Implement a queue in which push_rear(), pop_front(

Question

I came across this question: Implement a queue in which push_rear(), pop_front() and get_min() are all constant time operations.

I initially thought of using a min-heap data structure which has O(1) complexity for a get_min(). But push_rear() and pop_front() would be O(log(n)).

Does anyone know what would be the best way to implement such a queue which has O(1) push(), pop() and min()?

I googled about this, and wanted to point out this Algorithm Geeks thread. But it seems that none of the solutions follow constant time rule for all 3 methods: push(), pop() and min().

Thanks for all the suggestions.

Answer 1

You Can actually use a LinkedList to maintain the Queue.

Each element in LinkedList will be of Type

class LinkedListElement
{
   LinkedListElement next;
   int currentMin;
}

You can have two pointers One points to the Start and the other points to the End.

If you add an element to the start of the Queue. Examine the Start pointer and the node to insert. If node to insert currentmin is less than start currentmin node to insert currentmin is the minimum. Else update the currentmin with start currentmin.

Repeat the same for enque.

Answer 2

JavaScript implementation

(Credit to adamax's solution for the idea; I loosely based an implementation on it. Jump to the bottom to see fully commented code or read through the general steps below. Note that this finds the maximum value in O(1) constant time rather than the minimum value--easy to change up):

The general idea is to create two Stacks upon construction of the MaxQueue (I used a linked list as the underlying Stack data structure--not included in the code; but any Stack will do as long as it's implemented with O(1) insertion/deletion). One we'll mostly pop from (dqStack) and one we'll mostly push to (eqStack).

---

Insertion: O(1) worst case

For enqueue, if the MaxQueue is empty, we'll push the value to dqStack along with the current max value in a tuple (the same value since it's the only value in the MaxQueue); e.g.:

const m = new MaxQueue();

m.enqueue(6);

/*
the dqStack now looks like:
[6, 6] - [value, max]
*/

If the MaxQueue is not empty, we push just the value to eqStack;

m.enqueue(7);
m.enqueue(8);

/*
dqStack:         eqStack: 8
         [6, 6]           7 - just the value
*/

then, update the maximum value in the tuple.

/*
dqStack:         eqStack: 8
         [6, 8]           7
*/

---

Deletion: O(1) amortized

For dequeue we'll pop from dqStack and return the value from the tuple.

m.dequeue();
> 6

// equivalent to:
/*
const tuple = m.dqStack.pop() // [6, 8]
tuple[0];
> 6
*/

Then, if dqStack is empty, move all values in eqStack to dqStack, e.g.:

// if we build a MaxQueue
const maxQ = new MaxQueue(3, 5, 2, 4, 1);

/*
the stacks will look like:

dqStack:         eqStack: 1
                          4
                          2
         [3, 5]           5
*/

As each value is moved over, we'll check if it's greater than the max so far and store it in each tuple:

maxQ.dequeue(); // pops from dqStack (now empty), so move all from eqStack to dqStack
> 3

// as dequeue moves one value over, it checks if it's greater than the ***previous max*** and stores the max at tuple[1], i.e., [data, max]:
/*
dqStack: [5, 5] => 5 > 4 - update                          eqStack:
         [2, 4] => 2 < 4 - no update                         
         [4, 4] => 4 > 1 - update                            
         [1, 1] => 1st value moved over so max is itself            empty
*/

Because each value is moved to dqStack at most once, we can say that dequeue has O(1) amortized time complexity.

---

Finding the maximum value: O(1) worst case

Then, at any point in time, we can call getMax to retrieve the current maximum value in O(1) constant time. As long as the MaxQueue is not empty, the maximum value is easily pulled out of the next tuple in dqStack.

maxQ.getMax();
> 5

// equivalent to calling peek on the dqStack and pulling out the maximum value:
/*
const peekedTuple = maxQ.dqStack.peek(); // [5, 5]
peekedTuple[1];
> 5
*/

