Introduction

In this blog post, we will explore how to solve the LeetCode problem "2101" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a list of bombs. The range of a bomb is defined as the area where its effect can be felt. This area is in the shape of a circle with the center as the location of the bomb. The bombs are represented by a 0-indexed 2D integer array bombs where bombs[i] = [xi, yi, ri]. xi and yi denote the X-coordinate and Y-coordinate of the location of the ith bomb, whereas ri denotes the radius of its range. You may choose to detonate a single bomb. When a bomb is detonated, it will detonate all bombs that lie in its range. These bombs will further detonate the bombs that lie in their ranges. Given the list of bombs, return the maximum number of bombs that can be detonated if you are allowed to detonate only one bomb. Example 1: Input: bombs = [[2,1,3],[6,1,4]] Output: 2 Explanation: The above figure shows the positions and ranges of the 2 bombs. If we detonate the left bomb, the right bomb will not be affected. But if we detonate the right bomb, both bombs will be detonated. So the maximum bombs that can be detonated is max(1, 2) = 2. Example 2: Input: bombs = [[1,1,5],[10,10,5]] Output: 1 Explanation: Detonating either bomb will not detonate the other bomb, so the maximum number of bombs that can be detonated is 1. Example 3: Input: bombs = [[1,2,3],[2,3,1],[3,4,2],[4,5,3],[5,6,4]] Output: 5 Explanation: The best bomb to detonate is bomb 0 because: - Bomb 0 detonates bombs 1 and 2. The red circle denotes the range of bomb 0. - Bomb 2 detonates bomb 3. The blue circle denotes the range of bomb 2. - Bomb 3 detonates bomb 4. The green circle denotes the range of bomb 3. Thus all 5 bombs are detonated. Constraints: 1 <= bombs.length <= 100 bombs[i].length == 3 1 <= xi, yi, ri <= 105

Explanation

Here's a breakdown of the solution:

• Graph Representation: Model the bombs and their detonation relationships as a directed graph. An edge from bomb i to bomb j exists if detonating bomb i would also detonate bomb j (i.e., bomb j is within the radius of bomb i).
• Depth-First Search (DFS): For each bomb, perform a DFS to determine how many bombs would be detonated if that bomb were the starting point.

• Maximum Detonation Count: Keep track of the maximum number of bombs detonated from any starting bomb.

• Time & Space Complexity: O(n^2) time complexity due to the graph construction and DFS for each bomb. O(n^2) space complexity in the worst case for the graph representation (adjacency matrix). The DFS itself will use call stack space which can be O(n) in the worst case

Code

    def maximumDetonation(bombs):
    """
    Calculates the maximum number of bombs that can be detonated by detonating a single bomb.

    Args:
        bombs: A list of bombs, where each bomb is represented as [x, y, r].

    Returns:
        The maximum number of bombs that can be detonated.
    """

    n = len(bombs)
    adj = [[] for _ in range(n)]

    # Build the adjacency list representing the graph
    for i in range(n):
        for j in range(n):
            if i != j:
                x1, y1, r1 = bombs[i]
                x2, y2, _ = bombs[j]
                distance_squared = (x1 - x2)**2 + (y1 - y2)**2
                if distance_squared <= r1**2:
                    adj[i].append(j)

    def dfs(start_node, visited):
        """
        Performs Depth-First Search to count the number of reachable bombs.

        Args:
            start_node: The index of the bomb to start the DFS from.
            visited: A set to keep track of visited bombs.

        Returns:
            The number of reachable bombs from the starting bomb.
        """
        count = 0
        stack = [start_node]
        visited.add(start_node)
        count += 1
        while stack:
            node = stack.pop()
            for neighbor in adj[node]:
                if neighbor not in visited:
                    visited.add(neighbor)
                    stack.append(neighbor)
                    count += 1
        return count

    max_detonated = 0
    for i in range(n):
        max_detonated = max(max_detonated, dfs(i, set()))

    return max_detonated