Simple Python solution
IntuitionTo solve this problem, we need to remove some edges to ensure all nodes can still reach node 0 while keeping outgoing edges within the allowed limit (t

# Intuition
To solve this problem, we need to remove some edges to ensure all nodes can still reach node 0 while keeping outgoing edges within the allowed limit (`threshold`). At the same time, we aim to minimize the weight of the largest edge in the graph.

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# Approach
1. **Sort the edges by weight**: This allows us to control the maximum edge weight we include in the graph.
2. **Binary Search**: Use binary search on the maximum edge weight to find the smallest value that satisfies all conditions.
3. **Graph Validation**: For each candidate maximum weight:
   - Build the graph with only edges below the current weight.
   - Check if node 0 is reachable from all other nodes using BFS.
   - Ensure that each node has at most `threshold` outgoing edges by keeping only the smallest ones.

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# Complexity
- **Time complexity**:  
  - Building and checking the graph for connectivity and edge limits takes $O(V + E)$ per binary search step.  
  - Binary search runs for $O(\log E)$ unique weights.  
  - Total: $O((V + E) \cdot \log E)$.

- **Space complexity**:  
  - Storing the graph and visited nodes requires $O(V + E)$.  
  - Heap usage for pruning edges adds $O(V \cdot \text{threshold})$.  
  - Total: $O(V + E)$.

# Code
```python3 []
class Solution:
    def minMaxWeight(self, n: int, edges: List[List[int]], threshold: int) -> int:
        edges.sort(key=lambda x: x[2])
        weights = sorted(set(w for _, _, w in edges))

        def is_valid_graph(max_weight):
            graph = defaultdict(list)
            reverse_graph = defaultdict(list)

            for u, v, w in edges:
                if w <= max_weight:
                    graph[u].append(v)
                    reverse_graph[v].append(u)

            visited = [False] * n
            queue = deque([0])
            visited[0] = True

            while queue:
                node = queue.popleft()
                for neighbor in reverse_graph[node]:
                    if not visited[neighbor]:
                        visited[neighbor] = True
                        queue.append(neighbor)

            if not all(visited):
                return False

            for node in range(n):
                if len(graph[node]) > threshold:
                    min_heap = []
                    for neighbor in graph[node]:
                        heapq.heappush(min_heap, neighbor)
                        if len(min_heap) > threshold:
                            heapq.heappop(min_heap)
                    graph[node] = list(min_heap)

            return True

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

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

        return result

```
The `reverse_graph` is used to check if **all nodes can reach node 0**. 

- In a normal graph, edges go **outward** from a node.
- To check if all nodes can reach node 0, we need to traverse **inward** edges (i.e., reversed edges).

By building the `reverse_graph` (where the direction of every edge is flipped), we can easily perform a BFS starting from node 0 to ensure that every other node has a path to it. 

This simplifies connectivity checks, as it avoids needing to manually reverse edges during traversal.
['Graph', 'Python3']