Oct 08, 2026 by Offer

TikTok Interview Questions 2027: Frog Jump (LeetCode 403) Explained

For this TikTok interview question — Frog Jump (LeetCode 403) — use memoized DFS: from each stone try jumps of k−1, k, k+1 from the last jump k, memoizing failed (stone, jump) states, with a hash map from position to index for O(1) landing checks. Define the state before coding — it's the whole problem.

TikTok Interview Questions: What Frog Jump Tests

Frog Jump is commonly reported by candidates in TikTok's harder coding rounds because it tests whether you can define a search state that captures the problem's structure. The naive approaches — plain recursion (exponential blowup) or greedy (wrong) — both fail instructively. Interviewers are testing dynamic thinking: recognizing that the relevant state is (position, last jump length), and that memoizing failed states turns an exponential search into a tractable one.

TikTok Interview Questions: How to Solve Step by Step

Define the state. The frog's future depends on two things: which stone it is on and how far it last jumped. So the state is (stone index i, last jump k). Say this explicitly — it is the insight the interviewer is waiting for.
Build the position map. Map each stone's coordinate to its index in a hash map, so checking "does a stone exist at position x?" is O(1).
Write the DFS. From state (i, k), try jumps k-1, k, k+1 (skipping non-positive jumps): compute the landing position, look it up in the map, and recurse on the new state.
Memoize failures. Keep a set of (i, k) states already proven dead — return False immediately on revisit. This is what makes the solution pass.
Base cases. Reaching the last stone returns True; the first jump must be 1 unit (if stone[1] != stone[0] + 1, return False early).
State the complexity. Roughly O(n^2) states in the worst case with the memo, O(n^2) space — and say why the memo is what saves it.

Example line: "The trick is that position alone isn't enough state — the frog's options depend on its last jump, so I memoize on the pair (stone, jump length), which prunes the exponential search."

Common Mistakes

Forgetting the k-1 jump can be zero or negative. Always guard jump <= 0.
Memoizing only the position. Without the jump length in the key, the memo is wrong — different jumps from the same stone have different futures.
No early exit. Checking the first jump up front is a cheap, impressive optimization.

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FAQ

Can this be solved with DP instead of DFS? Yes — a DP table of reachable (stone, jump) pairs is equivalent; memoized DFS is usually cleaner to write.

Why not BFS? BFS works too, but the state space and memo logic are identical — pick whichever you write more reliably.

What is the hardest part under time pressure? Getting the memo key right; practice writing (index, k) pairs until it is automatic.

Should I code the position map or use binary search? The hash map is O(1) and simpler — mention binary search only as an alternative.

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