Python Beats 100% with Explanation (Better than Editorial)
Approach: BFS/Greedy Search with a Seen SetProblem Breakdown
You are givenncities labeled from0ton-1and a list ofconnections, whereconnections[i] = [a, b]repres
### `Approach: BFS/Greedy Search with a Seen Set`

#### `Problem Breakdown`
- You are given `n` cities labeled from `0` to `n-1` and a list of `connections`, where `connections[i] = [a, b]` represents a directed road from city `a` to city `b`.
- You must reorder some roads so that **every city can reach city `0`**.
- Return the **minimum number of roads that need to be reversed**.

---

### `Step-by-Step Breakdown`

1. **Initialize Data Structures**
   - `seen = {0}`: A set to track visited cities, starting with city `0`.
   - `ans = 0`: A counter to track the number of edges that need to be reversed.

2. **Iterate Until All Cities Are Visited**
   - Continue processing while `len(seen) < n` (until all cities are visited).
   - Create a temporary `check = []` list to store unprocessed connections.

3. **Process Each Connection**
   - If `path[1]` is already in `seen`, it means the road is **correctly directed** (`b → a`), so just mark `path[0]` as seen.
   - If `path[0]` is already in `seen`, it means we must **reverse the road** (`a → b`), so:
     - Mark `path[1]` as seen.
     - Increase `ans` since we are reversing the edge.
   - Otherwise, add the unprocessed connection to `check`.

4. **Reverse Remaining Paths in `connections`**
   - At the end of each iteration, update `connections = check[::-1]` to process remaining roads in the next loop.

5. **Return the Number of Reversals (`ans`)**
   - This ensures that all roads lead to city `0` using the minimum number of reversals.

---

### `Time and Space Complexity Analysis`

- **Time Complexity: `O(n²)` in the worst case**
  - We may check all `n` edges multiple times (`O(n)` iterations × `O(n)` operations per iteration).
  - However, in most cases, it is **closer to `O(n)`** since we gradually reduce the number of unprocessed edges.
- **Space Complexity: `O(n)`**
  - The `seen` set stores up to `n` cities.

---

### `Example Walkthrough`

#### `Input`
```python


from typing import List

class Solution:
    def minReorder(self, n: int, connections: List[List[int]]) -> int:
        # Initialize a set to track visited cities
        seen = {0}
        # Counter for the number of roads that need to be reversed
        ans = 0
        
        while len(seen) < n:
            check = []
            for path in connections:
                if path[1] in seen:
                    seen.add(path[0])
                elif path[0] in seen:
                    seen.add(path[1])
                    ans += 1
                else:
                    check.append(path)
            # Reverse the order of unprocessed connections for the next iteration
            connections = check[::-1]
        
        return ans





```


['Python3']