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

  • Difficulty Level : Easy
  • Last Updated : 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++




// 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
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 */
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))
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
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
?>
Javascript



<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>
Output: 
True
 
Time Complexity: O(N), where N is size of stack.
 

My Personal Notes
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Sanjit_Prasad
@Sanjit_Prasad
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