Understand
30-second explanation
Build the interview-ready mental model before comparing implementations.
Model the 100 squares as graph nodes. From any square you roll 1–6, so each square has up to six outgoing edges; if a destination square is the foot of a ladder or the mouth of a snake, you are immediately moved to its other end. Every roll costs one move, so BFS from square 1 gives the fewest rolls to reach square 100. The board is fixed at 100 squares, so it runs in O(1).
Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview contextHide interview context
What's being tested
Whether you can model a rules-based board as an implicit graph, see that equal-cost moves call for BFS, and correctly apply jumps to the landing square rather than double-counting them as extra moves.
Where candidates slip
Treating a ladder/snake as a separate move (it is free — part of landing), or applying the jump to the current square instead of the destination, or marking the mouth visited instead of the end.
How to stand out
Build a jump[] lookup so the snake/ladder remap is O(1), state that a roll is always one move so BFS (not Dijkstra) is right, and handle the overshoot and -1 cases explicitly.
Choose
Compare the available approaches
Use constraints and trade-offs to select the route you can defend in an interview.
Interactive decision treeInteractiveChoose the right approachFirst decision previewFewest moves to the goalAll moves same cost?YesNoExplore the full treeCollapse the tree
How to read it:From 6, rolling a 6 lands on 12; the ladder lifts you to 98 — all in a single move.
Learn & trace
Build each solution step by step
Move from the core idea to a visual trace, implementation, proof, and edge cases.
Approach 1 of 2
Recursive Move Search (Baseline)
1Hint 1Explore hintHide hint
The brute-force instinct: from a square, try all six rolls, recurse, and take 1 + the minimum of the results.
2Hint 2Explore hintHide hint
Now add a snake that sends square 20 back to square 5. What happens when your recursion goes 5 → 20 → 5 → 20 …?
3Hint 3Explore hintHide hint
Cycles mean plain recursion never terminates. You would need a visited guard — and once you add that, a queue (BFS) is the cleaner shape.
A · Core idea
The natural first idea is to treat 'minimum rolls from square s' as a recursive quantity: try each of the six rolls, recurse on where you land (after any jump), and return 1 + min over the six.
Why memoized recursion does not work here. Memoization assumes the dependency graph of subproblems is a DAG — that f(s) depends only on strictly 'smaller' states. Ladders and snakes break that. A snake from 20 → 5 means reaching 20 lets you go back to 5, so f(5) depends on f(20) which depends on f(5). The recursion has a cycle, and there is no base ordering to memoize against.
Why even a visited-guarded DFS is wrong for this question. You can stop infinite loops by refusing to revisit a square within the current path, but DFS explores depth-first, so the first time it reaches square 100 is not necessarily by the fewest rolls. You would have to explore all paths and take the minimum — exponential again.
The lesson it teaches. The cycles plus the 'fewest moves' objective are exactly the signature of a shortest-path problem on a graph with cycles. The right tool processes squares in increasing distance and marks each visited once — that is BFS. Use this baseline to explain why you abandon recursion, not as something you would submit.
B · Visual dry run
Interactive dry run
Select a frame to connect the decision with its visual state.
ladders = [], snakes = [[20,5]], asking f(5)How to read it:Square 20 snakes back to 5, so f(5) depends on f(20) depends on f(5) — recursion cannot terminate.
non-terminating — motivates BFSC · Code with teaching comments
Reference implementation preview// Illustrative only — loops on boards with snakes. int minRolls(int square, int[] jump, boolean[] inPath) { if (square == 100) return 0; inPath[square] = true; int best = Integer.MAX_VALUE;Explore the full codeHide the full code
D · Why it works and complexityExponential without memo / unsafe with cycles timeO(board) call stack spaceOpen the invariant and proofExplore the proofHide the proof
Snakes turn the state dependency into a cycle, which is precisely what kills naive recursion/DP. The moment you reach for a visited set to break the cycle, you have rediscovered BFS — so go there directly.
ComplexityExponential without memo / unsafe with cycles time, O(board) call stack space — Without memoization the recursion branches by 6 at each square — exponential. Memoization fails outright because snakes create cycles, so 'minimum rolls from square s' depends on itself. Bolting on a visited set turns it into a clumsy DFS that does not give shortest paths reliably.
