Introduction

In this blog post, we will explore how to solve the LeetCode problem "3419" 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 two integers, n and threshold, as well as a directed weighted graph of n nodes numbered from 0 to n - 1. The graph is represented by a 2D integer array edges, where edges[i] = [Ai, Bi, Wi] indicates that there is an edge going from node Ai to node Bi with weight Wi. You have to remove some edges from this graph (possibly none), so that it satisfies the following conditions: Node 0 must be reachable from all other nodes. The maximum edge weight in the resulting graph is minimized. Each node has at most threshold outgoing edges. Return the minimum possible value of the maximum edge weight after removing the necessary edges. If it is impossible for all conditions to be satisfied, return -1. Example 1: Input: n = 5, edges = [[1,0,1],[2,0,2],[3,0,1],[4,3,1],[2,1,1]], threshold = 2 Output: 1 Explanation: Remove the edge 2 -> 0. The maximum weight among the remaining edges is 1. Example 2: Input: n = 5, edges = [[0,1,1],[0,2,2],[0,3,1],[0,4,1],[1,2,1],[1,4,1]], threshold = 1 Output: -1 Explanation: It is impossible to reach node 0 from node 2. Example 3: Input: n = 5, edges = [[1,2,1],[1,3,3],[1,4,5],[2,3,2],[3,4,2],[4,0,1]], threshold = 1 Output: 2 Explanation: Remove the edges 1 -> 3 and 1 -> 4. The maximum weight among the remaining edges is 2. Example 4: Input: n = 5, edges = [[1,2,1],[1,3,3],[1,4,5],[2,3,2],[4,0,1]], threshold = 1 Output: -1 Constraints: 2 <= n <= 105 1 <= threshold <= n - 1 1 <= edges.length <= min(105, n * (n - 1) / 2). edges[i].length == 3 0 <= Ai, Bi < n Ai != Bi 1 <= Wi <= 106 There may be multiple edges between a pair of nodes, but they must have unique weights.

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Binary Search for Max Weight: The core idea is to binary search over the possible maximum edge weights. For each potential maximum weight, we construct a subgraph containing only edges with weights less than or equal to that value.
• Check Reachability and Outgoing Degree: We then check if node 0 is reachable from all other nodes in the subgraph, and if each node's outgoing degree is within the given threshold. If both conditions are met, we try a smaller maximum weight. Otherwise, we try a larger one.

• DFS for Reachability: Depth-First Search (DFS) is used to efficiently determine reachability from all nodes to node 0 within the constructed subgraph.

• Runtime Complexity: O(E log E + E log n), where E is the number of edges and n is the number of nodes. The E log E comes from sorting the edges. The E log n comes from the binary search and DFS within each binary search iteration. Storage Complexity: O(E + N) where N is the number of nodes and E the number of edges.

Code

    def solve():
    n, edges, threshold = n_val, edges_val, threshold_val

    def check(max_weight):
        adj = [[] for _ in range(n)]
        for u, v, w in edges:
            if w <= max_weight:
                adj[u].append(v)

        # Check outgoing degree constraint
        for i in range(n):
            if len(adj[i]) > threshold:
                return False

        # Check reachability
        reachable = [False] * n

        def dfs(start_node):
            reachable_nodes = set()
            visited = set()
            stack = [start_node]

            while stack:
                node = stack.pop()
                if node in visited:
                    continue
                visited.add(node)
                reachable_nodes.add(node)
                for neighbor in adj[node]:
                    stack.append(neighbor)
            return reachable_nodes

        for i in range(n):
            reachable_nodes = dfs(i)
            if 0 not in reachable_nodes:
                return False

        return True


    weights = sorted(list(set(w for _, _, w in edges)))

    if not weights:
        return -1

    left, right = 0, len(weights) - 1
    ans = -1

    while left <= right:
        mid = (left + right) // 2
        if check(weights[mid]):
            ans = weights[mid]
            right = mid - 1
        else:
            left = mid + 1

    return ans

#Example usages (For testing purposes)
n_val = 5
edges_val = [[1,0,1],[2,0,2],[3,0,1],[4,3,1],[2,1,1]]
threshold_val = 2
print(solve())

n_val = 5
edges_val = [[0,1,1],[0,2,2],[0,3,1],[0,4,1],[1,2,1],[1,4,1]]
threshold_val = 1
print(solve())

n_val = 5
edges_val = [[1,2,1],[1,3,3],[1,4,5],[2,3,2],[3,4,2],[4,0,1]]
threshold_val = 1
print(solve())

n_val = 5
edges_val = [[1,2,1],[1,3,3],[1,4,5],[2,3,2],[4,0,1]]
threshold_val = 1
print(solve())