Check if a queue can be sorted into another queue using a stack

Given a Queue consisting of first n natural numbers (in random order). The task is to check whether the given Queue elements can be arranged in increasing order in another Queue using a stack. The operation allowed are: 

1. Push and pop elements from the stack 
2. Pop (Or Dequeue) from the given Queue. 
3. Push (Or Enqueue) in the another Queue.

Examples :

Input : Queue[] = { 5, 1, 2, 3, 4 } 
Output : Yes 
Pop the first element of the given Queue i.e 5. 
Push 5 into the stack. 
Now, pop all the elements of the given Queue and push them to 
second Queue. 
Now, pop element 5 in the stack and push it to the second Queue. 
  
Input : Queue[] = { 5, 1, 2, 6, 3, 4 } 
Output : No 
Push 5 to stack. 
Pop 1, 2 from given Queue and push it to another Queue. 
Pop 6 from given Queue and push to stack. 
Pop 3, 4 from given Queue and push to second Queue. 
Now, from using any of above operation, we cannot push 5 
into the second Queue because it is below the 6 in the stack. 

Observe, second Queue (which will contain the sorted element) takes inputs (or enqueue elements) either from given Queue or Stack. So, the next expected (which will initially be 1) element must be present as a front element of a given Queue or top element of the Stack. So, simply simulate the process for the second Queue by initializing the expected element as 1. And check if we can get the expected element from the front of the given Queue or from the top of the Stack. If we cannot take it from either of them then pop the front element of the given Queue and push it in the Stack. 

Also, observe, that the stack must also be sorted at each instance i.e the element at the top of the stack must be the smallest in the stack. For eg. let x > y, then x will always be expected before y. So, x cannot be pushed before y in the stack. Therefore, we cannot push an element with a higher value on the top of the element having a lesser value.

Algorithm: 

1. Initialize the expected_element = 1 
2. Check if either front element of given Queue or top element of the stack have expected_element 
   1. If yes, increment expected_element by 1, repeat step 2. 
   2. Else, pop front of Queue and push it to the stack. If the popped element is greater than top of the Stack, return "No".

Below is the implementation of this approach: 

// CPP Program to check if a queue of first  // n natural number can be sorted using a stack #include <bits/stdc++.h> using namespace std; // Function to check if given queue element  // can be sorted into another queue using a // stack. bool checkSorted(int n, queue<int>& q) {  stack<int> st;  int expected = 1;  int fnt;  // while given Queue is not empty.  while (!q.empty()) {  fnt = q.front();  q.pop();  // if front element is the expected element  if (fnt == expected)  expected++;  else {  // if stack is empty, push the element  if (st.empty()) {  st.push(fnt);  }  // if top element is less than element which  // need to be pushed, then return false.  else if (!st.empty() && st.top() < fnt) {  return false;  }  // else push into the stack.  else  st.push(fnt);  }  // while expected element are coming from  // stack, pop them out.  while (!st.empty() && st.top() == expected) {  st.pop();  expected++;  }  }  // if the final expected element value is equal  // to initial Queue size and the stack is empty.  if (expected - 1 == n && st.empty())  return true;  return false; } // Driven Program int main() {  queue<int> q;  q.push(5);  q.push(1);  q.push(2);  q.push(3);  q.push(4);  int n = q.size();  (checkSorted(n, q) ? (cout << "Yes") :  (cout << "No"));  return 0; } 

// Java Program to check if a queue  // of first n natural number can // be sorted using a stack import java.io.*; import java.util.*; class GFG {  static Queue<Integer> q =   new LinkedList<Integer>();    // Function to check if given   // queue element can be sorted   // into another queue using a stack.  static boolean checkSorted(int n)  {  Stack<Integer> st =   new Stack<Integer>();  int expected = 1;  int fnt;    // while given Queue  // is not empty.  while (q.size() != 0)   {  fnt = q.peek();  q.poll();    // if front element is   // the expected element  if (fnt == expected)  expected++;    else  {  // if stack is empty,   // push the element  if (st.size() == 0)   {  st.push(fnt);  }    // if top element is less than   // element which need to be   // pushed, then return false.  else if (st.size() != 0 &&   st.peek() < fnt)   {  return false;  }    // else push into the stack.  else  st.push(fnt);  }    // while expected element are  // coming from stack, pop them out.  while (st.size() != 0 &&   st.peek() == expected)   {  st.pop();  expected++;  }  }    // if the final expected element   // value is equal to initial Queue  // size and the stack is empty.  if (expected - 1 == n &&   st.size() == 0)  return true;    return false;  }    // Driver Code  public static void main(String args[])  {  q.add(5);  q.add(1);  q.add(2);  q.add(3);  q.add(4);    int n = q.size();  if (checkSorted(n))  System.out.print("Yes");  else  System.out.print("No");  } } // This code is contributed by  // Manish Shaw(manishshaw1) 

