1670. Design Front Middle Back Queue

Problem

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 <= 10^9

• At most 1000 calls will be made to pushFront, pushMiddle, pushBack, popFront, popMiddle, and popBack.

Solution (Java)

class FrontMiddleBackQueue {
    private int[] queue = new int[1000];
    private int cur = -1;

    public FrontMiddleBackQueue() {
        // empty constructor
    }

    public void pushFront(int val) {
        cur++;
        for (int i = cur; i > 0; i--) {
            queue[i] = queue[i - 1];
        }
        queue[0] = val;
    }

    public void pushMiddle(int val) {
        if (cur < 0) {
            pushFront(val);
            return;
        }
        cur++;
        int mid = cur / 2;
        for (int i = cur; i > mid; i--) {
            queue[i] = queue[i - 1];
        }
        queue[mid] = val;
    }

    public void pushBack(int val) {
        if (cur < 0) {
            pushFront(val);
            return;
        }
        cur++;
        queue[cur] = val;
    }

    public int popFront() {
        if (cur < 0) {
            return -1;
        }
        int result = queue[0];
        for (int i = 0; i < cur; i++) {
            queue[i] = queue[i + 1];
        }
        cur--;
        return result;
    }

    public int popMiddle() {
        if (cur < 0) {
            return -1;
        }
        int mid = cur / 2;
        int result = queue[mid];
        for (int i = mid; i < cur; i++) {
            queue[i] = queue[i + 1];
        }
        cur--;
        return result;
    }

    public int popBack() {
        if (cur < 0) {
            return -1;
        }
        int result = queue[cur];
        cur--;
        return result;
    }
}

/**
 * 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();
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).