1670

1670. Design Front Middle Back Queue (Medium)

Design a queue that supports push and pop operations in the front, middle, and back.

Implement the FrontMiddleBack class:

• FrontMiddleBack() Initializes the queue.
• void pushFront(int val) Adds val to the front of the queue.
• void pushMiddle(int val) Adds val to the middle of the queue.
• void pushBack(int val) Adds val to the back of the queue.
• int popFront() Removes the front element of the queue and returns it. If the queue is empty, return -1.
• int popMiddle() Removes the middle element of the queue and returns it. If the queue is empty, return -1.
• int popBack() Removes the back element of the queue and returns it. If the queue is empty, return -1.

Notice that when there are two middle position choices, the operation is performed on the frontmost middle position choice. For example:

• Pushing 6 into the middle of [1, 2, 3, 4, 5] results in [1, 2, 6, 3, 4, 5].
• Popping the middle from [1, 2, 3, 4, 5, 6] returns 3 and results in [1, 2, 4, 5, 6].

 

Example 1:

Input:
["FrontMiddleBackQueue", "pushFront", "pushBack", "pushMiddle", "pushMiddle", "popFront", "popMiddle", "popMiddle", "popBack", "popFront"]
[[], [1], [2], [3], [4], [], [], [], [], []]
Output:
[null, null, null, null, null, 1, 3, 4, 2, -1]

Explanation:
FrontMiddleBackQueue q = new FrontMiddleBackQueue();
q.pushFront(1);   // [1]
q.pushBack(2);    // [1, 2]
q.pushMiddle(3);  // [1, 3, 2]
q.pushMiddle(4);  // [1, 4, 3, 2]
q.popFront();     // return 1 -> [4, 3, 2]
q.popMiddle();    // return 3 -> [4, 2]
q.popMiddle();    // return 4 -> [2]
q.popBack();      // return 2 -> []
q.popFront();     // return -1 -> [] (The queue is empty)

 

Constraints:

• 1 <= val <= 109
• At most 1000 calls will be made to pushFront, pushMiddle, pushBack, popFront, popMiddle, and popBack.

Related Topics:
Linked List, Design, Dequeue

Similar Questions:

• Design Circular Deque (Medium)
• Design Circular Queue (Medium)

Solution 1. Two Deques

Use two deque<int> a, b where a and b represent the first and second half of the numbers in queue, respectively.

Make sure a.size() be either b.size() or b.size() - 1 so that when we pushMiddle, always push to a.

a2b and b2a are two helper functions nudging numbers around to ensure the above criteria.

// OJ: https://leetcode.com/problems/design-front-middle-back-queue/
// Author: github.com/lzl124631x
// Time: O(1) for all
// Space: O(N)
class FrontMiddleBackQueue {
    deque<int> a, b;
    void a2b() {
        if (a.size() <= b.size()) return;
        b.push_front(a.back());
        a.pop_back();
    }
    void b2a() {
        if (b.size() <= a.size() + 1) return;
        a.push_back(b.front());
        b.pop_front();
    }
public:
    FrontMiddleBackQueue() {}
    void pushFront(int val) {
        a.push_front(val);
        a2b();
    }
    void pushMiddle(int val) {
        a.push_back(val);
        a2b();
    }
    void pushBack(int val) {
        b.push_back(val);
        b2a();
    }
    int popFront() {
        if (a.empty() && b.empty()) return -1;
        int ans;
        if (a.empty()) {
            ans = b.front();
            b.pop_front();
        } else {
            ans = a.front();
            a.pop_front();
            b2a();
        }
        return ans;
    }
    int popMiddle() {
        if (a.empty() && b.empty()) return -1;
        int ans;
        if (a.size() == b.size()) {
            ans = a.back();
            a.pop_back();
        } else {
            ans = b.front();
            b.pop_front();
        }
        return ans;
    }
    int popBack() {
        if (a.empty() && b.empty()) return -1;
        int ans = b.back();
        b.pop_back();
        a2b();
        return ans;
    }
};