Implement Stack using Queues | Queue in Data Structure & Algorithms | Interview Question Hello World

This is the video under the series of DATA STRUCTURE & ALGORITHM in a STACK & QUEUE Playlist. Now we are going to Learn to solve Questions on Queue in Data Structure from Leetcode Implement Stack using Queues.

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[DPCODE]

That's the wonderful explanation on the stack & queue. Bhaiya!!

[neelsoni]

Most Easiest Way To Understand Thank You SO MucH.

[Manish-wwlg]

bhaiya in your example you have taken element in stack 2, 3, 4(down to up) but it should be 4, 3, 2 (down to up) bcz how you have put 4, 3, 2 in q1.

[apna_Vaibhav]

very to the point explaination sir thank you

[sus_tha_coder]

great video... best channel for dsa problems.

[ankitkumar-rioh]

in question clearly written we have only use push from back and pop of queue ! so you had use front

[samridhrana]

done without watching video love you bhaiya

[pritishpattnaik]

bhaiya can we consider the rear as the top of stack ?

[memefulland]

done by myself in first go using 1 queue, thanks bro improving daily bcs of u:
class MyStack {
public:
queue<int>q1;
MyStack() {

}

void push(int x) {
q1.push(x);
}

int pop() {
int n=q1.size();
for(int i=0;i<n-1;i++)
{
int f=q1.front();
q1.pop();
q1.push(f);
}
int f=q1.front();
q1.pop();
return f;

}

int top() {
return q1.back();
}

bool empty() {
return q1.empty();
}
};

[code-nk]

1st method :
class MyStack {
queue<int> queue1;
queue<int> queue2;
// Move all elements except the last one to queue2 --otherwise queue2 will be similar to queue 1 -- so now last elem is remaining in queue1 we can perform desired operation on it and then swap both queues afterwards so -- queue1 does not remain empty
void transferexceptlast(){
while (queue1.size() > 1) {
queue2.push(queue1.front());
queue1.pop();
}
}
public:
MyStack() {

}


void push(int x) {
queue1.push(x);
}

// Removes the element on top of the stack and returns that element.
int pop() {
transferexceptlast();
int val = queue1.front();
queue1.pop();;
swap(queue1, queue2);
return val;
}

// Get the top element.
int top() {
transferexceptlast();
// The last element in queue1 is the top element of the stack
int topElement = queue1.front();
// Move the top element to queue2
queue2.push(topElement);
queue1.pop();
// Swap queue1 and queue2
swap(queue1, queue2);
return topElement;
}

// Returns whether the stack is empty.
bool empty() {
return queue1.empty();
}



};

2nd method :

class MyStack {
queue<int>q;
public:
//method is that : 👇👇
//When a new element is pushed, insert it at the end of the queue.
// Then, rotate all the elements (except the newly added elem) to the back of the queue, effectively bringing the newly added element to the front.
MyStack() {

}

void push(int x) {
q.push(x);
// rotate the elem to make the last elem as front elem
for(int i = 0; i<q.size()-1;++i){
q.push(q.front());
q.pop();
}

}

int pop() {
int topVal = q.front();
q.pop();
return topVal;

}

int top() {
return q.front();
}

bool empty() {
return q.empty();
}
};

[asurstark]

This code showing wrong answer
class MyStack {
public:
MyStack() {

}
queue<int> q1, q2;

void push(int x)
{
while(!q1.empty())
{
q2.push(q1.front());
q1.pop();
}
q1.push(x);
while(!q2.empty())
{
q1.push(q2.front());
q2.pop();
}
}

int pop()
{
int val = q1.front();
q2.pop();
return val;
}

int top()
{
return q1.front();

}

bool empty() {
return q1.empty();
}
};

/**
* Your MyStack object will be instantiated and called as such:
* MyStack* obj = new MyStack();
* obj->push(x);
* int param_2 = obj->pop();
* int param_3 = obj->top();
* bool param_4 = obj->empty();
*/

[whatthepeeps]

Bhaiuya isme apne shurum me bola tha ki elements stack ke ulte daalne pdenge but fir wo kiye bina kaise ye program solve hua??

[darshanusdad]

Sir please send c language code not c++
Please sir