Check if an array is stack sortable

Given an array of N distinct elements where elements are between 1 and N both inclusive, check if it is stack-sortable or not. An array A[] is said to be stack sortable if it can be stored in another array B[], using a temporary stack S. The operations that are allowed on array are:

1. Remove the starting element of array A[] and push it into the stack.
2. Remove the top element of the stack S and append it to the end of array B.

If all the element of A[] can be moved to B[] by performing these operations such that array B is sorted in ascending order, then array A[] is stack sortable. 

Examples:

Input : A[] = { 3, 2, 1 }
Output : YES
Explanation : 
Step 1: Remove the starting element of array A[] 
        and push it in the stack S. ( Operation 1)
        That makes A[] = { 2, 1 } ; Stack S = { 3 }
Step 2: Operation 1
        That makes A[] = { 1 } Stack S = { 3, 2 }
Step 3: Operation 1
        That makes A[] = {} Stack S = { 3, 2, 1 }
Step 4: Operation 2
        That makes Stack S = { 3, 2 } B[] = { 1 }
Step 5: Operation 2
        That makes Stack S = { 3 } B[] = { 1, 2 }
Step 6: Operation 2
        That makes Stack S = {} B[] = { 1, 2, 3 }
  
Input : A[] = { 2, 3, 1}
Output : NO

Given, array A[] is a permutation of [1, …, N], so let us suppose the initially B[] = {0}. Now we can observe that:

1. We can only push an element in the stack S if the stack is empty or the current element is less than the top of the stack.
2. We can only pop from the stack only if the top of the stack is as the array B[] will contain {1, 2, 3, 4, …, n}.

If we are not able to push the starting element of the array A[], then the given array is Not Stack Sortable. Below is the implementation of above idea: 

C++

// C++ implementation of above approach.
#include <bits/stdc++.h>
using namespace std;

// Function to check if A[] is
// Stack Sortable or Not.
bool check(int A[], int N)
{
    // Stack S
    stack<int> S;

    // Pointer to the end value of array B.
    int B_end = 0;

    // Traversing each element of A[] from starting
    // Checking if there is a valid operation
    // that can be performed.
    for (int i = 0; i < N; i++)
    {
        // If the stack is not empty
        if (!S.empty())
        {
            // Top of the Stack.
            int top = S.top();

            // If the top of the stack is
            // Equal to B_end+1, we will pop it
            // And increment B_end by 1.
            while (top == B_end + 1)
            {
                // if current top is equal to
                // B_end+1, we will increment
                // B_end to B_end+1
                B_end = B_end + 1;

                // Pop the top element.
                S.pop();

                // If the stack is empty We cannot
                // further perform this operation.
                // Therefore break
                if (S.empty())
                {
                    break;
                }

                // Current Top
                top = S.top();
            }

            // If stack is empty
            // Push the Current element
            if (S.empty()) {
                S.push(A[i]);
            }
            else
            {
                top = S.top();

                // If the Current element of the array A[]
                // if smaller than the top of the stack
                // We can push it in the Stack.
                if (A[i] < top)
                {
                    S.push(A[i]);
                }
                // Else We cannot sort the array
                // Using any valid operations.
                else
                {
                    // Not Stack Sortable
                    return false;
                }
            }
        }
        else
        {
            // If the stack is empty push the current
            // element in the stack.
            S.push(A[i]);
        }
    }

    // Stack Sortable
    return true;
}

// Driver's Code
int main()
{
    int A[] = { 4, 1, 2, 3 };
    int N = sizeof(A) / sizeof(A[0]);
    check(A, N)? cout<<"YES": cout<<"NO";
    return 0;
}

Java

// Java implementation of above approach.
import java.util.Stack;

class GFG {

// Function to check if A[] is
// Stack Sortable or Not.
    static boolean check(int A[], int N) {
        // Stack S
        Stack<Integer> S = new Stack<Integer>();

        // Pointer to the end value of array B.
        int B_end = 0;

        // Traversing each element of A[] from starting
        // Checking if there is a valid operation
        // that can be performed.
        for (int i = 0; i < N; i++) {
            // If the stack is not empty
            if (!S.empty()) {
                // Top of the Stack.
                int top = S.peek();

                // If the top of the stack is
                // Equal to B_end+1, we will pop it
                // And increment B_end by 1.
                while (top == B_end + 1) {
                    // if current top is equal to
                    // B_end+1, we will increment
                    // B_end to B_end+1
                    B_end = B_end + 1;

                    // Pop the top element.
                    S.pop();

                    // If the stack is empty We cannot
                    // further perform this operation.
                    // Therefore break
                    if (S.empty()) {
                        break;
                    }

                    // Current Top
                    top = S.peek();
                }

                // If stack is empty
                // Push the Current element
                if (S.empty()) {
                    S.push(A[i]);
                } else {
                    top = S.peek();

                    // If the Current element of the array A[]
                    // if smaller than the top of the stack
                    // We can push it in the Stack.
                    if (A[i] < top) {
                        S.push(A[i]);
                    } // Else We cannot sort the array
                    // Using any valid operations.
                    else {
                        // Not Stack Sortable
                        return false;
                    }
                }
            } else {
                // If the stack is empty push the current
                // element in the stack.
                S.push(A[i]);
            }
        }

