Implement a Queue using Stack

This is the opposite of inserting a Stack using 2 queues. Here we are going to implement a Queue using Stacks. The main thing to remember here is that A queue has operations on both ends - Front and rear, and follow FIFO structure. 

Queue can be implemented using 2 stacks, Either we can make changes in the enqueue function or in the dequeue function, for this particular problem we are going to make changes to the Enqueue function so that->
  • while stack1 is not empty, push everything from stack1 to stack2
  • push x to stack1 
  • push everything back to stack1
  • time complexity will be O(n)
  • for dequeue() -> if stack1 is not empty, pop item from stack1 and return it 
  • time complexity for dequeue is O(1)   
Code-->

class queue{
    stack<int> s1, s2;
    
    void enqueue(int x)
    {
        while(!s1.empty())
            s2.push(s1.top());
            s1.pop():
    
       s1.push(x);
        while(!s2.empty())
            s1.push(s2.top());
            s2.pop();
    }
    
    void dequeue()
    {
        if(s1.empty())        
            cout<<Queue is empty";
            exit(0);
        int x= s1.top();
        s1.pop();
        return x;
    }
};

Popular posts from this blog

Finding the Subarrays (Hackerearth)

Image
Finding subarrays can become a cumbersome task to handle if we don't know the right approach for it. In this particular question we will be using a new method.  Pair is used to combine together two values which may be different in type. Pair provides a way to store two heterogeneous objects as a single unit. Such as vector<first attribute, second attribute>. To access these attributes we use keywords 'first' and 'second' Approach - Input values and find total sum of all terms Create a loop and calculate average term by term, use logic ->  sum= first element this_average=sum/elements in the sum SZ= Elements left in array= size-elements in sum other_average=(total-sum)/SZ Use if loop to find whether this_average is bigger than other_average or not. If true, input the (i+1)th and (j+1)th element in the pair, Increment counter by 1 Sort the pair and output it. Code- #include< bits/stdc++.h > using namespace std ; int main () {       ...
Read more

Palindrome Index (Hackerrank)

Image
Here we have another Hackerrank question on String manipulation, this one involving the classic Palindrome index.  Question-  Approach - Pretty simple, we just need to check values one by one, iterating a loop from index 0 being compared to index n, where n is the last value in the index, narrowing it down to inner values one by one. The point where the elements do not stay equal is the point where element needs to be removed in order to make the string into a palindrome  The steps are -  initialise i = 0; j= string.length()-1 //last value in the string Run a loop from i to middle of the string, i increments while j decrements, bringing us a section to focus on within the string Run an if condition which checks whether character at i is equal to j or not If it is not equal, we check if the element after character at i is equal to j or not If yes, we check whether the element before j is equal to character at i and character after i is subsequently equal to element at...
Read more

Sherlock and Array (Hackerrank)

Image
Sometimes it's better to stop and first formulate a way to solve a problem before jumping right into it. Why? Well because it gives you more time to better understand a problem and secondly, in some cases even a simpler way to solve a problem.  Remember kids! When solving algorithms, Math is your friend! I realized that a little late into this question and when i did, it was just a 5 min snippet. Question-  Approach - It's quite simple and straight forward.  Instead of dividing array into two halves, getting their sums, checking whether the sums are equal and trying to find out if an element which causes sum to be unequal exist or not, here's a better approach. array - [5,6,8,11] , here we have 5+6=11 in LHS and 11 in RHS with 8 being an extra element. Imagine LHS and RHS as same, ex - x+y+x= sum. i.e 2x= sum-y. Here we just need to check if this equality holds, if it does we print YES and add value to x, else we print NO Code- 
Read more