Check if a queue can be sorted into another queue using a stack
Last Updated :
17 Aug, 2022
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:
- Push and pop elements from the stack
- Pop (Or Dequeue) from the given Queue.
- 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:
- Initialize the expected_element = 1
- Check if either front element of given Queue or top element of the stack have expected_element
- If yes, increment expected_element by 1, repeat step 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:
C++
#include <bits/stdc++.h>
using namespace std;
bool checkSorted(int n, queue<int>& q)
{
stack<int> st;
int expected = 1;
int fnt;
while (!q.empty()) {
fnt = q.front();
q.pop();
if (fnt == expected)
expected++;
else {
if (st.empty()) {
st.push(fnt);
}
else if (!st.empty() && st.top() < fnt) {
return false;
}
else
st.push(fnt);
}
while (!st.empty() && st.top() == expected) {
st.pop();
expected++;
}
}
if (expected - 1 == n && st.empty())
return true;
return false;
}
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
import java.io.*;
import java.util.*;
class GFG
{
static Queue<Integer> q =
new LinkedList<Integer>();
static boolean checkSorted(int n)
{
Stack<Integer> st =
new Stack<Integer>();
int expected = 1;
int fnt;
while (q.size() != 0)
{
fnt = q.peek();
q.poll();
if (fnt == expected)
expected++;
else
{
if (st.size() == 0)
{
st.push(fnt);
}
else if (st.size() != 0 &&
st.peek() < fnt)
{
return false;
}
else
st.push(fnt);
}
while (st.size() != 0 &&
st.peek() == expected)
{
st.pop();
expected++;
}
}
if (expected - 1 == n &&
st.size() == 0)
return true;
return false;
}
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");
}
}
|
Python3
from queue import Queue
def checkSorted(n, q):
st = []
expected = 1
fnt = None
while (not q.empty()):
fnt = q.queue[0]
q.get()
if (fnt == expected):
expected += 1
else:
if (len(st) == 0):
st.append(fnt)
elif (len(st) != 0 and st[-1] < fnt):
return False
else:
st.append(fnt)
while (len(st) != 0 and
st[-1] == expected):
st.pop()
expected += 1
if (expected - 1 == n and len(st) == 0):
return True
return False
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")
|
C#
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{
static bool checkSorted(int n,
ref Queue<int> q)
{
Stack<int> st = new Stack<int>();
int expected = 1;
int fnt;
while (q.Count != 0)
{
fnt = q.Peek();
q.Dequeue();
if (fnt == expected)
expected++;
else
{
if (st.Count == 0)
{
st.Push(fnt);
}
else if (st.Count != 0 &&
st.Peek() < fnt)
{
return false;
}
else
st.Push(fnt);
}
while (st.Count != 0 &&
st.Peek() == expected)
{
st.Pop();
expected++;
}
}
if (expected - 1 == n &&
st.Count == 0)
return true;
return false;
}
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");
}
}
|
Javascript
<script>
let q = [];
function checkSorted(n)
{
let st = [];
let expected = 1;
let fnt;
while (q.length != 0)
{
fnt = q[0];
q.shift();
if (fnt == expected)
expected++;
else
{
if (st.length == 0)
{
st.push(fnt);
}
else if (st.length != 0 &&
st[st.length - 1] < fnt)
{
return false;
}
else
st.push(fnt);
}
while (st.length != 0 &&
st[st.length - 1] == expected)
{
st.pop();
expected++;
}
}
if ((expected - 1) == n && st.length == 0)
return true;
return false;
}
q.push(5);
q.push(1);
q.push(2);
q.push(3);
q.push(4);
let n = q.length;
if (checkSorted(n))
document.write("Yes");
else
document.write("No");
</script>
|
Complexity Analysis:
- Time Complexity: O(n)
- Space Complexity: O(n)
Video Contributed by Parul Shandilya
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