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 7Input: 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.
- 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,
- 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 <bits/stdc++.h>using namespace std;// Function to check if it is possible// count the number of stepsvoid 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 Codeint main(){ int sa[] = { 4, 3, 1, 2, 5 }; int n = sizeof(sa) / sizeof(sa[0]); countSteps(sa, n); return 0;} |
Java
// Java program for Sudo// Placement | playing with stacksimport 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<Integer> sc = new Stack<Integer>(); // 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 stacksfrom typing import List# Function to check if it is possible# count the number of stepsdef 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 Codeif __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 stacksusing 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
<script>// Javascript program for Sudo// Placement | playing with stacks// Function to check if it is possible// count the number of stepsfunction countSteps(sa, n){ // Another stack let sc = []; // variables to count transfers let required = 1, transfer = 0; // iterate in the stack in reverse order for (let 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.length!=0 && sc[sc.length-1] < sa[i]) { document.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.length!=0 && sc[sc.length-1] == required) { required++; sc.pop(); transfer++; } } // print the steps document.write("YES " + transfer+"<br>");}// Driver Codelet sa=[4, 3, 1, 2, 5];let n = sa.length;countSteps(sa, n);// This code is contributed by rag2127</script> |
Output
YES 7
Time Complexity: O(n2)
Auxiliary Space: O(n)
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