Leetcode #1670: Design Front Middle Back Queue
In this guide, we solve Leetcode #1670 Design Front Middle Back Queue in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Design a queue that supports push and pop operations in the front, middle, and back. Implement the FrontMiddleBack class: FrontMiddleBack() Initializes the queue.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Queue, Array, Linked List, Data Stream
Intuition
We need level-order exploration or shortest-step expansion, which maps directly to a queue.
BFS guarantees the first time you reach a node is the shortest in unweighted graphs.
Approach
Initialize the queue with starting nodes and expand outward layer by layer.
Track visited nodes to avoid cycles and redundant work.
Steps:
- Initialize a queue with start nodes.
- Pop, process, and enqueue neighbors.
- Track visited nodes.
Example
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)
Python Solution
class FrontMiddleBackQueue:
def __init__(self):
self.q1 = deque()
self.q2 = deque()
def pushFront(self, val: int) -> None:
self.q1.appendleft(val)
self.rebalance()
def pushMiddle(self, val: int) -> None:
self.q1.append(val)
self.rebalance()
def pushBack(self, val: int) -> None:
self.q2.append(val)
self.rebalance()
def popFront(self) -> int:
if not self.q1 and not self.q2:
return -1
if self.q1:
val = self.q1.popleft()
else:
val = self.q2.popleft()
self.rebalance()
return val
def popMiddle(self) -> int:
if not self.q1 and not self.q2:
return -1
if len(self.q1) == len(self.q2):
val = self.q1.pop()
else:
val = self.q2.popleft()
self.rebalance()
return val
def popBack(self) -> int:
if not self.q2:
return -1
val = self.q2.pop()
self.rebalance()
return val
def rebalance(self):
if len(self.q1) > len(self.q2):
self.q2.appendleft(self.q1.pop())
if len(self.q2) > len(self.q1) + 1:
self.q1.append(self.q2.popleft())
# Your FrontMiddleBackQueue object will be instantiated and called as such:
# obj = FrontMiddleBackQueue()
# obj.pushFront(val)
# obj.pushMiddle(val)
# obj.pushBack(val)
# param_4 = obj.popFront()
# param_5 = obj.popMiddle()
# param_6 = obj.popBack()
Complexity
The time complexity is O(V+E). The space complexity is , where is the number of elements in the queue.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.