# Sudo Placement[1.3] | Playing with Stacks

• Difficulty Level : Expert
• Last Updated : 19 Jul, 2021

You are given 3 stacks, A(Input Stack), B(Auxiliary Stack) and C(Output Stack). Initially stack A contains numbers from 1 to N, you need to transfer all the numbers from stack A to stack C in sorted order i.e in the end, the stack C should have smallest element at the bottom and largest at top. You can use stack B i.e at any time you can push/pop elements to stack B also. At the end stack A, B should be empty.

Examples:

Input: A = {4, 3, 1, 2, 5}
Output: Yes 7

Input: A = {3, 4, 1, 2, 5}
Output: No

Approach: Iterate from the bottom of the given stack. Initialize required as the bottom most element in stackC at the end i.e., 1. Follow the given below algorithm to solve the above problem.

• if the stack element is equal to the required element, then the number of transfers will be one which is the count of transferring from A to C.
• if it is not equal to the required element, then check if it is possible to transfer it by comparing it with the topmost element in the stack.
1. If the topmost element in stackC is greater than the stackA[i] element, then it is not possible to transfer it in a sorted way,
2. else push the element to stackC and increment transfer.
• Iterate in the stackC and pop out the top most element until it is equal to the required and increment required and transfer in every steps.

Below is the implementation of the above approach:

## C++

 `// C++ program for``// Sudo Placement | playing with stacks``#include ``using` `namespace` `std;` `// Function to check if it is possible``// count the number of steps``void` `countSteps(``int` `sa[], ``int` `n)``{` `    ``// Another stack``    ``stack<``int``> sc;` `    ``// variables to count transfers``    ``int` `required = 1, transfer = 0;` `    ``// iterate in the stack in reverse order``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// if the last element has to be``        ``// inserted by removing elements``        ``// then count the number of steps``        ``if` `(sa[i] == required) {``            ``required++;``            ``transfer++;``        ``}``        ``else` `{``            ``// if stack is not empty and top element``            ``// is smaller than current element``            ``if` `(!sc.empty() && sc.top() < sa[i]) {``                ``cout << ``"NO"``;``                ``return``;``            ``}``            ``// push into stack and count operation``            ``else` `{` `                ``sc.push(sa[i]);``                ``transfer++;``            ``}``        ``}``        ``// stack not empty, then pop the top element``        ``// pop out all elements till is it equal to required``        ``while` `(!sc.empty() && sc.top() == required) {``            ``required++;``            ``sc.pop();``            ``transfer++;``        ``}``    ``}` `    ``// print the steps``    ``cout << ``"YES "` `<< transfer;``}` `// Driver Code``int` `main()``{``    ``int` `sa[] = { 4, 3, 1, 2, 5 };``    ``int` `n = ``sizeof``(sa) / ``sizeof``(sa);``    ``countSteps(sa, n);``    ``return` `0;``}`

