Implement Stack using Queues
Last Updated :
25 Mar, 2025
Implement a stack using queues. The stack should support the following operations:
- Push(x): Push an element onto the stack.
- Pop(): Pop the element from the top of the stack and return it.
A Stack can be implemented using two queues. Let Stack to be implemented be 's' and queues used to implement are 'q1' and 'q2'.
Stack 's' can be implemented in two ways:
By making push() operation costly - Push in O(n) and Pop() in O(1)
The idea is to keep newly entered element at the front of 'q1' so that pop operation dequeues from 'q1'. 'q2' is used to move every new element in front of 'q1'.
Follow the below steps to implement the push(s, x) operation:
- Enqueue x to q2.
- One by one dequeue everything from q1 and enqueue to q2.
- Swap the queues of q1 and q2.
Follow the below steps to implement the pop(s) operation:
- Dequeue an item from q1 and return it.
C++
/* Program to implement a stack using
two queue */
#include <bits/stdc++.h>
using namespace std;
class Stack {
// Two inbuilt queues
queue<int> q1, q2;
public:
void push(int x)
{
// Push x first in empty q2
q2.push(x);
// Push all the remaining
// elements in q1 to q2.
while (!q1.empty()) {
q2.push(q1.front());
q1.pop();
}
// swap the names of two queues
swap(q1, q2);
}
void pop()
{
// if no elements are there in q1
if (q1.empty())
return;
q1.pop();
}
int top()
{
if (q1.empty())
return -1;
return q1.front();
}
int size() { return q1.size(); }
};
// Driver code
int main()
{
Stack s;
s.push(1);
s.push(2);
s.push(3);
cout << "current size: " << s.size() << endl;
cout << s.top() << endl;
s.pop();
cout << s.top() << endl;
s.pop();
cout << s.top() << endl;
cout << "current size: " << s.size() << endl;
return 0;
}
/* Java Program to implement a stack using
two queue */
import java.util.*;
class GfG {
static class Stack {
// Two inbuilt queues
static Queue<Integer> q1
= new LinkedList<Integer>();
static Queue<Integer> q2
= new LinkedList<Integer>();
// To maintain current number of
// elements
static int curr_size;
static void push(int x)
{
// Push x first in empty q2
q2.add(x);
// Push all the remaining
// elements in q1 to q2.
while (!q1.isEmpty()) {
q2.add(q1.peek());
q1.remove();
}
// swap the names of two queues
Queue<Integer> q = q1;
q1 = q2;
q2 = q;
}
static void pop()
{
// if no elements are there in q1
if (q1.isEmpty())
return;
q1.remove();
}
static int top()
{
if (q1.isEmpty())
return -1;
return q1.peek();
}
static int size() { return q1.size(); }
}
// driver code
public static void main(String[] args)
{
Stack s = new Stack();
s.push(1);
s.push(2);
s.push(3);
System.out.println("current size: " + s.size());
System.out.println(s.top());
s.pop();
System.out.println(s.top());
s.pop();
System.out.println(s.top());
System.out.println("current size: " + s.size());
}
}
// This code is contributed by Prerna
# Program to implement a stack using
# two queue
from _collections import deque
class Stack:
def __init__(self):
# Two inbuilt queues
self.q1 = deque()
self.q2 = deque()
def push(self, x):
# Push x first in empty q2
self.q2.append(x)
# Push all the remaining
# elements in q1 to q2.
while (self.q1):
self.q2.append(self.q1.popleft())
# swap the names of two queues
self.q1, self.q2 = self.q2, self.q1
def pop(self):
# if no elements are there in q1
if self.q1:
self.q1.popleft()
def top(self):
if (self.q1):
return self.q1[0]
return None
def size(self):
return len(self.q1)
# Driver Code
if __name__ == '__main__':
s = Stack()
s.push(1)
s.push(2)
s.push(3)
print("current size: ", s.size())
print(s.top())
s.pop()
print(s.top())
s.pop()
print(s.top())
print("current size: ", s.size())
# This code is contributed by PranchalK
// C# Program to implement a stack using two queues
using System;
using System.Collections.Generic;
class GfG
{
static class Stack
{
// Two inbuilt queues
static Queue<int> q1 = new Queue<int>();
static Queue<int> q2 = new Queue<int>();
// To maintain current number of elements
static int curr_size;
static void Push(int x)
{
// Push x first in empty q2
q2.Enqueue(x);
// Push all the remaining elements in q1 to q2.
