Trie-guided backtracking
Find a Dictionary on the Tile Board
01. The question
Given {"board":[[letters]],"words":[...]}, return all listed words traceable through side-sharing tiles without reusing a tile in one route. Return unique found words alphabetically.
| r / c | 0 | 1 | 2 |
|---|---|---|---|
| 0 | c | a | t |
| 1 | r | r | e |
Follow only prefixes from the shared dictionary trie. Backtracking releases each route’s tiles.
Before revealing the idea, predict the result and describe what information you would need to track.
02. Worked example
Input{"board":[["c","a","t"],["r","r","e"]],"words":["cat","care","car","tree"]}
Output["car","care","cat"]
- Build a shared prefix trie.
- Follow board letters through trie children.
- Record terminal words and release tiles on return.
Try your code
Define solve(data, parameter="") and return your result. Your code runs locally, not on a server.
Edit your test cases
Use 1-10 objects with data and an optional string parameter. Expected outputs are computed from the lesson. An optional expected field must be correct.
Input contract & reference code
Example data: {"board":[["c","a","t"],["r","r","e"]],"words":["cat","care","car","tree"]}. Parameter: "". Numeric parameters arrive as strings; convert them with int(parameter) when needed.
The example and generated cases use the question’s exact variant. This checks behavior on a small test batch—not performance on every possible input.
Python downloads only when you run code, using compressed runtime assets. Code is not saved or uploaded. Enable JavaScript to run it.
Printed output
Five-second execution limit per batch. Standard-library Python only; additional packages and network access are unavailable. Stop ends execution without blocking the lesson.
03. The idea
Try sketching your own approach first. What would you track, and why would it work?
Reveal the idea
- Build a shared prefix trie.
- Follow board letters through trie children.
- Record terminal words and release tiles on return.
04. Watch it happen
Use Step to move slowly, or Play to follow the whole trace. Change a preset or load your own input.
Input & settings
Red = current item / blue = active window or cells / green = completed work. A value docks at its actual destination; the original input stays unchanged. Positions start at 0.
Working state / what changes
Ready to begin
Enable JavaScript to step through the animation. The worked example and solution below remain readable without it.
Python / why it changes
The line behind the step
def solve(data, parameter=""):
board, words = data["board"], data["words"]
root = {}
for word in words:
node = root
for letter in word:
node = node.setdefault(letter, {})
node["$"] = word
used, found = set(), set()
rows, cols = len(board), len(board[0])
def visit(r, c, parent):
if not (0 <= r < rows and 0 <= c < cols) or (r, c) in used:
return
letter = board[r][c]
if letter not in parent:
return
node = parent[letter]
if "$" in node:
found.add(node["$"])
used.add((r, c))
for dr, dc in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
visit(r + dr, c + dc, node)
used.remove((r, c))
for r in range(rows):
for c in range(cols):
visit(r, c, root)
return sorted(found)How to use this walkthrough
Hover or focus a control for its label. Play forms the input and follows values into the working state. Step shows one change; Back and the slider restore exact snapshots. Reduced motion uses instant updates. Saved snapshots and sorted heap displays are teaching overhead, not algorithm complexity.
05. Python solution
def solve(data, parameter=""):
board, words = data["board"], data["words"]
root = {}
for word in words:
node = root
for letter in word:
node = node.setdefault(letter, {})
node["$"] = word
used, found = set(), set()
rows, cols = len(board), len(board[0])
def visit(r, c, parent):
if not (0 <= r < rows and 0 <= c < cols) or (r, c) in used:
return
letter = board[r][c]
if letter not in parent:
return
node = parent[letter]
if "$" in node:
found.add(node["$"])
used.add((r, c))
for dr, dc in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
visit(r + dr, c + dc, node)
used.remove((r, c))
for r in range(rows):
for c in range(cols):
visit(r, c, root)
return sorted(found)06. Cost & edge cases
- Time
- O(dictionary letters + RC × 4^L) conservative bound
- Space
- O(dictionary letters + L)
- Validate the exact input format and bounds before tracing.
- Equal labels do not collapse vertex identity; missing nodes and invalid pointers are explicit.