DFS backtracking
Follow a Letter-Tile Route
01. The question
Find whether a requested lowercase word can be traced on a letter board using side-sharing cells. A cell may be used only once in a route. Return true or false; diagonal steps are not allowed.
| r / c | 0 | 1 | 2 |
|---|---|---|---|
| 0 | c | a | t |
| 1 | r | r | e |
A valid route is [0,0] → [0,1] → [1,1] → [1,2]. Each tile is used once.
Before revealing the idea, predict the result and describe what information you would need to track.
02. Worked example
Inputboard = [["c","a","t"],["r","r","e"]], word = "care"
Outputtrue
- Start with c at [0,0].
- Follow a, then r below it, then e to the right.
- When a branch fails, release its tile before trying another branch.
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: [["c","a","t"],["r","r","e"]]. Parameter: "care". 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
- Try every matching start cell.
- Mark the current route, then explore orthogonal neighbors for the next letter.
- Undo the mark on return so other routes can use the tile.
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 tile_route(board, word):
if not word:
return True
rows, cols = len(board), len(board[0])
used = set()
def search(r, c, index):
if not (0 <= r < rows and 0 <= c < cols):
return False
if (r, c) in used or board[r][c] != word[index]:
return False
used.add((r, c))
found = index == len(word) - 1
if not found:
for dr, dc in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
if search(r + dr, c + dc, index + 1):
found = True
break
used.remove((r, c))
return found
for r in range(rows):
for c in range(cols):
if search(r, c, 0):
return True
return FalseHow 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 tile_route(board, word):
if not word:
return True
rows, cols = len(board), len(board[0])
used = set()
def search(r, c, index):
if not (0 <= r < rows and 0 <= c < cols):
return False
if (r, c) in used or board[r][c] != word[index]:
return False
used.add((r, c))
found = index == len(word) - 1
if not found:
for dr, dc in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
if search(r + dr, c + dc, index + 1):
found = True
break
used.remove((r, c))
return found
for r in range(rows):
for c in range(cols):
if search(r, c, 0):
return True
return False06. Cost & edge cases
- Time
- O(RC × 4^L) conservative upper bound
- Space
- O(min(L, RC)) for the route and recursion
- An empty word returns true.
- Returning to the same tile is forbidden.
- Board is not modified; failed paths release their marks.