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Challenge

Problem Statement - Implement a Queue using two Stacks

I implemented dequeue() in \$O(1)\$ at the cost of enqueue() in \$O(n)\$

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <stack>
using namespace std;

struct Queue{
    stack<int> s1,s2;

    int dequeue(){
        int x = s1.top();
        s1.pop();
        return x;
    }

    void peek(){
        cout<< s1.top()<<"\n";
    }

    void enqueue(int data){
        while(!s1.empty()){
            s2.push(s1.top());
            s1.pop();
        }

        s1.push(data);

        while(!s2.empty()){
            s1.push(s2.top());
            s2.pop();
        }
    }
};


int main() {
    int t; cin>>t;
    Queue q;

    while(t--){
        int k; cin>>k;
        switch(k){
            case 1:
                int x; cin>>x;
                q.enqueue(x);
                break;
            case 2:
                q.dequeue();
                break;
            case 3:
                q.peek();
        }
    }
    return 0;
}

There are 15 test cases. This code runs properly for the first 5 test cases then I get a 'Time Limit Exceeded' message for the rest of test cases.

This means the complexity of my program is too much.

So I tried reversing it, I made enqueue() as \$O(1)\$ and dequeue() as \$O(n)\$ but even that did not change anything.

What should I do to reduce time complexity of this code?

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  • 5
    \$\begingroup\$ You might get more votes and an answer if you include a description of the programming challenge. \$\endgroup\$ – pacmaninbw Aug 3 at 18:04
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Algorithmic complexity for combined operations

Containers are interesting elements in most programming languages: they have an internal state, namely their elements and therefore their size.

This introduces an additional state compared to usual algorithmic asymptotical complexity analysis. For example, a naive implemented std::vector::push_back will yield \$\mathcal O(n^2)\$ complexity if push_back increases the capacity only by one:

// pseudo-code, also very bad performance and no error handling, do not use!
void vector::push_back(T value){
    if(_size == _capacity) {
         // has to move all elements and is therefore O(n)
         reserve(_capacity + 1);
    }
    _data[_size++] = value;
}

While trivial, this code has a severe problem: we have to copy all elements in ever iteration. If we use push_back in a loop, we end up with a quadratic complexity.

The real push_back has therefore some tricks up in its sleeves, as the standard dictates that it will have \$O(1)\$ amortized complexity. More about that later in this review.

However, this small introduction should give you a hint about the central flaw in your current approach.

A hidden quadratic term

I implemented dequeue() in \$O(1)\$ at the cost of enqueue() in \$O(n)\$

For this task, it's much more important to see the cost on both functions for \$k\$ enqueued elements, not a single one.

So let's envision the queue {1,2,3,4}. How many steps do we need?

to enqueue 1: 
  1. put `1`  in `s1`
to enqueue 2:
  1. move `1` from `s1` to `s2`           
  2. put `2` in `s1`
  3. move `1` from `s2` to `s1`
- to enqueue 3:
  1. move `1` from `s1` to `s2`
  2. move `2` from `s1` to `s2`
  3. put `3` in `s1`
  4. move `2` from `s2` to `s1`
  5. move `1` from `s2` to `s1`
- to enqueue 4:
  1. move `1` from `s1` to `s2`
  2. move `2` from `s1` to `s2`
  3. move `3` from `s1` to `s2`
  4. put `4` in `s1`
  5. move `3` from `s2` to `s1`
  6. move `2` from `s2` to `s1`
  7. move `1` from `s2` to `s1`

We see that the \$n\$th element will take \$2(n-1)\$ swaps. Therefore, if we insert a total of \$n\$ elements into our queue, we end up with \$\mathcal O(n^2)\$ to complete all enqueues.

We need to do better.

A better queue with two stacks

So let's get back to the drawing board. What do we need?

  1. We need to enqueue
  2. We need to dequeue
  3. We need to peek.

None of those terms indicates that we need to have all data in one stack. So first of all, we need to use better names, because s1 and s2 are pretty bland and non-descriptive:

class Queue{
    stack<int> in;
    stack<int> out;

    void flip();   // < new

public:
    void enqueue(int);
    int  dequeue();
    int  peek();
};

I'll implement the methods outside of the class declaration to keep the code segments short, but you're free to place them inline again. Note that I switched from a struct to a class, because its Queues job to make sure that in and out are handled correct; no one else should be able to change them.

What will we use in and out for? Well, as long as we have elements in out, we will use them for dequeue and peek. And whatever gets enqueued gets pushed right ontop of in, no questions asked.

The critical part is how to get an element from in to out, right? I let you think about that for some paragraphs, but you may also stop here and try it yourself; the declaration above contains a clue.

So, let's have a look at my proposal for the definition of enqueue(), peek and dequeue():

void Queue::enqueue(int value) {
    in.push(value);
}

int Queue::peek() {
    if (out.empty()) {
        flip();
    }
    return out.top();
}

int Queue::dequeue() {
    if (out.empty()) {
        flip();
    }
    int value = out.top();
    out.pop();
    return value;
}

Note that the additional private method flip is yet missing. However, I will state the envisioned complexity:

  • enqueue is \$\mathcal O(1)\$ (amortized)
  • dequeue is \$\mathcal O(1)\$ (amortized)
  • peek is \$\mathcal O(1)\$ (amortized)

Intrigued? Great. So let's check flip:

void Queue::flip() {
    while(not in.empty()) {
        out.push(in.top());
        in.pop();
    }
}

flip takes all elements in in and moves them to out. It thereby changes the order of the elements, so that the top in in will be the last to get moved out of out. Note that we call flip only when out is empty.

