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Find common elements of Stack and Queue

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  • Last Updated : 12 Jul, 2022

Given a stack of M elements and a queue of N elements in sorted order. The task is to find out the common elements of the stack and the queue.

Examples:

Input: stack = [1, 3, 5, 7], queue = [1, 2, 5, 9]
Output: 5, 1
Explanation: 1 and 5 is present in both stack and queue.

Input: stack = [1, 3], queue = [2, 4]
Output: Not Found
Explanation: There is no common element.

 

Approach: The given problem can be solved with the help of the following idea:

As both are sorted, the top element of the stack will be the maximum and the front of the queue will be the minimum. So reverse any of them and compare the elements in top of stack and front of queue to find the common elements.

Follow the illustration below for a better understanding.

Illustration:

Say, stack = [1, 3, 5, 7] where 7 is at the top and 
the queue = [1, 2, 5, 9] where 1 is at the front.

Say we are reversing the queue. Do the following to reverse the queue:

  • In first step:
    One by one pop the element from the queue(i.e., all the elements of queue) and push into stack.
            => After First Iteration, Stack = [1, 3, 5, 7, 1] and Queue = [2, 5, 9]
            => After Second Iteration, Stack = [1, 3, 5, 7, 1, 2] and Queue = [5, 9]
            => After Third Iteration, Stack = [1, 3, 5, 7, 1, 2, 5] and Queue = [9]
            => After Fourth Iteration, Stack = [1, 3, 5, 7, 1, 2, 5, 9] and Queue = []
  • In second step:
           => One by one pop the element from the stack(i.e., coming from queue)  and push into queue.
           => After First Iteration, Stack = [1, 3, 5, 7, 1, 2, 5] and Queue = [9]
           => After Second Iteration, Stack = [1, 3, 5, 7, 1, 2] and Queue = [9, 5]
           => After Third Iteration, Stack = [1, 3, 5, 7, 1] and Queue = [9, 5, 2]
           => After Fourth Iteration, Stack = [1, 3, 5, 7] and Queue = [9, 5, 2, 1]

Now the following for finding the common elements.

1st Step:
        => stack top < queue front.
        => Pop queue front.
        => So stack is [1, 3, 5, 7] and queue [5, 2, 1]

2nd step:
        => stack top > queue front
        => Pop stack top
        => So stack [1, 3, 5] and queue [5, 2, 1]

3rd step:
        => stack top = queue front
        => Pop stack top and queue front
        => So stack [1, 3] and queue [2, 1]
        => Common elements [5]

4th step:
        => stack top > queue front
        => Pop stack top
        => So stack [1] and queue [2, 1]

5th Step:
        => stack top < queue front.
        => Pop queue front.
        => So stack is [1] and queue [1]

6th step:
        => stack top = queue front
        => Pop stack top and queue front
        => So stack [] and queue []
        => Common elements [5, 1].

Follow the below steps to solve the problem:

  • Reverse the Queue.
  • Traverse stack and queue while stack and queue do not become empty.
    • If top of stack = front of queue that is a common element.
    • Else if, top of stack > front of queue, pop the top element of the stack.
    • Else, top of stack < front of queue, pop the front element of the stack.
  • Print the common elements.

Below is the implementation of the above approach:

C++




// C++ code to implement the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find common element
// of stack and queue
vector<int> findCommonElement(stack<int>& St,
                              queue<int>& Q)
{
    // Initialize size of queue Q to 0
    int Size = 0;
    vector<int> v;
 
    // Put every element of queue into stack
    // and calculate size of queue
    while (!Q.empty()) {
        St.push(Q.front());
        Q.pop();
        Size++;
    }
 
    // Put extra element of stack into queue
    // again extra element of stack is the
    // element coming from queue. Now, the
    // queue is reverse
    while (Size != 0) {
        Q.push(St.top());
        St.pop();
        Size--;
    }
 
    // Traverse while stack and queue is not
    // empty
    while (!St.empty() && !Q.empty()) {
        // Top element of stack
        int a = St.top();
 
        // Front element of queue
        int b = Q.front();
 