while (q1.Count > 0)
{
q2.Enqueue(q1.Dequeue());
}
// swap the names of two queues
Queue<int> q = q1;
q1 = q2;
q2 = q;
}
static void Pop()
{
// if no elements are there in q1
if (q1.Count == 0)
return;
q1.Dequeue();
}
static int Top()
{
if (q1.Count == 0)
return -1;
return q1.Peek();
}
static int Size() { return q1.Count; }
}
// driver code
public static void Main(string[] args)
{
Stack s = new Stack();
s.Push(1);
s.Push(2);
s.Push(3);
Console.WriteLine("current size: " + s.Size());
Console.WriteLine(s.Top());
s.Pop();
Console.WriteLine(s.Top());
s.Pop();
Console.WriteLine(s.Top());
Console.WriteLine("current size: " + s.Size());
}
}
/*Javascript Program to implement a stack using
two queue */
// Two inbuilt queues
class Stack {
constructor() {
this.q1 = [];
this.q2 = [];
}
push(x) {
// Push x first in isEmpty q2
this.q2.push(x);
// Push all the remaining
// elements in q1 to q2.
while (this.q1.length != 0) {
this.q2.push(this.q1[0]);
this.q1.shift();
}
// swap the names of two queues
this.q = this.q1;
this.q1 = this.q2;
this.q2 = this.q;
}
pop() {
// if no elements are there in q1
if (this.q1.length == 0)
return;
this.q1.shift();
}
top() {
if (this.q1.length == 0)
return -1;
return this.q1[0];
}
size() {
console.log(this.q1.length);
}
isEmpty() {
// return true if the queue is empty.
return this.q1.length == 0;
}
front() {
return this.q1[0];
}
}
// Driver code
let s = new Stack();
s.push(1);
s.push(2);
s.push(3);
console.log("current size: ");
s.size();
console.log(s.top());
s.pop();
console.log(s.top());
s.pop();
console.log(s.top());
console.log("current size: ");
s.size();
// This code is contributed by adityamaharshi21
Outputcurrent size: 3
3
2
1
current size: 1
Time Complexity:
- Push operation: O(n), As all the elements need to be popped out from the Queue (q1) and push them back to Queue (q2).
- Pop operation: O(1), As we need to remove the front element from the Queue.
Auxiliary Space: O(n), As we use two queues for the implementation of a Stack.
By making pop() operation costly - Push in O(1) and Pop() in O(n)
The new element is always enqueued to q1. In pop() operation, if q2 is empty then all the elements except the last, are moved to q2. Finally, the last element is dequeued from q1 and returned.
Follow the below steps to implement the push(s, x) operation:
- Enqueue x to q1 (assuming the size of q1 is unlimited).
Follow the below steps to implement the pop(s) operation:
- One by one dequeue everything except the last element from q1 and enqueue to q2.
- Dequeue the last item of q1, the dequeued item is the result, store it.
- Swap the names of q1 and q2
- Return the item stored in step 2.
C++
// Program to implement a stack
// using two queue
#include <bits/stdc++.h>
using namespace std;
class Stack {
queue<int> q1, q2;
public:
void pop()
{
if (q1.empty())
return;
// Leave one element in q1 and
// push others in q2.
while (q1.size() != 1) {
q2.push(q1.front());
q1.pop();
}
// Pop the only left element
// from q1
q1.pop();
// swap the names of two queues
swap(q1, q2);
}
void push(int x) { q1.push(x); }
int top()
{
if (q1.empty())
return -1;
while (q1.size() != 1) {
q2.push(q1.front());
q1.pop();
}
// last pushed element
int temp = q1.front();
// to empty the auxiliary queue after
// last operation
q1.pop();
// push last element to q2
q2.push(temp);
// swap the two queues names
queue<int> q = q1;
q1 = q2;
q2 = q;
return temp;
}
int size() { return q1.size(); }
};
// Driver code
int main()
{
Stack s;
s.push(1);
s.push(2);
s.push(3);
cout << "current size: " << s.size() << endl;
cout << s.top() << endl;
s.pop();
cout << s.top() << endl;
s.pop();
cout << s.top() << endl;
cout << "current size: " << s.size() << endl;
return 0;
}
/* Java Program to implement a stack
using two queue */
import java.util.*;
class Stack {
Queue<Integer> q1 = new LinkedList<>(),
q2 = new LinkedList<>();
void remove()
{
if (q1.isEmpty())
return;
// Leave one element in q1 and
// push others in q2.