Amortized analysis

*"Wait a second! That's \$\mathcal O(n)\$" I hear you say. And that's completely correct. However, how often do we need to call flip? Or, rather more important, how often do elements get moved?

The answer on the latter question is: exactly one time from in to out. At no point will they move back. The number of flip calls is much trickier, though, but it doesn't really matter. At worst, flip may need to flip all elements, for example if we use it as follows:

for(int i = 0; i < 100; ++i) {
    queue.enqueue(i);
}
for(int i = 0; i < 100; ++i) {
    queue.dequeue(); // first dequeue will flip
}

However, only the first dequeue will flip. All others will yield the element immediately. Therefore, if we enqueue \$n\$ elements and then dequeue all of them, we end up with dequeue()'s complexity as:

$$\frac{n\mathcal O(1) + \mathcal O(n)}{n} = \mathcal O(1).$$

We can compare that with your original variant: $$\frac{n\mathcal O(n)}{n} = \mathcal O(n)$$

This is why your code exceeded the time limit.

Further review on code

There are some other findings that don't need as much detail as the algorithm, but nonetheless can be improved:

  • there are some unused includes (<vector>)
  • using namespace std is considered bad practice
  • the names are misleading or at least not self-descriptive
  • peek() usually returns a value and does not print
  • the struct Queue should have been a class Queue due to the assumption that the elements are always in the right order.
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  • \$\begingroup\$ Thanks. This answer dives in a lot deeper than the question. I am reading the links you provided. Later I used simple arrays to implement stacks rather than STL stack and the solution got accepted. \$\endgroup\$ – KshitijV97 Aug 4 at 6:43
  • 1
    \$\begingroup\$ You might consider std::vector as an alternative to an array-based stack, . std::stack usually uses std::vector underneath. You might as well cut out the middle-man and use std::vector and std::vector::reverse to make sure that push_back won't allocate for the first ~3k elements or so. \$\endgroup\$ – Zeta Aug 4 at 9:15
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You might want to look at the discussion tab for why the code is timing out. Basically the code should reverse the input stack only when peek or dequeue is called and not in enqueue.

The rest of answer is a review of the code as posted and it ignores the fact that HackerRank supplied some of the code, such as the includes and the using namespace std. Issues such as readability and maintainability are beyond the scope of HackerRank but are considerations when writing good code.

Avoid Using Namespace STD
If you are coding professionally you probably should get out of the habit of using the "using namespace std;" statement. The code will more clearly define where cout and other functions are coming from (std::cin, std::cout). As you start using namespaces in your code it is better to identify where each function comes from because there may be function name collisions from different namespaces. The function cout you may override within your own classes. This stack overflow question discusses this in more detail.

Magic Numbers
Numeric constants in code are sometimes referred to as Magic Numbers, because there is no obvious meaning for them. The values for the variable k are defined by the problem, but it might be better to use symbolic constants rather than raw numbers in the switch statement. That would make the code easier to read and maintain. C++ provides a couple of methods for this, there could be an enum, or they could be defined as constants using const or constexpr. Any of these would make the code more readable. There is a discussion of this on stackoverflow.

Use Descriptive Variable Names
The variables s1 and s2 are not very clear, and if they weren't in a std::stack declarations I really would have no idea what they were. Since this is a queue problem it might be better to name them front and rear to represent what they are used for. It is very hard to maintain code with variable names such as s1, s2, q, t, k and x. As an example for k I might use queryIndex.

Since the restrictions on all input indicates there will be no negative numbers it might be better to use unsigned rather than int.

Prefer to Not Include What Isn't Necessary
It is much better to only include what is necessary in a source file. There are currently 6 files included but only two are necessary (iostream and stack) for this implementation. Including cstdio is bad because it might lead the programmer to use C I/O statements rather the C++ ones. Including only what is needed improves compile times, because the contents of all the includes are part of the compilation. It could lead to other problems if you are implementing your own class rather than using a C++ container class (such as std::queue from #include <queue>).

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <stack>

Prefer One Declaration Per Line
Maintaining code is easier when one can find the declarations for variables. It would be easier to find where s2 is declared if it was on a separate line.

    std::stack<int> s1, s2;

Versus

    std::stack<int> s1;
    std::stack<int> s2;

Remember you may win a lottery and may not be the one maintaining the code.

The same reasoning applies to 2 statements on a line such as

    int k; cin>>k;
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One thing not already mentioned in the answers: When you have a member function which does not change the state of the object, then make it const:

// no
void peek(){ cout<< s1.top()<<"\n"; }

// better
void peek() const { cout<< s1.top()<<"\n"; }

// even better
int peek() const { return s1.top(); }
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  • 2
    \$\begingroup\$ Excellent addition, completely overlooked that. \$\endgroup\$ – Zeta Aug 4 at 9:16

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