---

Code

class MaxQueue {
  constructor(...data) {
    // create a dequeue Stack from which we'll pop
    this.dqStack = new Stack();
    // create an enqueue Stack to which we'll push
    this.eqStack = new Stack();
    // if enqueueing data at construction, iterate through data and enqueue each
    if (data.length) for (const datum of data) this.enqueue(datum);
  }
  enqueue(data) { // O(1) constant insertion time
    // if the MaxQueue is empty,
    if (!this.peek()) {
      // push data to the dequeue Stack and indicate it's the max;
      this.dqStack.push([data, data]); // e.g., enqueue(8) ==> [data: 8, max: 8]
    } else {
      // otherwise, the MaxQueue is not empty; push data to enqueue Stack
      this.eqStack.push(data);
      // save a reference to the tuple that's next in line to be dequeued
      const next = this.dqStack.peek();
      // if the enqueueing data is > the max in that tuple, update it
      if (data > next[1]) next[1] = data;
    }
  }
  moveAllFromEqToDq() { // O(1) amortized as each value will move at most once
    // start max at -Infinity for comparison with the first value
    let max = -Infinity;
    // until enqueue Stack is empty,
    while (this.eqStack.peek()) {
      // pop from enqueue Stack and save its data
      const data = this.eqStack.pop();
      // if data is > max, set max to data
      if (data > max) max = data;
      // push to dequeue Stack and indicate the current max; e.g., [data: 7: max: 8]
      this.dqStack.push([data, max]);
    }
  }
  dequeue() { // O(1) amortized deletion due to calling moveAllFromEqToDq from time-to-time
    // if the MaxQueue is empty, return undefined
    if (!this.peek()) return;
    // pop from the dequeue Stack and save it's data
    const [data] = this.dqStack.pop();
    // if there's no data left in dequeue Stack, move all data from enqueue Stack
    if (!this.dqStack.peek()) this.moveAllFromEqToDq();
    // return the data
    return data;
  }
  peek() { // O(1) constant peek time
    // if the MaxQueue is empty, return undefined
    if (!this.dqStack.peek()) return;
    // peek at dequeue Stack and return its data
    return this.dqStack.peek()[0];
  }
  getMax() { // O(1) constant time to find maximum value
    // if the MaxQueue is empty, return undefined
    if (!this.peek()) return;
    // peek at dequeue Stack and return the current max
    return this.dqStack.peek()[1];
  }
}

Answer 3

Java Implementation

import java.io.*;
import java.util.*;

public class queueMin {
    static class stack {

        private Node<Integer> head;

        public void push(int data) {
            Node<Integer> newNode = new Node<Integer>(data);
            if(null == head) {
                head = newNode;
            } else {
                Node<Integer> prev = head;
                head = newNode;
                head.setNext(prev);
            }
        }

        public int pop() {
            int data = -1;
            if(null == head){
                System.out.println("Error Nothing to pop");
            } else {
                data = head.getData();
                head = head.getNext();
            }

            return data;
        }

        public int peek(){
            if(null == head){
                System.out.println("Error Nothing to pop");
                return -1;
            } else {
                return head.getData();
            }
        }

        public boolean isEmpty(){
            return null == head;
        }
    }

    static class stackMin extends stack {
        private stack s2;

        public stackMin(){
            s2 = new stack();
        }

        public void push(int data){
            if(data <= getMin()){
                s2.push(data);
            }

            super.push(data);
        }

        public int pop(){
            int value = super.pop();
            if(value == getMin()) {
                s2.pop();
            }
            return value;
        }

        public int getMin(){
            if(s2.isEmpty()) {
                return Integer.MAX_VALUE;
            }
            return s2.peek();
        }
    }

     static class Queue {

        private stackMin s1, s2;

        public Queue(){
            s1 = new stackMin();
            s2 = new stackMin();
        }

        public  void enQueue(int data) {
            s1.push(data);
        }

        public  int deQueue() {
            if(s2.isEmpty()) {
                while(!s1.isEmpty()) {
                    s2.push(s1.pop());
                }
            }

            return s2.pop();
        }

        public int getMin(){
            return Math.min(s1.isEmpty() ? Integer.MAX_VALUE : s1.getMin(), s2.isEmpty() ? Integer.MAX_VALUE : s2.getMin());
        }

