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