Introduction

In this blog post, we will explore how to solve the LeetCode problem "882" 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 an undirected graph (the "original graph") with n nodes labeled from 0 to n - 1. You decide to subdivide each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge. The graph is given as a 2D array of edges where edges[i] = [ui, vi, cnti] indicates that there is an edge between nodes ui and vi in the original graph, and cnti is the total number of new nodes that you will subdivide the edge into. Note that cnti == 0 means you will not subdivide the edge. To subdivide the edge [ui, vi], replace it with (cnti + 1) new edges and cnti new nodes. The new nodes are x1, x2, ..., xcnti, and the new edges are [ui, x1], [x1, x2], [x2, x3], ..., [xcnti-1, xcnti], [xcnti, vi]. In this new graph, you want to know how many nodes are reachable from the node 0, where a node is reachable if the distance is maxMoves or less. Given the original graph and maxMoves, return the number of nodes that are reachable from node 0 in the new graph. Example 1: Input: edges = [[0,1,10],[0,2,1],[1,2,2]], maxMoves = 6, n = 3 Output: 13 Explanation: The edge subdivisions are shown in the image above. The nodes that are reachable are highlighted in yellow. Example 2: Input: edges = [[0,1,4],[1,2,6],[0,2,8],[1,3,1]], maxMoves = 10, n = 4 Output: 23 Example 3: Input: edges = [[1,2,4],[1,4,5],[1,3,1],[2,3,4],[3,4,5]], maxMoves = 17, n = 5 Output: 1 Explanation: Node 0 is disconnected from the rest of the graph, so only node 0 is reachable. Constraints: 0 <= edges.length <= min(n * (n - 1) / 2, 104) edges[i].length == 3 0 <= ui < vi < n There are no multiple edges in the graph. 0 <= cnti <= 104 0 <= maxMoves <= 109 1 <= n <= 3000

Explanation

Here's a breakdown of the solution approach and the Python code:

• High-Level Approach:

  • Build an adjacency list representing the subdivided graph. Calculate distances from node 0 using Dijkstra's algorithm.
  • Count reachable original nodes (distance <= maxMoves).
  • For each edge, calculate how many of the new nodes on that edge are reachable given the remaining moves after reaching one of the original nodes connected by the edge.

• Complexity:

  • Runtime: O(E log V), where E is the number of edges in the subdivided graph and V is the number of nodes in the subdivided graph. In the worst case, E can be O(number of edges in original graph max cnti), and V can be O(n + number of edges in original graph max cnti), so the complexity can be expressed as O(number of edges max_cnt log (n + number of edges * max_cnt)).
  • Storage: O(V + E) to store the graph and distances.

• Python Code:

Code

    import heapq

def reachableNodes(edges, maxMoves, n):
    """
    Calculates the number of reachable nodes from node 0 in the subdivided graph.

    Args:
        edges: A list of edges, where each edge is a list [u, v, cnt], indicating an edge between nodes u and v
               in the original graph and cnt is the number of new nodes to subdivide the edge into.
        maxMoves: The maximum number of moves allowed.
        n: The number of nodes in the original graph.

    Returns:
        The number of reachable nodes from node 0 in the subdivided graph.
    """

    adj = {i: [] for i in range(n)}
    for u, v, cnt in edges:
        adj[u].append((v, cnt + 1))  # cnt + 1 is the edge weight (number of nodes)
        adj[v].append((u, cnt + 1))

    dist = {i: float('inf') for i in range(n)}
    dist[0] = 0
    pq = [(0, 0)]  # (distance, node)

    while pq:
        d, u = heapq.heappop(pq)

        if d > dist[u]:
            continue

        for v, weight in adj[u]:
            if dist[u] + weight < dist[v]:
                dist[v] = dist[u] + weight
                heapq.heappush(pq, (dist[v], v))

    reachable_nodes = 0
    for d in dist.values():
        if d <= maxMoves:
            reachable_nodes += 1

    reachable_new_nodes = 0
    for u, v, cnt in edges:
        dist_u = dist[u] if u in dist else float('inf')
        dist_v = dist[v] if v in dist else float('inf')

        reach_u = max(0, maxMoves - dist_u)
        reach_v = max(0, maxMoves - dist_v)

        reachable_new_nodes += min(cnt, reach_u + reach_v)

    return reachable_nodes + reachable_new_nodes