# Python Program to check if a queue of first  # n natural number can be sorted using a stack  from queue import Queue # Function to check if given queue element  # can be sorted into another queue using a  # stack.  def checkSorted(n, q): st = [] expected = 1 fnt = None # while given Queue is not empty.  while (not q.empty()): fnt = q.queue[0] q.get() # if front element is the  # expected element  if (fnt == expected): expected += 1 else: # if stack is empty, put the element  if (len(st) == 0): st.append(fnt) # if top element is less than element which  # need to be puted, then return false.  elif (len(st) != 0 and st[-1] < fnt): return False # else put into the stack.  else: st.append(fnt) # while expected element are coming  # from stack, pop them out.  while (len(st) != 0 and st[-1] == expected): st.pop() expected += 1 # if the final expected element value is equal  # to initial Queue size and the stack is empty.  if (expected - 1 == n and len(st) == 0): return True return False # Driver Code if __name__ == '__main__': q = Queue() q.put(5) q.put(1) q.put(2) q.put(3) q.put(4) n = q.qsize() if checkSorted(n, q): print("Yes") else: print("No") # This code is contributed by PranchalK 

// C# Program to check if a queue  // of first n natural number can // be sorted using a stack using System; using System.Linq; using System.Collections.Generic; class GFG {  // Function to check if given   // queue element can be sorted   // into another queue using a stack.  static bool checkSorted(int n,   ref Queue<int> q)  {  Stack<int> st = new Stack<int>();  int expected = 1;  int fnt;    // while given Queue  // is not empty.  while (q.Count != 0)   {  fnt = q.Peek();  q.Dequeue();    // if front element is   // the expected element  if (fnt == expected)  expected++;    else   {  // if stack is empty,   // push the element  if (st.Count == 0)   {  st.Push(fnt);  }    // if top element is less than   // element which need to be   // pushed, then return false.  else if (st.Count != 0 &&   st.Peek() < fnt)   {  return false;  }    // else push into the stack.  else  st.Push(fnt);  }    // while expected element are  // coming from stack, pop them out.  while (st.Count != 0 &&   st.Peek() == expected)   {  st.Pop();  expected++;  }  }  // if the final expected element   // value is equal to initial Queue  // size and the stack is empty.  if (expected - 1 == n &&   st.Count == 0)  return true;    return false;  }    // Driver Code  static void Main()  {  Queue<int> q = new Queue<int>();  q.Enqueue(5);  q.Enqueue(1);  q.Enqueue(2);  q.Enqueue(3);  q.Enqueue(4);    int n = q.Count;  if (checkSorted(n, ref q))  Console.Write("Yes");  else  Console.Write("No");  } } // This code is contributed by  // Manish Shaw(manishshaw1) 

// JavaScript Program to check if a queue // of first n natural numbers can be sorted using a stack let q = []; // Function to check if given queue elements  // can be sorted into another queue using a stack. function checkSorted(n) {  let st = [];  let expected = 1;  let fnt;  // While given queue is not empty  while (q.length !== 0) {  fnt = q.shift(); // Remove front element  // If front element is the expected element  if (fnt === expected) {  expected++;  } else {  // If stack is empty, push the element  if (st.length === 0) {  st.push(fnt);  }  // If top element is smaller than current element, return false  else if (st[st.length - 1] < fnt) {  return false;  } else {  st.push(fnt);  }  }  // While expected elements are coming from stack, pop them out  while (st.length !== 0 && st[st.length - 1] === expected) {  st.pop();  expected++;  }  }  // If the final expected element value is equal to initial queue size  // and the stack is empty, return true  return (expected - 1) === n && st.length === 0; } // Example queue q.push(5); q.push(1); q.push(2); q.push(3); q.push(4); let n = q.length; console.log(checkSorted(n) ? "Yes" : "No"); 

Yes

Complexity Analysis:

• Time Complexity: O(n)
• Space Complexity: O(n)