while (q1.size() != 1) {
q2.add(q1.peek());
q1.remove();
}
// Pop the only left element
// from q1
q1.remove();
// swap the names of two queues
Queue<Integer> q = q1;
q1 = q2;
q2 = q;
}
void add(int x) { q1.add(x); }
int top()
{
if (q1.isEmpty())
return -1;
while (q1.size() != 1) {
q2.add(q1.peek());
q1.remove();
}
// last pushed element
int temp = q1.peek();
// to empty the auxiliary queue after
// last operation
q1.remove();
// push last element to q2
q2.add(temp);
// swap the two queues names
Queue<Integer> q = q1;
q1 = q2;
q2 = q;
return temp;
}
int size() { return q1.size(); }
// Driver code
public static void main(String[] args)
{
Stack s = new Stack();
s.add(1);
s.add(2);
s.add(3);
System.out.println("current size: " + s.size());
System.out.println(s.top());
s.remove();
System.out.println(s.top());
s.remove();
System.out.println(s.top());
System.out.println("current size: " + s.size());
}
}
// This code is contributed by Princi Singh
# Program to implement a stack using
# two queue
from _collections import deque
class Stack:
def __init__(self):
# Two inbuilt queues
self.q1 = deque()
self.q2 = deque()
def push(self, x):
self.q1.append(x)
def pop(self):
# if no elements are there in q1
if (not self.q1):
return
# Leave one element in q1 and push others in q2
while(len(self.q1) != 1):
self.q2.append(self.q1.popleft())
# swap the names of two queues
self.q1, self.q2 = self.q2, self.q1
def top(self):
# if no elements are there in q1
if (not self.q1):
return
# Leave one element in q1 and push others in q2
while(len(self.q1) != 1):
self.q2.append(self.q1.popleft())
# Pop the only left element from q1 to q2
top = self.q1[0]
self.q2.append(self.q1.popleft())
# swap the names of two queues
self.q1, self.q2 = self.q2, self.q1
return top
def size(self):
return len(self.q1)
# Driver Code
if __name__ == '__main__':
s = Stack()
s.push(1)
s.push(2)
s.push(3)
print("current size: ", s.size())
print(s.top())
s.pop()
print(s.top())
s.pop()
print(s.top())
print("current size: ", s.size())
# This code is contributed by jainlovely450
using System;
using System.Collections.Generic;
class Stack {
Queue<int> q1 = new Queue<int>();
Queue<int> q2 = new Queue<int>();
void Remove() {
if (q1.Count == 0)
return;
// Leave one element in q1 and
// push others in q2.
while (q1.Count != 1) {
q2.Enqueue(q1.Dequeue());
}
// Pop the only left element
// from q1
q1.Dequeue();
// swap the names of two queues
Queue<int> temp = q1;
q1 = q2;
q2 = temp;
}
void Add(int x) { q1.Enqueue(x); }
int Top() {
if (q1.Count == 0)
return -1;
while (q1.Count != 1) {
q2.Enqueue(q1.Dequeue());
}
// last pushed element
int temp = q1.Peek();
// to empty the auxiliary queue after
// last operation
q1.Dequeue();
// push last element to q2
q2.Enqueue(temp);
// swap the two queues names
Queue<int> q = q1;
q1 = q2;
q2 = q;
return temp;
}
int Size() { return q1.Count; }
// Driver code
public static void Main(string[] args) {
Stack s = new Stack();
s.Add(1);
s.Add(2);
s.Add(3);
Console.WriteLine("current size: " + s.Size());
Console.WriteLine(s.Top());
s.Remove();
Console.WriteLine(s.Top());
s.Remove();
Console.WriteLine(s.Top());
Console.WriteLine("current size: " + s.Size());
}
}
/*Javascript Program to implement a stack using
two queue */
// Two inbuilt queues
class Stack {
constructor() {
this.q1 = [];
this.q2 = [];
}
pop()
{
if (this.q1.length == 0)
return;
// Leave one element in q1 and
// push others in q2.
