Leetcode #1117: Building H2O
In this guide, we solve Leetcode #1117 Building H2O 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
There are two kinds of threads: oxygen and hydrogen. Your goal is to group these threads to form water molecules.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Concurrency
Intuition
The constraints allow a direct scan that keeps only the essential state.
By translating the requirements into a clean loop, the logic stays easy to reason about.
Approach
Iterate through the data once, updating the state needed to compute the answer.
Return the final state after the traversal is complete.
Steps:
- Parse the input.
- Iterate and update state.
- Return the computed answer.
Example
Input: water = "HOH"
Output: "HHO"
Explanation: "HOH" and "OHH" are also valid answers.
Python Solution
from threading import Semaphore
class H2O:
def __init__(self):
self.h = Semaphore(2)
self.o = Semaphore(0)
def hydrogen(self, releaseHydrogen: "Callable[[], None]") -> None:
self.h.acquire()
# releaseHydrogen() outputs "H". Do not change or remove this line.
releaseHydrogen()
if self.h._value == 0:
self.o.release()
def oxygen(self, releaseOxygen: "Callable[[], None]") -> None:
self.o.acquire()
# releaseOxygen() outputs "O". Do not change or remove this line.
releaseOxygen()
self.h.release(2)
Complexity
The time complexity is O(n). The space complexity is O(1) to O(n).
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.