leetcode/docs/(algorithms)/breadth-first-search.mdx

---
title: Breadth-First Search
---

# Breadth-First Search (BFS)

## The idea

You want a mango seller. You ask your friends first.
No luck? You ask friends-of-friends. Then their friends.
You search in **rings**, closest first.

That is BFS. It answers two questions:
1. **Is there a path** from A to B?
2. **What is the shortest path?** (unweighted — fewest hops)

BFS is *always* shortest path on unweighted graphs. First time you reach a node = fewest possible steps.

## The picture

```mermaid
flowchart TD
    YOU((You)) --> A((Alice))
    YOU --> B((Bob))
    A --> C((Claire))
    A --> D((Dan))
    B --> E((Eve))
    E --> M((Mango seller ✓))

    style YOU fill:#1565c0,color:#fff
    style A fill:#6a1b9a,color:#fff
    style B fill:#6a1b9a,color:#fff
    style C fill:#ef6c00,color:#fff
    style D fill:#ef6c00,color:#fff
    style E fill:#ef6c00,color:#fff
    style M fill:#2e7d32,color:#fff
```

Blue = start. Purple = ring 1. Orange = ring 2. Green = found at ring 3.

## The two tools

1. **Queue** — first in, first out. This keeps the rings in order.
2. **Visited set** — never check the same node twice. Without it, cycles loop forever.

## The template

```python
from collections import deque

def bfs(start, graph: dict) -> None:
    queue = deque([start])
    visited = {start}

    while queue:
        level_size = len(queue)          # snapshot = one full ring
        for _ in range(level_size):
            node = queue.popleft()
            # process node here
            for nxt in graph.get(node, []):
                if nxt not in visited:
                    visited.add(nxt)     # mark WHEN queued, not when popped
                    queue.append(nxt)
```

## Multi-source BFS (LC 994 — Rotting Oranges)

Start with *all* rotten oranges in the queue at once.
Each ring = one minute of rot spreading.

```python
from collections import deque

def oranges_rotting(grid: list[list[int]]) -> int:
    rows, cols = len(grid), len(grid[0])
    queue = deque()
    fresh = 0

    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == 2:
                queue.append((r, c))
            elif grid[r][c] == 1:
                fresh += 1

    minutes = 0
    dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)]

    while queue and fresh > 0:
        for _ in range(len(queue)):
            r, c = queue.popleft()
            for dr, dc in dirs:
                nr, nc = r + dr, c + dc
                if 0 <= nr < rows and 0 <= nc < cols and grid[nr][nc] == 1:
                    grid[nr][nc] = 2
                    fresh -= 1
                    queue.append((nr, nc))
        minutes += 1

    return minutes if fresh == 0 else -1
```

## Complexity

| | Time | Space |
|---|---|---|
| BFS | O(V + E) | O(V) |

V = nodes, E = edges. On a grid: O(rows × cols).

## Close-out ritual

Before you submit, say out loud:
1. Time and space complexity.
2. One edge case trace: empty graph, start = target, or unreachable target.

## Plan problems

**Tree BFS A-set:** LC 102 · 199 · 1448
**Graph BFS A-set:** LC 994 · 127
**B-set:** LC 103 · 637 · 909 · 433 · 286