## Java

 `// Java program for Sudo``// Placement | playing with stacks``import` `java.util.*;` `class` `GFG``{` `    ``// Function to check if it is possible``    ``// count the number of steps``    ``static` `void` `countSteps(``int` `sa[], ``int` `n)``    ``{` `        ``// Another stack``        ``Stack sc = ``new` `Stack();` `        ``// variables to count transfers``        ``int` `required = ``1``, transfer = ``0``;` `        ``// iterate in the stack in reverse order``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{` `            ``// if the last element has to be``            ``// inserted by removing elements``            ``// then count the number of steps``            ``if` `(sa[i] == required)``            ``{``                ``required++;``                ``transfer++;``            ``}``            ``else``            ``// if stack is not empty and top element``            ``// is smaller than current element``            ``if` `(!sc.empty() && sc.peek() < sa[i])``            ``{``                ``System.out.print(``"NO"``);``                ``return``;``            ``}``            ` `            ``// push into stack and count operation``            ``else``            ``{` `                ``sc.push(sa[i]);``                ``transfer++;``            ``}``            ``// stack not empty, then pop the top element``            ``// pop out all elements till is it equal to required``            ``while` `(!sc.empty() && sc.peek() == required)``            ``{``                ``required++;``                ``sc.pop();``                ``transfer++;``            ``}``        ``}` `        ``// print the steps``        ``System.out.println(``"YES "` `+ transfer);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `sa[] = {``4``, ``3``, ``1``, ``2``, ``5``};``        ``int` `n = sa.length;``        ``countSteps(sa, n);``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 program for``# Sudo Placement | playing with stacks``from` `typing ``import` `List` `# Function to check if it is possible``# count the number of steps``def` `countSteps(sa: ``List``[``int``], n: ``int``) ``-``> ``None``:``    ` `    ``# Another stack``    ``sc ``=` `[]` `    ``# Variables to count transfers``    ``required ``=` `1``    ``transfer ``=` `0``    ` `    ``# Iterate in the stack in reverse order``    ``for` `i ``in` `range``(n):``        ` `        ``# If the last element has to be``        ``# inserted by removing elements``        ``# then count the number of steps``        ``if` `(sa[i] ``=``=` `required):``            ``required ``+``=` `1``            ``transfer ``+``=` `1``            ` `        ``else``:``            ` `            ``# If stack is not empty and top element``            ``# is smaller than current element``            ``if` `(sc ``and` `sc[``-``1``] < sa[i]):``                ``print``(``"NO"``)``                ``return` `            ``# push into stack and count operation``            ``else``:``                ``sc.append(sa[i])``                ``transfer ``+``=` `1` `        ``# stack not empty, then pop the top``        ``# element pop out all elements till``        ``# is it equal to required``        ``while` `(sc ``and` `sc[``-``1``] ``=``=` `required):``            ``required ``+``=` `1``            ``sc.pop()``            ``transfer ``+``=` `1` `    ``# Print the steps``    ``print``(``"YES {}"``.``format``(transfer))` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``sa ``=` `[ ``4``, ``3``, ``1``, ``2``, ``5` `]``    ``n ``=` `len``(sa)``    ` `    ``countSteps(sa, n)` `# This code is contributed by sanjeev2552`

## C#

 `// C# program for Sudo``// Placement | playing with stacks``using` `System;``using` `System.Collections.Generic;   ``    ` `public` `class` `GFG``{`` ` `    ``// Function to check if it is possible``    ``// count the number of steps``    ``static` `void` `countSteps(``int` `[]sa, ``int` `n)``    ``{`` ` `        ``// Another stack``        ``Stack<``int``> sc = ``new` `Stack<``int``>();`` ` `        ``// variables to count transfers``        ``int` `required = 1, transfer = 0;`` ` `        ``// iterate in the stack in reverse order``        ``for` `(``int` `i = 0; i < n; i++)``        ``{`` ` `            ``// if the last element has to be``            ``// inserted by removing elements``            ``// then count the number of steps``            ``if` `(sa[i] == required)``            ``{``                ``required++;``                ``transfer++;``            ``}``            ``else``            ``// if stack is not empty and top element``            ``// is smaller than current element``            ``if` `(sc.Count!=0 && sc.Peek() < sa[i])``            ``{``                ``Console.Write(``"NO"``);``                ``return``;``            ``}``             ` `            ``// push into stack and count operation``            ``else``            ``{`` ` `                ``sc.Push(sa[i]);``                ``transfer++;``            ``}``            ``// stack not empty, then pop the top element``            ``// pop out all elements till is it equal to required``            ``while` `(sc.Count!=0 && sc.Peek() == required)``            ``{``                ``required++;``                ``sc.Pop();``                ``transfer++;``            ``}``        ``}`` ` `        ``// print the steps``        ``Console.WriteLine(``"YES "` `+ transfer);``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `[]sa = {4, 3, 1, 2, 5};``        ``int` `n = sa.Length;``        ``countSteps(sa, n);``    ``}``}``// This code has been contributed by 29AjayKumar`

## Javascript

 ``
Output:
`YES 7`

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