while (this.q1.length != 1){
this.q2.push(this.q1[0]);
this.q1.shift();
}
// Pop the only left element
// from q1f
this.q1.shift();
// swap the names of two queues
this.q = this.q1;
this.q1 = this.q2;
this.q2 = this.q;
}
push(x) {
// if no elements are there in q1
this.q1.push(x);
}
top() {
if (this.q1.length == 0)
return -1;
while (this.q1.length != 1) {
this.q2.push(this.q1[0]);
this.q1.shift();
}
// last pushed element
let temp = this.q1[0];
// to empty the auxiliary queue after
// last operation
this.q1.shift();
// push last element to q2
this.q2.push(temp);
// swap the two queues names
this.q = this.q1;
this.q1 = this.q2;
this.q2 = this.q;
return temp;
}
size() {
console.log(this.q1.length);
}
isEmpty() {
// return true if the queue is empty.
return this.q1.length == 0;
}
front() {
return this.q1[0];
}
}
// Driver code
let s = new Stack();
s.push(1);
s.push(2);
s.push(3);
console.log("current size: ");
s.size();
console.log(s.top());
s.pop();
console.log(s.top());
s.pop();
console.log(s.top());
console.log("current size: ");
s.size();
// This code is contributed by Susobhan Akhuli
Outputcurrent size: 3
3
2
1
current size: 1
Time Complexity:
- Push operation: O(1), As, on each push operation the new element is added at the end of the Queue.
- Pop operation: O(n), As, on each pop operation, all the elements are popped out from the Queue (q1) except the last element and pushed into the Queue (q2).
Auxiliary Space: O(n) since 2 queues are used.
Using single queue and Recursion Stack
Using only one queue and make the queue act as a Stack in modified way of the above discussed approach.
Follow the below steps to implement the idea:
- The idea behind this approach is to make one queue and push the first element in it.
- After the first element, we push the next element and then push the first element again and finally pop the first element.
- So, according to the FIFO rule of the queue, the second element that was inserted will be at the front and then the first element as it was pushed again later and its first copy was popped out.
- So, this acts as a Stack and we do this at every step i.e. from the initial element to the second last element, and the last element will be the one that we are inserting and since we will be pushing the initial elements after pushing the last element, our last element becomes the first element.
C++
#include <bits/stdc++.h>
using namespace std;
// Stack Class that acts as a queue
class Stack {
queue<int> q;
public:
void push(int data)
{
int s = q.size();
// Push the current element
q.push(data);
// Pop all the previous elements and put them after
// current element
for (int i = 0; i < s; i++) {
// Add the front element again
q.push(q.front());
// Delete front element
q.pop();
}
}
void pop()
{
if (q.empty())
cout << "No elements\n";
else
q.pop();
}
int top() { return (q.empty()) ? -1 : q.front(); }
int size() { return q.size(); }
bool empty() { return (q.empty()); }
};
int main()
{
Stack st;
st.push(1);
st.push(2);
st.push(3);
cout << "current size: " << st.size() << "\n";
cout << st.top() << "\n";
st.pop();
cout << st.top() << "\n";
st.pop();
cout << st.top() << "\n";
cout << "current size: " << st.size();
return 0;
}
import java.util.*;
class Stack {
// One queue
Queue<Integer> q1 = new LinkedList<Integer>();
void push(int x)
{
// Get previous size of queue
int s = q1.size();
// Push the current element
q1.add(x);
// Pop all the previous elements and put them after
// current element
for (int i = 0; i < s; i++) {
q1.add(q1.remove());
}
}
void pop()
{
// if no elements are there in q1