E · Edge cases and pitfalls3Trying to memoize across a cycle — there is no valid subproblem ordering.Explore edge casesHide edge cases
Pitfalls
- Trying to memoize across a cycle — there is no valid subproblem ordering.
- Assuming DFS finds the shortest path — it does not visit in distance order.
- Applying the jump to the current square instead of the landing square.
Approach 2 of 2 · Optimal
BFS over the Board Graph (Optimal)
1Hint 1Explore hintHide hint
Each square is a node; a single roll is an edge to one of the next six squares (after any jump). Every edge costs one move.
2Hint 2Explore hintHide hint
Precompute jump[s] = the snake/ladder end if s starts one, else s itself. Then a move from u to u+roll lands on jump[u+roll].
3Hint 3Explore hintHide hint
BFS from square 1. The first time square 100 is dequeued (or marked), its distance is the minimum number of rolls. Use a visited array to handle the cycles snakes create.
A · Core idea
BFS is the exact fit: equal-cost moves, an implicit graph, and cycles handled by a visited set.
Modeling the board as a graph. Each square 1…100 is a node. From square u, a die roll of r ∈ {1,…,6} moves you to u + r. If u + r is the foot of a ladder or the mouth of a snake, you are carried to its other end. Precomputing jump[s] — the endpoint if s starts a snake/ladder, otherwise s — makes that redirect an O(1) lookup. So the real neighbor of u for roll r is jump[u + r].
Why a roll is one edge, not the distance you travel. Whether you advance one square or six, or then climb a long ladder, it all happens in a single turn. So every edge in this graph has weight one. With uniform weights, BFS visits squares in increasing number of rolls, and the first time it reaches square 100 is by the fewest rolls — no priority queue needed.
Why the visited set is essential. Snakes send you backward, creating cycles. Without a visited marker, BFS could enqueue the same square endlessly. Marking each square the first time it is reached (its shortest-roll distance) both guarantees termination and keeps the work O(board).
Why the loop is iterative. BFS runs off a FIFO queue with no recursion, so there is no call-stack depth to worry about — the cycle-laden board cannot overflow the stack the way the recursive baseline threatened to.
Handling overshoot and unreachable. A roll with u + r > 100 cannot be played, so skip it. If the queue empties without ever marking square 100, the goal is unreachable on this board and the answer is -1.
B · Visual dry run
Interactive dry run
Select a frame to connect the decision with its visual state.
ladders = [[4,100]], snakes = []Interactive diagram
BFS layers from square 1
dequeue 1 (d 0)With the ladder 4→100, square 100 is reached in BFS layer 1.
1C · Code with teaching comments
D · Why it works and complexityO(1) timeO(1) spaceOpen the invariant and proofExplore the proofHide the proof
The whole problem collapses once you see squares as nodes and rolls as equal-weight edges, with jump[] folding each snake/ladder into an O(1) 'apply on arrival' redirect. BFS then reads off the answer as the level of square 100.
ComplexityO(1) time, O(1) space — The board is fixed at 100 squares, each with at most 6 outgoing edges, so BFS does at most 100 × 6 = 600 constant-time steps. With a fixed-size board that is O(1) time and space. (For a general N×N board it would be O(N²).)
E · Edge cases and pitfalls7ladders = [], snakes = [] — No jumps; BFS finds 100 at ⌈99/6⌉ = 17 rolls.Explore edge casesHide edge cases
Edge cases handled
ladders = [], snakes = []No jumps; BFS finds 100 at ⌈99/6⌉ = 17 rolls.a snake at 100's only approach forming a trapIf every path into 100 lands on a snake mouth, dist[100] stays -1 → return -1.ladder foot reachable on the first rolljump applied immediately; square 100 can be reached in a single move (answer 1).
Pitfalls
- Counting a ladder or snake as an extra move — it is free, part of landing.
- Applying the jump to the source square instead of the landing square.
- Marking the snake/ladder mouth visited instead of its endpoint, so the real target is re-enqueued.
- Forgetting to skip rolls that overshoot 100.
Practice
Try the solution yourself
Edit the optimal implementation, test another input, and observe the result.
Editable code previewpublic int quickestWayUp(int[][] ladders, int[][] snakes) { int[] jump = new int[101]; for (int i = 1; i <= 100; i++) jump[i] = i; for (int[] l : ladders) jump[l[0]] = l[1]; for (int[] s : snakes) jump[s[0]] = s[1];Explore the playgroundClose the playground
Avoid mistakes
Recognize the interview traps
Understand why common attempts fail before reviewing the corrected implementation.
