If you are a software engineer or a computer science student, you might have heard of Leetcode. Leetcode is a popular platform that offers coding challenges to help programmers improve their problem-solving skills. One of the most challenging problems on Leetcode is the “Busy Intersection” problem. In this article, we will discuss how to tackle this problem and provide some tips to help you solve it efficiently.
Understanding the Problem
The Busy Intersection problem is a classic concurrency problem that involves multiple threads and synchronization. The problem statement is as follows:
You are given an array of incoming cars and outgoing cars at a busy intersection. Each car has a unique ID and a direction (either “in” or “out”). Your task is to implement a function that will determine the maximum number of cars that were present at the intersection at any given time.
Approach
To solve this problem, we need to keep track of the number of cars at the intersection at any given time. We can do this by using a variable called “count.” We can iterate through the array of cars and increment or decrement the count based on whether the car is incoming or outgoing. However, since this problem involves concurrency, we need to ensure that our solution is thread-safe.
Using Semaphores
One way to ensure thread-safety is by using semaphores. A semaphore is a synchronization primitive that allows us to control access to a shared resource. In this case, the shared resource is the “count” variable. We can use two semaphores, one for incoming cars and one for outgoing cars. We can initialize both semaphores with a value of 1.
When an incoming car arrives, it needs to acquire the incoming semaphore before incrementing the count. Similarly, when an outgoing car arrives, it needs to acquire the outgoing semaphore before decrementing the count. This ensures that only one car can access the count variable at a time.
Using Mutexes
Another way to ensure thread-safety is by using mutexes. A mutex is a synchronization primitive that allows us to protect a shared resource from concurrent access. In this case, the shared resource is the “count” variable. We can use a mutex to ensure that only one thread can access the count variable at a time.
When an incoming car arrives, it needs to acquire the mutex before incrementing the count. Similarly, when an outgoing car arrives, it needs to acquire the mutex before decrementing the count. This ensures that only one car can access the count variable at a time.
Using Atomic Operations
Atomic operations are another way to ensure thread-safety. An atomic operation is an operation that is guaranteed to be executed as a single, indivisible step. In this case, we can use an atomic integer to represent the count variable. When an incoming car arrives, we can use an atomic increment operation to increment the count. Similarly, when an outgoing car arrives, we can use an atomic decrement operation to decrement the count.
How to Solve the Busy Intersection Problem on Hackerrank
If you are a programmer, you must have heard of Hackerrank. It is a platform that provides coding challenges to help you improve your programming skills. One of the challenges on Hackerrank is the busy intersection problem. This problem requires you to find the maximum number of cars that can pass through a busy intersection without colliding. In this article, we will discuss how to solve the busy intersection problem on Hackerrank.
Understanding the Problem | How to Solve the Busy Intersection Problem on Hackerrank
Before we dive into the solution, let’s first understand the problem statement. The problem statement goes as follows: There is a busy intersection in your city. The intersection has several roads leading to it, and each road has a traffic signal. The traffic signals are not synchronized, which means that they change at different times. You are given the time when each signal changes from red to green. You need to find the maximum number of cars that can pass through the intersection without colliding. The input for this problem consists of two arrays, one containing the time when each signal changes from red to green, and the other containing the duration of each green signal. The output should be an integer representing the maximum number of cars that can pass through the intersection without colliding.
Problem Solving Approach
To solve this problem, we need to find the time when each car will reach the intersection and check if it can pass through without colliding with another car. We can do this by keeping track of the time when each signal changes from red to green and calculating the time when each car will reach the intersection. We can use two arrays to keep track of the time when each signal changes from red to green and the duration of each green signal. We can then iterate over these arrays and calculate the time when each car will reach the intersection. To calculate the time when each car will reach the intersection, we need to know the distance between the car and the intersection and the speed of the car. We can assume that all cars are traveling at a constant speed and that the distance between each car and the intersection is known. Once we have calculated the time when each car will reach the intersection, we can check if it can pass through without colliding with another car. If a car can pass through without colliding, we can increment the count of cars that have passed through the intersection.
Complexity Analysis
The time complexity of this solution is O(n^2), where n is the length of the arrival array. This is because we are iterating over all possible pairs of cars. The space complexity of this solution is O(1), as we are not using any extra space.
Optimized Approach
The above approach has a time complexity of O(n^2), which is not efficient for large inputs. We can optimize this approach by using a priority queue to keep track of the cars that can pass through the intersection without colliding. We can initialize a priority queue with the first car and its arrival time. We can then iterate over the remaining cars and check if they can pass through without colliding with the car at the front of the priority queue. If a car can pass through without colliding, we can add it to the priority queue. If a car cannot pass through without colliding, we can discard it.
Code
Let’s take a look at the code to solve the busy intersection problem on Hackerrank using a priority queue. “`python import heapq def busyIntersection(arrival, duration): n = len(arrival) count = 0 pq = [] heapq.heappush(pq, arrival[0] + duration[0]) for i in range(1, n): if arrival[i] > pq[0]: heapq.heappop(pq) else: count += 1 heapq.heappush(pq, arrival[i] + duration[i]) return count + len(pq) “` In this code, we first initialize the count to 0 and create an empty priority queue. We then add the first car and its arrival time to the priority queue. We then iterate over the remaining cars and check if they can pass through without colliding with the car at the front of the priority queue. If a car can pass through without colliding, we add it to the priority queue. If a car cannot pass through without colliding, we discard it. At the end of the iteration, we return the count of cars that have passed through the intersection plus the length of the priority queue.
Conclusion
The Busy Intersection problem is a challenging concurrency problem that requires careful synchronization to ensure thread-safety. We discussed three approaches to solve this problem: using semaphores, using mutexes, and using atomic operations. Each approach has its advantages and disadvantages, and the choice of approach depends on the specific requirements of the problem.
In summary, solving the Busy Intersection problem requires a good understanding of concurrency and synchronization. By using the right approach and following best practices, you can efficiently solve this problem and improve your problem-solving skills.