if (q1.isEmpty())
return;
q1.remove();
}
int top()
{
if (q1.isEmpty())
return -1;
return q1.peek();
}
int size() { return q1.size(); }
// driver code
public static void main(String[] args)
{
Stack s = new Stack();
s.push(1);
s.push(2);
s.push(3);
System.out.println("current size: " + s.size());
System.out.println(s.top());
s.pop();
System.out.println(s.top());
s.pop();
System.out.println(s.top());
System.out.println("current size: " + s.size());
}
}
// This code is contributed by Vishal Singh Shekhawat
from _collections import deque
# Stack Class that acts as a queue
class Stack:
def __init__(self):
self.q = deque()
# Push operation
def push(self, data):
# Get previous size of queue
s = len(self.q)
# Push the current element
self.q.append(data)
# Pop all the previous elements and put them after
# current element
for i in range(s):
self.q.append(self.q.popleft())
# Removes the top element
def pop(self):
if (not self.q):
print("No elements")
else:
self.q.popleft()
# Returns top of stack
def top(self):
if (not self.q):
return
return self.q[0]
def size(self):
return len(self.q)
if __name__ == '__main__':
st = Stack()
st.push(1)
st.push(2)
st.push(3)
print("current size: ", st.size())
print(st.top())
st.pop()
print(st.top())
st.pop()
print(st.top())
print("current size: ", st.size())
using System;
using System.Collections;
class GfG {
public class Stack
{
// One inbuilt queue
public Queue q = new Queue();
public void push(int x)
{
// Get previous size of queue
int s = q.Count;
// Push the current element
q.Enqueue(x);
// Pop all the previous elements and put them
// afte current element
for (int i = 0; i < s; i++) {
// Add the front element again
q.Enqueue(q.Peek());
// Delete front element
q.Dequeue();
}
}
// Removes the top element
public void pop()
{
// if no elements are there in q
if (q.Count == 0)
Console.WriteLine("No elements");
else
q.Dequeue();
}
// Returns top of stack
public int top()
{
if (q.Count == 0)
return -1;
return (int)q.Peek();
}
public int size() { return q.Count; }
};
// Driver code
public static void Main(String[] args)
{
Stack st = new Stack();
st.push(1);
st.push(2);
st.push(3);
Console.WriteLine("current size: " + st.size());
Console.WriteLine(st.top());
st.pop();
Console.WriteLine(st.top());
st.pop();
Console.WriteLine(st.top());
Console.WriteLine("current size: " + st.size());
}
}
// This code is contributed by Susobhan Akhuli
// One inbuilt queue
class Stack {
constructor() {
this.q = [];
}
// Push operation
push(data) {
// Get previous size of queue
let s = this.q.length;
// Push the current element
this.q.push(data);
// Pop all the previous elements and put them after
// current element
for (let i = 0; i < s; i++) {
// Add the front element again
this.q.push(this.q[0]);
// Delete front element
this.q.shift();
}
}
// Removes the top element
pop() {
// if no elements are there in q1
if (this.q.length == 0)
console.log("No elements");
else
this.q.shift();
}
top() {
if (this.q.length == 0)
return -1;
return this.q[0];
}
size() {
console.log(this.q.length);
}
isEmpty() {
// return true if the queue is empty.
return this.q.length == 0;
}
front() {
return this.q[0];
}
}
// Driver code
let st = new Stack();
st.push(1);
st.push(2);
st.push(3);
console.log("current size: ");
st.size();
console.log(st.top());
st.pop();
console.log(st.top());
st.pop();
console.log(st.top());
console.log("current size: ");
st.size();
// This code is contributed by Susobhan Akhuli
Outputcurrent size: 3
3
2
1
current size: 1
Time Complexity:
- Push operation: O(n)
- Pop operation: O(1)
Auxiliary Space: O(n) since 1 queue is used.