        // Stack Sortable
        return true;
    }

// Driver's Code
    public static void main(String[] args) {

        int A[] = {4, 1, 2, 3};
        int N = A.length;

        if (check(A, N)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }

    }
}
//This code is contributed by PrinciRaj1992

Python3

# Python implementation of above approach


def check(A, N):

    # Stack S
    S = []

    # Pointer to the end value of array B.
    B_end = 0

    # Traversing each element of A[] from starting
    # Checking if there is a valid operation
    # that can be performed.
    for i in range(N):

        # if Stack is not empty
        if len(S) != 0:
            # top of the stack
            top = S[-1]

            # If the top of the stack is
            # Equal to B_end+1, we will pop it
            # And increment B_end by 1.
            while top == B_end + 1:

                # if current top is equal to
                # B_end+1, we will increment
                # B_end to B_end+1
                B_end = B_end + 1

                # Pop the top element
                S.pop()

                # If the stack is empty We cannot
                # further perform this operation.
                # Therefore break
                if len(S) == 0:
                    break

                # Current top
                top = S[-1]

            # If stack is empty
            # Push the Current element
            if len(S) == 0:
                S.append(A[i])
            else:
                top = S[-1]

                # If the Current element of the array A[]
                # if smaller than the top of the stack
                # We can push it in the Stack.
                if A[i] < top:
                    S.append(A[i])

                # Else We cannot sort the array
                # Using any valid operations.
                else:
                    # Not Stack Sortable
                    return False

        else:
            # If the stack is empty push the current
            # element in the stack.
            S.append(A[i])

    return True


# Driver's Function
if __name__ == "__main__":
    A = [4, 1, 2, 3]
    N = len(A)
    if check(A, N):
        print("YES")
    else:
        print("NO")

C#

using System;
using System.Collections.Generic;

class GFG
{

  // Function to check if A[] is
  // Stack Sortable or Not.
  static bool check(int[] A, int N)
  {

    // Stack S
    Stack<int> S = new Stack<int>();

    // Pointer to the end value of array B.
    var B_end = 0;

    // Traversing each element of A[] from starting
    // Checking if there is a valid operation
    // that can be performed.
    for (int i = 0; i < N; i++)
    {

      // If the stack is not empty
      if (S.Count != 0)
      {

        // Top of the Stack.
        int top = S.Peek();

        // If the top of the stack is
        // Equal to B_end+1, we will pop it
        // And increment B_end by 1.
        while (top == B_end + 1)
        {

          // if current top is equal to
          // B_end+1, we will increment
          // B_end to B_end+1
          B_end = B_end + 1;

          // Pop the top element.
          S.Pop();

          // If the stack is empty We cannot
          // further perform this operation.
          // Therefore break
          if (S.Count == 0) {
            break;
          }

          // Current Top
          top = S.Peek();
        }

        // If stack is empty
        // Push the Current element
        if (S.Count == 0) {
          S.Push(A[i]);
        }
        else {
          top = S.Peek();

          // If the Current element of the array
          // A[] if smaller than the top of the
          // stack We can push it in the Stack.
          if (A[i] < top) {
            S.Push(A[i]);
          }
          else
          {

            // Not Stack Sortable
            return false;
          }
        }
      }
      else
      {

        // If the stack is empty push the current
        // element in the stack.
        S.Push(A[i]);
      }
    }

    // Stack Sortable
    return true;
  }

  // Driver's Code
  public static void Main(string[] args)
  {
    int[] A = {4, 1, 2, 3};
    int N = A.Length;
    if (check(A, N)) {
      Console.WriteLine("YES");
    }
    else {
      Console.WriteLine("NO");
    }
  }
}

// This code is contributed by aadityaburujwale.

Javascript

// JS implementation of above approach.


// Function to check if A[] is
// Stack Sortable or Not.
function check(A, N) {
    // Stack S
    let S = [];

    // Pointer to the end value of array B.
    let B_end = 0;

    // Traversing each element of A[] from starting
    // Checking if there is a valid operation
    // that can be performed.
    for (let i = 0; i < N; i++) {
        // If the stack is not empty
        if (S.length != 0) {
            // Top of the Stack.
            let top = S[0];

            // If the top of the stack is
            // Equal to B_end+1, we will pop it
            // And increment B_end by 1.
            while (top == B_end + 1) {
                // if current top is equal to
                // B_end+1, we will increment
                // B_end to B_end+1
                B_end = B_end + 1;

                // Pop the top element.
                S.shift();

                // If the stack is empty We cannot
                // further perform this operation.
                // Therefore break
                if (S.length == 0) {
                    break;
                }

                // Current Top
                top = S[0];
            }

            // If stack is empty
            // Push the Current element
            if (S.length != 0) {
                S.push(A[i]);
            }
            else {
                top = S[0];

                // If the Current element of the array A[]
                // if smaller than the top of the stack
                // We can push it in the Stack.
                if (A[i] < top) {
                    S.push(A[i]);
                }
                // Else We cannot sort the array
                // Using any valid operations.
                else {
                    // Not Stack Sortable
                    return false;
                }
            }
        }
        else {
            // If the stack is empty push the current
            // element in the stack.
            S.push(A[i]);
        }
    }

    // Stack Sortable
    return true;
}

// Driver's Code
let A = [4, 1, 2, 3];
let N = A.length;
check(A, N) ? console.log("YES") : console.log("NO");


// This code is contributed by adityamaharshi21

Output:

YES

Time Complexity: O(N)

Auxiliary Space: O(N) because using stack