1116. Print Zero Even Odd LeetCode Solution

Last updated: February 22, 2025

In this guide, you will get 1116. Print Zero Even Odd LeetCode Solution with the best time and space complexity. The solution to Print Zero Even Odd problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Print Zero Even Odd solution in C++
Print Zero Even Odd solution in Java
Print Zero Even Odd solution in Python
Additional Resources

1116. Print Zero Even Odd LeetCode Solution image

Problem Statement of Print Zero Even Odd

You have a function printNumber that can be called with an integer parameter and prints it to the console.

For example, calling printNumber(7) prints 7 to the console.

You are given an instance of the class ZeroEvenOdd that has three functions: zero, even, and odd. The same instance of ZeroEvenOdd will be passed to three different threads:

Thread A: calls zero() that should only output 0’s.
Thread B: calls even() that should only output even numbers.
Thread C: calls odd() that should only output odd numbers.

Modify the given class to output the series “010203040506…” where the length of the series must be 2n.
Implement the ZeroEvenOdd class:

ZeroEvenOdd(int n) Initializes the object with the number n that represents the numbers that should be printed.
void zero(printNumber) Calls printNumber to output one zero.
void even(printNumber) Calls printNumber to output one even number.
void odd(printNumber) Calls printNumber to output one odd number.

Example 1:

Input: n = 2
Output: “0102”
Explanation: There are three threads being fired asynchronously.
One of them calls zero(), the other calls even(), and the last one calls odd().
“0102” is the correct output.

Example 2:

Input: n = 5
Output: “0102030405”

Constraints:

1 <= n <= 1000

Complexity Analysis

Time Complexity: Google AdSense
Space Complexity: Google Analytics

1116. Print Zero Even Odd LeetCode Solution in C++

// LeetCode doesn't support C++20 yet, so we don't have std::counting_semaphore
// or binary_semaphore.
#include <semaphore.h>

class ZeroEvenOdd {
 public:
  ZeroEvenOdd(int n) : n(n) {
    sem_init(&zeroSemaphore, /*pshared=*/0, /*value=*/1);
    sem_init(&evenSemaphore, /*pshared=*/0, /*value=*/0);
    sem_init(&oddSemaphore, /*pshared=*/0, /*value=*/0);
  }

  ~ZeroEvenOdd() {
    sem_destroy(&zeroSemaphore);
    sem_destroy(&evenSemaphore);
    sem_destroy(&oddSemaphore);
  }

  // printNumber(x) outputs "x", where x is an integer.
  void zero(function<void(int)> printNumber) {
    for (int i = 0; i < n; ++i) {
      sem_wait(&zeroSemaphore);
      printNumber(0);
      sem_post(&(i % 2 == 0 ? oddSemaphore : evenSemaphore));
    }
  }

  void even(function<void(int)> printNumber) {
    for (int i = 2; i <= n; i += 2) {
      sem_wait(&evenSemaphore);
      printNumber(i);
      sem_post(&zeroSemaphore);
    }
  }

  void odd(function<void(int)> printNumber) {
    for (int i = 1; i <= n; i += 2) {
      sem_wait(&oddSemaphore);
      printNumber(i);
      sem_post(&zeroSemaphore);
    }
  }

 private:
  const int n;
  sem_t zeroSemaphore;
  sem_t evenSemaphore;
  sem_t oddSemaphore;
};
/* code provided by PROGIEZ */

1116. Print Zero Even Odd LeetCode Solution in Java

class ZeroEvenOdd {
  public ZeroEvenOdd(int n) {
    this.n = n;
  }

  // printNumber.accept(x) outputs "x", where x is an integer.
  public void zero(IntConsumer printNumber) throws InterruptedException {
    for (int i = 0; i < n; ++i) {
      zeroSemaphore.acquire();
      printNumber.accept(0);
      (i % 2 == 0 ? oddSemaphore : evenSemaphore).release();
    }
  }

  public void even(IntConsumer printNumber) throws InterruptedException {
    for (int i = 2; i <= n; i += 2) {
      evenSemaphore.acquire();
      printNumber.accept(i);
      zeroSemaphore.release();
    }
  }

  public void odd(IntConsumer printNumber) throws InterruptedException {
    for (int i = 1; i <= n; i += 2) {
      oddSemaphore.acquire();
      printNumber.accept(i);
      zeroSemaphore.release();
    }
  }

  private int n;
  private Semaphore zeroSemaphore = new Semaphore(1);
  private Semaphore evenSemaphore = new Semaphore(0);
  private Semaphore oddSemaphore = new Semaphore(0);
}
// code provided by PROGIEZ

1116. Print Zero Even Odd LeetCode Solution in Python

from threading import Semaphore


class ZeroEvenOdd:
  def __init__(self, n):
    self.n = n
    self.zeroSemaphore = Semaphore(1)
    self.evenSemaphore = Semaphore(0)
    self.oddSemaphore = Semaphore(0)

  # printNumber(x) outputs "x", where x is an integer.
  def zero(self, printNumber: 'Callable[[int], None]') -> None:
    for i in range(self.n):
      self.zeroSemaphore.acquire()
      printNumber(0)
      (self.oddSemaphore if i & 2 == 0 else self.evenSemaphore).release()

  def even(self, printNumber: 'Callable[[int], None]') -> None:
    for i in range(2, self.n + 1, 2):
      self.evenSemaphore.acquire()
      printNumber(i)
      self.zeroSemaphore.release()

  def odd(self, printNumber: 'Callable[[int], None]') -> None:
    for i in range(1, self.n + 1, 2):
      self.oddSemaphore.acquire()
      printNumber(i)
      self.zeroSemaphore.release()
 # code by PROGIEZ

Additional Resources

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