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Stack implementation in different language
Stack in C++ STLIn C++, stack container follows LIFO (Last In First Out) order of insertion and deletion. It means that most recently inserted element is removed first and the first inserted element will be removed last. This is done by inserting and deleting elements at only one end of the stack which is generally
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Stack Class in JavaThe Java Collection framework provides a Stack class, which implements a Stack data structure. The class is based on the basic principle of LIFO (last-in-first-out). Besides the basic push and pop operations, the class also provides three more functions, such as empty, search, and peek. The Stack cl
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Stack in PythonA stack is a linear data structure that stores items in a Last-In/First-Out (LIFO) or First-In/Last-Out (FILO) manner. In stack, a new element is added at one end and an element is removed from that end only. The insert and delete operations are often called push and pop. The functions associated wi
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C# Stack with ExamplesIn C# a Stack is a Collection that follows the Last-In-First-Out (LIFO) principle. It is used when we need last-in, first-out access to items. In C# there are both generic and non-generic types of Stacks. The generic stack is in the System.Collections.Generic namespace, while the non-generic stack i
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Implementation of Stack in JavaScriptA stack is a fundamental data structure in computer science that follows the Last In, First Out (LIFO) principle. This means that the last element added to the stack will be the first one to be removed. Stacks are widely used in various applications, such as function call management, undo mechanisms
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Stack in ScalaA stack is a data structure that follows the last-in, first-out(LIFO) principle. We can add or remove element only from one end called top. Scala has both mutable and immutable versions of a stack. Syntax : import scala.collection.mutable.Stack var s = Stack[type]() // OR var s = Stack(val1, val2, v
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Some questions related to Stack implementation
Design and Implement Special Stack Data StructureDesign a Data Structure SpecialStack that supports all the stack operations like push(), pop(), isEmpty(), isFull() and an additional operation getMin() which should return minimum element from the SpecialStack. All these operations of SpecialStack must be O(1). To implement SpecialStack, you should
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Implement two Stacks in an ArrayCreate a data structure twoStacks that represent two stacks. Implementation of twoStacks should use only one array, i.e., both stacks should use the same array for storing elements. Following functions must be supported by twoStacks.push1(int x) --> pushes x to first stack push2(int x) --> pus
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Implement Stack using QueuesImplement a stack using queues. The stack should support the following operations:Push(x): Push an element onto the stack.Pop(): Pop the element from the top of the stack and return it.A Stack can be implemented using two queues. Let Stack to be implemented be 's' and queues used to implement are 'q
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Implement k stacks in an arrayGiven an array of size n, the task is to implement k stacks using a single array. We mainly need to perform the following type of queries on the stack.push(x, i) : This operations pushes the element x into stack i pop(i) : This operation pops the top of stack iHere i varies from 0 to k-1Naive Approa
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Design a stack that supports getMin() in O(1) timeDesign a Data Structure SpecialStack that supports all the stack operations like push(), pop(), peek() and an additional operation getMin() which should return minimum element from the SpecialStack. All these operations of SpecialStack must have a time complexity of O(1). Example: Input: queries = [
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Implement a stack using single queueWe are given a queue data structure, the task is to implement a stack using a single queue.Also Read: Stack using two queuesThe idea is to keep the newly inserted element always at the front of the queue, preserving the order of previous elements by appending the new element at the back and rotating
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How to implement stack using priority queue or heap?How to Implement stack using a priority queue(using min heap)? Asked In: Microsoft, Adobe. Solution: In the priority queue, we assign priority to the elements that are being pushed. A stack requires elements to be processed in the Last in First Out manner. The idea is to associate a count that dete
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Create a customized data structure which evaluates functions in O(1)Create a customized data structure such that it has functions :- GetLastElement(); RemoveLastElement(); AddElement() GetMin() All the functions should be of O(1) Question Source : amazon interview questions Approach : create a custom stack of type structure with two elements, (element, min_till_now)
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Implement Stack and Queue using DequeDeque also known as double ended queue, as name suggests is a special kind of queue in which insertions and deletions can be done at the last as well as at the beginning. A link-list representation of deque is such that each node points to the next node as well as the previous node. So that insertio
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Easy problems on Stack