        // Push the common element
        // in vector if a = b
        if (a == b)
            v.push_back(a);
 
        // Else pop the larger value
        // from its container
        (a > b) ? St.pop() : Q.pop();
    }
    return v;
}
 
// Driver Code
int main()
{
    stack<int> St;
    queue<int> Q;
 
    // Fill element into stack
    St.push(1);
    St.push(3);
    St.push(5);
    St.push(7);
 
    // Fill element into queue
    Q.push(1);
    Q.push(2);
    Q.push(5);
    Q.push(9);
 
    // Find common element if exists
    vector<int> v = findCommonElement(St, Q);
 
    if (v.size() == 0)
        cout << "Not Found" << endl;
    for (auto i : v)
        cout << i << " ";
    return 0;
}

Java




// Java code to implement the above approach
import java.util.ArrayList;
 
class GFG
{
   
    // Function to find common element
    // of stack and queue
    static ArrayList<Integer>
    findCommonElement(ArrayList<Integer> St,
                      ArrayList<Integer> Q)
    {
 
        // Initialize size of queue Q to 0
        int Size = 0;
        ArrayList<Integer> v = new ArrayList<Integer>();
 
        // Put every element of queue into stack
        // and calculate size of queue
        while (Q.size() != 0) {
            St.add(Q.get(0));
            Q.remove(0);
            Size++;
        }
 
        // Put extra element of stack into queue
        // again extra element of stack is the
        // element coming from queue. Now, the
        // queue is reverse
        while (Size != 0) {
            Q.add(St.get(St.size() - 1));
            St.remove(St.size() - 1);
            Size--;
        }
 
        // Traverse while stack and queue is not
        // empty
        while (St.size() != 0 && Q.size() != 0) {
 
            // Top element of stack
            int a = St.get(St.size() - 1);
 
            // Front element of queue
            int b = Q.get(0);
 
            // Push the common element
            // in vector if a = b
            if (a == b)
                v.add(a);
 
            // Else pop the larger value
            // from its container
            if (a > b)
                St.remove(St.size() - 1);
            else
                Q.remove(0);
        }
        return v;
    }
    public static void main(String[] args)
    {
        // Driver Code
        ArrayList<Integer> St = new ArrayList<Integer>();
        ArrayList<Integer> Q = new ArrayList<Integer>();
 
        // Fill element into stack
        St.add(1);
        St.add(3);
        St.add(5);
        St.add(7);
 
        // Fill element into queue
        Q.add(1);
        Q.add(2);
        Q.add(5);
        Q.add(9);
 
        // Find common element if exists
        ArrayList<Integer> v = findCommonElement(St, Q);
 
        if (v.size() == 0)
            System.out.print("Not Found");
        for (int i = 0; i < v.size(); i++)
            System.out.print(v.get(i) + " ");
    }
}
 
// this code is contributed by phasing17

Python3




# Python code to implement the above approach
 
# Function to find common element
# of stack and queue
def findCommonElement(St, Q):
 
    # Initialize size of queue Q to 0
    Size = 0
    v = []
 
    # Put every element of queue into stack
    # and calculate size of queue
    while len(Q) != 0:
        St.append(Q[0])
        Q = Q[1:]
        Size += 1
 
    # Put extra element of stack into queue
    # again extra element of stack is the
    # element coming from queue. Now, the
    # queue is reverse
    while (Size != 0):
        Q.append(St[len(St) - 1])
        St.pop()
        Size -= 1
 
    # Traverse while stack and queue is not
    # empty
    while (len(St) != 0 and len(Q) != 0):
             
        # Top element of stack
        a = St[len(St) - 1]
 
        # Front element of queue
        b = Q[0]
 
        # append the common element
        # in vector if a = b
        if (a == b):
            v.append(a)
 
        # Else pop the larger value
        # from its container
        if (a > b):
            St.pop()
        else:
            Q = Q[1:]
    return v
 
# Driver Code
 
St = []
Q = []
 
