();            j++;        }    }         return j == len;} var pushed = [1, 2, 3, 4, 5];   var popped = [4, 5, 3, 2, 1];   var len = pushed.length;      document.write( (validateStackSequence(pushed, popped, len)  ? "True" : "False"));  // This code is contributed by SoumikMondal </script>
True
 
Time Complexity: O(N), where N is size of stack. 
rituraj_jain
princiraj1992
Rajput-Ji
Mithun Kumar
SoumikMondal
Arrays
Python Programs
Stack
Arrays
Stack
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