1Counting the jump as an extra roll
A ladder or snake fires automatically when you land — it does not consume a turn. Adding 1 for the jump inflates the answer.
Explore the code comparisonHide the code comparison
Counting the jump as an extra roll
A ladder or snake fires automatically when you land — it does not consume a turn. Adding 1 for the jump inflates the answer.
Don't write this
nxt = u + roll
d = dist[u] + 1
if jump[nxt] != nxt:
d += 1 # wrong: the jump is freeWrite this instead
nxt = jump[u + roll] # land + jump, all one move
d = dist[u] + 12Jumping from the wrong square
Snakes and ladders act on the square you LAND on, not the square you roll from. Remapping u before adding the roll skips the actual rule.
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Jumping from the wrong square
Snakes and ladders act on the square you LAND on, not the square you roll from. Remapping u before adding the roll skips the actual rule.
Don't write this
int from = jump[u]; // wrong: remap source
int next = from + roll;Write this instead
int next = jump[u + roll]; // remap the landing square3Missing the visited guard
A snake sends you backward, so without marking squares visited the same square is enqueued repeatedly and BFS never terminates.
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Missing the visited guard
A snake sends you backward, so without marking squares visited the same square is enqueued repeatedly and BFS never terminates.
Don't write this
for roll in range(1, 7):
nxt = jump[u + roll]
queue.append(nxt) # re-enqueues cycled squares foreverWrite this instead
for roll in range(1, 7):
nxt = jump[u + roll]
if dist[nxt] == -1:
dist[nxt] = dist[u] + 1
queue.append(nxt)Revise
Carry the pattern into the interview
Pattern to remember
BFS over an implicit board graph; apply jumps when landing
Code pattern to memorize
jump[square] = ladder/snake end, else square itself
BFS from 1, dist[1] = 0
for each square u dequeued:
for roll in 1..6:
next = jump[u + roll] # take the jump if there is one
if next <= 100 and unvisited:
dist[next] = dist[u] + 1; enqueue
return dist[100] or -1Rules to remember
- 1Each roll is one move regardless of whether you advance 1 or 6 — so all edges weigh the same and BFS is exact.
- 2Apply the snake/ladder jump to the landing square, not to the square you rolled from.
- 3A landing square that is a snake/ladder mouth has no separate visit — you go straight to its end; mark the end visited.
- 4Ignore rolls that overshoot 100 (
u + roll > 100); you can only finish by landing exactly or being carried by a ladder. - 5Return
-1if square 100 is never reached — snakes can trap some boards.
Use this pattern when
- Shortest number of moves on a board / grid where each move has the same cost
- The graph is **implicit** — edges come from rules (dice rolls), not an edge list
- Teleports / portals that relocate you on arrival (ladders, snakes, warps)
- Fixed, small state space where BFS explores everything once
Choose another pattern when
- Moves have differing costs → Dijkstra, not BFS
- You need the count of distinct paths, not the shortest → DP over the DAG
- The board is enormous and structured → look for a closed-form or math shortcut
- You must maximize score along the way → this is an optimization DP, not pure shortest path
Interview takeaway
Snakes and Ladders is shortest-path in disguise: squares are nodes, dice rolls are equal-weight edges, and ladders/snakes are 'apply on arrival' redirects. Recognizing the implicit graph turns a board game into a textbook BFS.
Continue
Transfer the pattern to another problem
Use a direct variation or pattern sibling to prove the idea is reusable.
Transferable patternBFS over an Implicit Board Graph. Recognize it when moves cost the same and the graph is described by rules (dice rolls, neighbor steps) with 'apply on arrival' redirects like ladders or portals. A jump[] table makes the redirect O(1); BFS reads off the fewest moves. The same shape powers Word Ladder, Open the Lock, and grid shortest paths.
Practice queueDirect variations and pattern siblings
Direct follow-ups
- Snakes and Ladders (N×N board)Generalize to an N×N boustrophedon board — same BFS, with index-to-coordinate mapping.
- BFS: Shortest Reach in a GraphThe explicit-graph cousin: uniform-weight single-source BFS without the board remap.
- Word LadderAnother implicit graph — words are nodes, single-letter changes are equal-weight edges.
- Open the LockComing soonBFS over lock states with forbidden 'deadends' acting like blocked squares.