Breadth-First Search Questions at Rippling: What to Expect

Breadth-First Search (BFS) matters at Rippling because their platform models complex, interconnected data—employee hierarchies, permission flows, and device management networks. Traversing these relationships efficiently to find shortest paths or nearest connections is a daily engineering task. With 3 out of 22 total coding questions focused on BFS, Rippling clearly prioritizes assessing your ability to reason about graphs, levels, and state transitions systematically. Mastering BFS demonstrates you can handle the layered data dependencies central to their product.

What to Expect — Types of Problems

Rippling’s BFS problems typically fall into two categories. First, explicit graph traversal on grid or node structures. You might be asked to find the shortest path in a maze, the distance to the nearest gate in an office floor plan, or the number of steps to change a system state. Second, level-order or serialization problems on trees, which are essentially BFS on a directed acyclic graph. This could involve printing a tree level-by-level, serializing a company org chart for their UI, or finding the minimum depth of a hierarchy.

The key is recognizing when "shortest path" or "level-by-level" is required. If the problem mentions "minimum steps," "nearest," "level order," or involves exploring all possibilities one step at a time, BFS is your primary tool.

How to Prepare — Study Tips with One Code Example

Focus on the core BFS pattern: a queue and a visited set. Practice writing it from memory. Always clarify if the graph is implicit (like a grid) or explicit (adjacency list). For tree problems, you often omit the visited set since trees are acyclic.

A critical pattern is BFS for shortest path in an unweighted grid. This appears frequently in variations. The template below finds the shortest distance from a starting point to a target in a 2D grid, avoiding obstacles.

from collections import deque

def shortest_path(grid, start, target):
    rows, cols = len(grid), len(grid[0])
    directions = [(1,0), (-1,0), (0,1), (0,-1)]
    queue = deque([(start[0], start[1], 0)])  # (row, col, distance)
    visited = set([(start[0], start[1])])

    while queue:
        r, c, dist = queue.popleft()
        if (r, c) == target:
            return dist
        for dr, dc in directions:
            nr, nc = r + dr, c + dc
            if (0 <= nr < rows and 0 <= nc < cols and
                grid[nr][nc] != '#' and  # '#' represents an obstacle
                (nr, nc) not in visited):
                visited.add((nr, nc))
                queue.append((nr, nc, dist + 1))
    return -1  # Target not reachable

Recommended Practice Order

1. Fundamentals: Binary Tree Level Order Traversal, Number of Islands.
2. Shortest Path: Rotting Oranges, Shortest Path in Binary Matrix.
3. State Search: Word Ladder, Open the Lock.
4. Rippling-Style: Practice problems tagged with "Rippling" and "BFS" on platforms like CodeJeet to acclimate to their problem framing.

Practice Breadth-First Search at Rippling