Infix to Postfix ExpressionWrite a program to convert an Infix expression to Postfix form.Infix expression: The expression of the form "a operator b" (a + b) i.e., when an operator is in-between every pair of operands.Postfix expression: The expression of the form "a b operator" (ab+) i.e., When every pair of operands is foll
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Prefix to Infix ConversionInfix : An expression is called the Infix expression if the operator appears in between the operands in the expression. Simply of the form (operand1 operator operand2). Example : (A+B) * (C-D)Prefix : An expression is called the prefix expression if the operator appears in the expression before the
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Prefix to Postfix ConversionGiven a Prefix expression, convert it into a Postfix expression. Conversion of Prefix expression directly to Postfix without going through the process of converting them first to Infix and then to Postfix is much better in terms of computation and better understanding the expression (Computers evalu
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Postfix to Prefix ConversionPostfix: An expression is called the postfix expression if the operator appears in the expression after the operands. Simply of the form (operand1 operand2 operator). Example : AB+CD-* (Infix : (A+B) * (C-D) )Prefix : An expression is called the prefix expression if the operator appears in the expre
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Postfix to InfixPostfix to infix conversion involves transforming expressions where operators follow their operands (postfix notation) into standard mathematical expressions with operators placed between operands (infix notation). This conversion improves readability and understanding.Infix expression: The expressi
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Convert Infix To Prefix NotationGiven an infix expression consisting of operators (+, -, *, /, ^) and operands (lowercase characters), the task is to convert it to a prefix expression.Infix Expression: The expression of type a 'operator' b (a+b, where + is an operator) i.e., when the operator is between two operands.Prefix Express
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Valid Parentheses in an ExpressionGiven a string s representing an expression containing various types of brackets: {}, (), and [], the task is to determine whether the brackets in the expression are balanced or not. A balanced expression is one where every opening bracket has a corresponding closing bracket in the correct order.Exa
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Arithmetic Expression EvaluationThe stack organization is very effective in evaluating arithmetic expressions. Expressions are usually represented in what is known as Infix notation, in which each operator is written between two operands (i.e., A + B). With this notation, we must distinguish between ( A + B )*C and A + ( B * C ) b
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Evaluation of Postfix ExpressionGiven a postfix expression, the task is to evaluate the postfix expression. A Postfix expression is of the form "a b operator" ("a b +") i.e., a pair of operands is followed by an operator.Examples:Input: arr = ["2", "3", "1", "*", "+", "9", "-"]Output: -4Explanation: If the expression is converted
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How to Reverse a Stack using RecursionWrite a program to reverse a stack using recursion, without using any loop.Example: Input: elements present in stack from top to bottom 4 3 2 1Output: 1 2 3 4Input: elements present in stack from top to bottom 1 2 3Output: 3 2 1The idea of the solution is to hold all values in Function Call Stack un
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Reverse individual wordsGiven string str, we need to print the reverse of individual words.Examples: Input: Hello WorldOutput: olleH dlroWExplanation: Each word in "Hello World" is reversed individually, preserving the original order, resulting in "olleH dlroW".Input: Geeks for GeeksOutput: skeeG rof skeeG[Expected Approac
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Reverse a String using StackGiven a string str, the task is to reverse it using stack. Example:Input: s = "GeeksQuiz"Output: ziuQskeeGInput: s = "abc"Output: cbaAlso read: Reverse a String – Complete Tutorial.As we all know, stacks work on the principle of first in, last out. After popping all the elements and placing them bac
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Reversing a QueueYou are given a queue Q, and your task is to reverse the elements of the queue. You are only allowed to use the following standard queue operations:enqueue(x): Add an item x to the rear of the queue.dequeue(): Remove an item from the front of the queue.empty(): Check if the queue is empty or not.Exa
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Intermediate problems on Stack
How to create mergeable stack?Design a stack with the following operations. push(Stack s, x): Adds an item x to stack s pop(Stack s): Removes the top item from stack s merge(Stack s1, Stack s2): Merge contents of s2 into s1. Time Complexity of all above operations should be O(1). If we use array implementation of the stack, then
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The Stock Span ProblemThe stock span problem is a financial problem where we have a series of daily price quotes for a stock denoted by an array arr[] and the task is to calculate the span of the stock's price for all days. The span of the stock's price on ith day represents the maximum number of consecutive days leading
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Next Greater Element (NGE) for every element in given ArrayGiven an array arr[] of integers, the task is to find the Next Greater Element for each element of the array in order of their appearance in the array. Note: The Next Greater Element for an element x is the first greater element on the right side of x in the array. Elements for which no greater elem
