Design Front Middle Back Queue

Description

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.

Code

Time Complexity: $O(1)$, Space Complexity: $O(N)$

• C++ std::list

class FrontMiddleBackQueue {
    list<int> List;
    list<int>::iterator mid;
    int n = 0;
public:
    FrontMiddleBackQueue() {
        
    }
    
    void pushFront(int val) {
        List.push_front(val);
        if(n == 0) {
            mid = List.begin();
        } else if(n % 2 == 1) {
            mid = prev(mid);
        } 
        n++;
    }
    
    void pushMiddle(int val) {
        if(n == 0) {
            List.push_back(val);
            mid = List.begin();
        } else if(n % 2 == 0) {
            List.insert(next(mid), val);
            mid = next(mid);
        } else {
            List.insert(mid, val);
            mid = prev(mid);
        }
        n++;
    }
    
    void pushBack(int val) {
        List.push_back(val);
        if(n == 0) {
            mid = List.begin();
        } else if(n % 2 == 0) {
            mid = next(mid);
        } 
        n++;
    }
    
    int popFront() {
        if (n==0) return -1;
        int ret = List.front();
 
        if(n % 2 == 0) {
            mid = next(mid);
        }
 
        List.pop_front();
        n--;
        return ret;
    }
    
    int popMiddle() {
        if (n==0) return -1;
        int ret = *mid;
        list<int>::iterator new_mid;
        if(n % 2 == 0) {
            new_mid = next(mid);
        } else {
            new_mid = prev(mid);
        }
        List.erase(mid);
        n--;
        mid = new_mid;
        return ret;
        
    }
    
    int popBack() {
        if (n==0) return -1;
        int ret = List.back();
 
        if(n % 2 == 1) {
            mid = prev(mid);
        }
 
        List.pop_back();
        n--;
        return ret;
        
    }
};
 
/**
 * Your FrontMiddleBackQueue object will be instantiated and called as such:
 * FrontMiddleBackQueue* obj = new FrontMiddleBackQueue();
 * obj->pushFront(val);
 * obj->pushMiddle(val);
 * obj->pushBack(val);
 * int param_4 = obj->popFront();
 * int param_5 = obj->popMiddle();
 * int param_6 = obj->popBack();
 */

Source

• Design Front Middle Back Queue - LeetCode