    }



   static class Node<T> {
        private T data;
        private T min;
        private Node<T> next;

        public Node(T data){
            this.data = data;
            this.next = null;
        }


        public void setNext(Node<T> next){
            this.next = next;
        }

        public T getData(){
            return this.data;
        }

        public Node<T> getNext(){
            return this.next;
        }

        public void setMin(T min){
            this.min = min;
        }

        public T getMin(){
            return this.min;
        }
    }

    public static void main(String args[]){
       try {
           FastScanner in = newInput();
           PrintWriter out = newOutput();
          // System.out.println(out);
           Queue q = new Queue();
           int t = in.nextInt();
           while(t-- > 0) {
               String[] inp = in.nextLine().split(" ");
               switch (inp[0]) {
                   case "+":
                       q.enQueue(Integer.parseInt(inp[1]));
                       break;
                   case "-":
                       q.deQueue();
                       break;
                   case "?":
                       out.println(q.getMin());
                   default:
                       break;
               }
           }
           out.flush();
           out.close();

       } catch(IOException e){
          e.printStackTrace();
       }
    }

    static class FastScanner {
        static BufferedReader br;
        static StringTokenizer st;

        FastScanner(File f) {
            try {
                br = new BufferedReader(new FileReader(f));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }
        public FastScanner(InputStream f) {
            br = new BufferedReader(new InputStreamReader(f));
        }
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDoulbe() {
            return Double.parseDouble(next());
        }
    }

    static FastScanner newInput() throws IOException {
        if (System.getProperty("JUDGE") != null) {
            return new FastScanner(new File("input.txt"));
        } else {
            return new FastScanner(System.in);
        }
    }
    static PrintWriter newOutput() throws IOException {
        if (System.getProperty("JUDGE") != null) {
            return new PrintWriter("output.txt");
        } else {
            return new PrintWriter(System.out);
        }
    }
}

Answer 4

Use one deque (A) to store the elements and another deque (B) to store the minimums.

When x is enqueued, push_back it to A and keep pop_backing B until the back of B is smaller than x, then push_back x to B.

when dequeuing A, pop_front A as return value, and if it is equal to the front of B, pop_front B as well.

when getting the minimum of A, use the front of B as return value.

dequeue and getmin are obviously O(1). For the enqueue operation, consider the push_back of n elements. There are n push_back to A, n push_back to B and at most n pop_back of B because each element will either stay in B or being popped out once from B. Over all there are O(3n) operations and therefore the amortized cost is O(1) as well for enqueue.

Lastly the reason this algorithm works is that when you enqueue x to A, if there are elements in B that are larger than x, they will never be minimums now because x will stay in the queue A longer than any elements in B (a queue is FIFO). Therefore we need to pop out elements in B (from the back) that are larger than x before we push x into B.

from collections import deque


class MinQueue(deque):
    def __init__(self):
        deque.__init__(self)
        self.minq = deque()

    def push_rear(self, x):
        self.append(x)
        while len(self.minq) > 0 and self.minq[-1] > x:
            self.minq.pop()
        self.minq.append(x)

    def pop_front(self):
        x = self.popleft()
        if self.minq[0] == x:
            self.minq.popleft()
        return(x)

    def get_min(self):
        return(self.minq[0])

Answer 5

You can implement a stack with O(1) pop(), push() and get_min(): just store the current minimum together with each element. So, for example, the stack [4,2,5,1] (1 on top) becomes [(4,4), (2,2), (5,2), (1,1)].

Then you can use two stacks to implement the queue. Push to one stack, pop from another one; if the second stack is empty during the pop, move all elements from the first stack to the second one.

E.g for a pop request, moving all the elements from first stack [(4,4), (2,2), (5,2), (1,1)], the second stack would be [(1,1), (5,1), (2,1), (4,1)]. and now return top element from second stack.

To find the minimum element of the queue, look at the smallest two elements of the individual min-stacks, then take the minimum of those two values. (Of course, there's some extra logic here is case one of the stacks is empty, but that's not too hard to work around).

It will have O(1) get_min() and push() and amortized O(1) pop().

Answer 6

Solutions to this question, including code, can be found here: http://discuss.joelonsoftware.com/default.asp?interview.11.742223.32