# Fill element into stack
St.append(1)
St.append(3)
St.append(5)
St.append(7)
 
# Fill element into queue
Q.append(1)
Q.append(2)
Q.append(5)
Q.append(9)
 
# Find common element if exists
v = findCommonElement(St, Q)
 
if (len(v) == 0):
    print("Not Found")
for i in v:
    print(i,end=" ")
 
# This code is contributed by shinjanpatra

C#




// C# code to implement the above approach
using System;
using System.Collections.Generic;
 
public class GFG
{
   
    // Function to find common element
    // of stack and queue
    public static List<int> findCommonElement(List<int> St,
                                              List<int> Q)
    {
 
        // Initialize size of queue Q to 0
        int Size = 0;
        List<int> v = new List<int>();
 
        // Put every element of queue into stack
        // and calculate size of queue
        while (Q.Count != 0) {
            St.Add(Q[0]);
            Q.RemoveAt(0);
            Size++;
        }
 
        // Put extra element of stack into queue
        // again extra element of stack is the
        // element coming from queue. Now, the
        // queue is reverse
        while (Size != 0) {
            Q.Add(St[St.Count - 1]);
            St.RemoveAt(St.Count - 1);
            Size--;
        }
 
        // Traverse while stack and queue is not
        // empty
        while (St.Count != 0 && Q.Count != 0) {
 
            // Top element of stack
            int a = St[St.Count - 1];
 
            // Front element of queue
            int b = Q[0];
 
            // Push the common element
            // in vector if a = b
            if (a == b)
                v.Add(a);
 
            // Else pop the larger value
            // from its container
            if (a > b)
                St.RemoveAt(St.Count - 1);
            else
                Q.RemoveAt(0);
        }
        return v;
    }
    public static void Main(string[] args)
    {
 
        // Driver Code
 
        List<int> St = new List<int>();
        List<int> Q = new List<int>();
 
        // Fill element into stack
        St.Add(1);
        St.Add(3);
        St.Add(5);
        St.Add(7);
 
        // Fill element into queue
        Q.Add(1);
        Q.Add(2);
        Q.Add(5);
        Q.Add(9);
 
        // Find common element if exists
        List<int> v = findCommonElement(St, Q);
 
        if (v.Count == 0)
            Console.WriteLine("Not Found");
        foreach(var ele in v) Console.Write(ele + " ");
    }
}
 
//This code is contributed by phasing17

Javascript




<script>
    // JavaScript code to implement the above approach
 
    // Function to find common element
    // of stack and queue
    const findCommonElement = (St, Q) => {
     
        // Initialize size of queue Q to 0
        let Size = 0;
        let v = [];
 
        // Put every element of queue into stack
        // and calculate size of queue
        while (Q.length != 0) {
            St.push(Q[0]);
            Q.shift();
            Size++;
        }
 
        // Put extra element of stack into queue
        // again extra element of stack is the
        // element coming from queue. Now, the
        // queue is reverse
        while (Size != 0) {
            Q.push(St[St.length - 1]);
            St.pop();
            Size--;
        }
 
        // Traverse while stack and queue is not
        // empty
        while (St.length != 0 && Q.length != 0)
        {
         
            // Top element of stack
            let a = St[St.length - 1];
 
            // Front element of queue
            let b = Q[0];
 
            // Push the common element
            // in vector if a = b
            if (a == b)
                v.push(a);
 
            // Else pop the larger value
            // from its container
            (a > b) ? St.pop() : Q.shift();
        }
        return v;
    }
 
    // Driver Code
 
    let St = [];
    let Q = [];
 
    // Fill element into stack
    St.push(1);
    St.push(3);
    St.push(5);
    St.push(7);
 
    // Fill element into queue
    Q.push(1);
    Q.push(2);
    Q.push(5);
    Q.push(9);
 
    // Find common element if exists
    let v = findCommonElement(St, Q);
 
    if (v.length == 0)
        document.write("Not Found");
    for (let i in v)
        document.write(`${v[i]} `);
 
    // This code is contributed by rakeshsahni
 
</script>

Output

5 1 

Time Complexity: O(M+N) where M =size of the queue and N = size of the stack
Auxiliary Space: O(M) to reverse elements of queue 


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