How do you solve Print Zero Even Odd on LeetCode?

Print Zero Even Odd (LeetCode #1116) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: The brute force approach involves using a simple loop to print zeros, even numbers, and odd numbers in the correct order. Each thread will wait for its turn to print, which can lead to inefficiencies.

What is the time complexity of Print Zero Even Odd?

The optimal solution for Print Zero Even Odd runs in undefined time and uses undefined space. undefined

What algorithm or data structure is used to solve Print Zero Even Odd?

Print Zero Even Odd uses the following concepts: Concurrency. The recommended patterns to study are: .

#1116

Print Zero Even Odd

Medium

Concurrency

LeetCode ↗

Approaches

💡

Intuition

Time Space

The brute force approach involves using a simple loop to print zeros, even numbers, and odd numbers in the correct order. Each thread will wait for its turn to print, which can lead to inefficiencies.

⚙️

Algorithm

3 steps

1Step 1: Create a loop that runs 2n times.
2Step 2: In each iteration, print a zero followed by either an even or odd number based on the iteration count.
3Step 3: Use flags or conditions to ensure that the correct thread prints at the right time.

solution.py28 lines

1import threading
2
3class ZeroEvenOdd:
4    def __init__(self, n):
5        self.n = n
6        self.lock = threading.Lock()
7        self.count = 0
8
9    def zero(self, printNumber):
10        for _ in range(self.n):
11            self.lock.acquire()
12            printNumber(0)
13            self.count += 1
14            self.lock.release()
15
16    def even(self, printNumber):
17        for i in range(2, self.n * 2 + 1, 2):
18            while self.count % 2 != 1:
19                pass
20            printNumber(i)
21            self.count += 1
22
23    def odd(self, printNumber):
24        for i in range(1, self.n * 2, 2):
25            while self.count % 2 != 0:
26                pass
27            printNumber(i)
28            self.count += 1

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.