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Next Greater Frequency ElementGiven an array, for each element find the value of the nearest element to the right which is having a frequency greater than that of the current element. If there does not exist an answer for a position, then make the value '-1'.Examples: Input: arr[] = [2, 1, 1, 3, 2, 1]Output: [1, -1, -1, 2, 1, -1
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Maximum product of indexes of next greater on left and rightGiven an array arr[1..n], for each element at position i (1 <= i <= n), define the following:left(i) is the closest index j such that j < i and arr[j] > arr[i]. If no such j exists, then left(i) = 0.right(i) is the closest index k such that k > i and arr[k] > arr[i]. If no such k e
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Iterative Tower of HanoiThe Tower of Hanoi is a mathematical puzzle with three poles and stacked disks of different sizes. The goal is to move all disks from the source pole to the destination pole using an auxiliary pole, following two rules:Only one disk can be moved at a time.A larger disk cannot be placed on a smaller
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Sort a stack using a temporary stackGiven a stack of integers, sort it in ascending order using another temporary stack.Examples: Input: [34, 3, 31, 98, 92, 23]Output: [3, 23, 31, 34, 92, 98]Explanation: After Sorting the given array it would be look like as [3, 23, 31, 34, 92, 98]Input: [3, 5, 1, 4, 2, 8]Output: [1, 2, 3, 4, 5, 8] Ap
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Reverse a stack without using extra space in O(n)Reverse a Stack without using recursion and extra space. Even the functional Stack is not allowed. Examples: Input : 1->2->3->4 Output : 4->3->2->1 Input : 6->5->4 Output : 4->5->6 We have discussed a way of reversing a stack in the below post.Reverse a Stack using Recu
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Delete middle element of a stackGiven a stack with push(), pop(), and empty() operations, The task is to delete the middle element of it without using any additional data structure.Input: s = [10, 20, 30, 40, 50]Output: [50, 40, 20, 10]Explanation: The bottom-most element will be 10 and the top-most element will be 50. Middle elem
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Check if a queue can be sorted into another queue using a stackGiven 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
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Check if an array is stack sortableGiven an array arr[] of n distinct elements, where each element is between 1 and n (inclusive), determine if it is stack-sortable.Note: An array a[] is considered stack-sortable if it can be rearranged into a sorted array b[] using a temporary stack stk with the following operations:Remove the first
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Largest Rectangular Area in a HistogramGiven a histogram represented by an array arr[], where each element of the array denotes the height of the bars in the histogram. All bars have the same width of 1 unit.Task is to find the largest rectangular area possible in a given histogram where the largest rectangle can be made of a number of c
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Maximum of minimums of every window size in a given arrayGiven an integer array arr[] of size n, the task is to find the maximum of the minimums for every window size in the given array, where the window size ranges from 1 to n.Example:Input: arr[] = [10, 20, 30]Output: [30, 20, 10]Explanation: First element in output indicates maximum of minimums of all
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Find index of closing bracket for a given opening bracket in an expressionGiven a string with brackets. If the start index of the open bracket is given, find the index of the closing bracket. Examples: Input : string = [ABC[23]][89] index = 0 Output : 8 The opening bracket at index 0 corresponds to closing bracket at index 8.Recommended PracticeClosing bracket indexTry It
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Maximum difference between nearest left and right smaller elementsGiven an array of integers, the task is to find the maximum absolute difference between the nearest left and the right smaller element of every element in the array. Note: If there is no smaller element on right side or left side of any element then we take zero as the smaller element. For example f
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Delete consecutive same words in a sequenceGiven an array of n strings arr[]. The task is to determine the number of words remaining after pairwise destruction. If two consecutive words in the array are identical, they cancel each other out. This process continues until no more eliminations are possible. Examples: Input: arr[] = ["gfg", "for
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Check mirror in n-ary treeGiven two n-ary trees, determine whether they are mirror images of each other. Each tree is described by e edges, where e denotes the number of edges in both trees. Two arrays A[]and B[] are provided, where each array contains 2*e space-separated values representing the edges of both trees. Each edg
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Reverse a number using stackGiven a number , write a program to reverse this number using stack.Examples: Input : 365Output : 563Input : 6899Output : 9986We have already discussed the simple method to reverse a number in this post. In this post we will discuss about how to reverse a number using stack.The idea to do this is to
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Reversing the first K elements of a QueueGiven an integer k and a queue of integers, The task is to reverse the order of the first k elements of the queue, leaving the other elements in the same relative order.Only following standard operations are allowed on the queue. enqueue(x): Add an item x to rear of queuedequeue(): Remove an item fr
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