# Check if an array is stack sortable

• Difficulty Level : Easy
• Last Updated : 01 Jun, 2022

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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}.
3. 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 ``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 perfom 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);``    ``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 S = ``new` `Stack();`` ` `        ``// 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 perfom 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 perfom 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"``)`

Output:

```YES
```

Time Complexity: O(N).

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