QA interview puzzles: the 20 that actually get asked

The puzzle, precisely

Three switches outside a closed room control three bulbs inside. Flip whatever you like, but you may open the door only once. Map every switch to its bulb.

How to think it through — out loud

The trap is treating on/off as the only signal — two states can't distinguish three bulbs in one visit. So find a second observable. A bulb that has been on recently is warm. Turn switch 1 on for ten minutes, then off. Turn switch 2 on. Enter: the lit bulb is switch 2, the dark-but-warm bulb is switch 1, the dark-and-cold bulb is switch 3.

The answer

Heat is the second signal: lit = switch 2, warm-off = switch 1, cold-off = switch 3.

The trap

Answering 'impossible with one visit' — the interviewer is checking whether you look for extra observables before declaring a dead end.

The puzzle, precisely

Three sealed jars: apples only, oranges only, and a mix. All three labels are wrong. You may draw a single fruit from a single jar, then relabel everything correctly.

How to think it through — out loud

Pick from the jar labelled Mixed — since its label is wrong, it's pure. Draw an apple? That jar is Apples. Now the jar labelled Oranges can't be oranges (label wrong) and can't be apples (taken) — it's Mixed. The last jar is Oranges. One draw, three labels, because you chose the jar whose wrong label guaranteed a pure result.

The answer

Draw from 'Mixed' — it's pure. That fruit names the jar; the other two rotate by elimination.

The trap

Drawing from Apples or Oranges first — a mixed jar could give you either fruit, and one draw tells you nothing certain.

The puzzle, precisely

You don't know which guard is which. You may ask exactly one question to exactly one of them, then must choose a door.

How to think it through — out loud

Any direct question ('is this the good door?') gives an answer you can't calibrate — you don't know who you asked. So build a question that passes through both minds: 'If I asked the other guard which door is safe, what would they say?' Truth-teller honestly reports the liar's lie. Liar lies about the truth-teller's truth. Both name the wrong door — so take the opposite.

The answer

Ask either guard what the OTHER would say, then take the opposite door. Both answers point wrong by construction.

The trap

Spending the question identifying who lies — you get the identity and no door, and the question is gone.

The puzzle, precisely

All doors start closed. On pass k you toggle every k-th door. After 100 passes, which doors are open?

How to think it through — out loud

Door n gets toggled once per divisor of n — door 12 by passes 1, 2, 3, 4, 6, 12. Divisors pair up (2 with 6, 3 with 4), so most doors get an even number of toggles and end closed. The exception: perfect squares, where one divisor pairs with itself (6×6 for 36) — odd toggles, door ends open.

The answer

Doors 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 — the perfect squares. Ten doors.

The trap

Starting to simulate pass by pass on the whiteboard — the interviewer wants the divisor insight, and simulation runs out of board.

The puzzle, precisely

An egg that survives a drop is reusable. Once both eggs break, you must already know the answer. Minimise the worst-case number of drops.

How to think it through — out loud

One egg alone forces floor-by-floor. Two eggs: use the first to bracket in jumps, the second to walk the bracket. Equal jumps of 10 give worst case 19. The refinement: after each first-egg drop you've spent one attempt, so make each next jump one floor smaller — then every path costs the same. Start at floor n where n + (n−1) + … + 1 ≥ 100 → n = 14. Drop at 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100.

The answer

14 drops worst case — first egg at floor 14, then jumps shrinking by one each time.

The trap

Stopping at the √100 = 10-jump answer (worst case 19) — the shrinking-step idea is the senior half of the puzzle.

The puzzle, precisely

Seven balls weigh the same; one is heavier. Using a two-pan balance exactly twice, identify the heavy ball.

How to think it through — out loud

The instinct is halving: 4v4, then 2v2, then 1v1 — three weighings. But a balance gives three outcomes: left, right, balanced. Use all three. Weigh 3 v 3, leaving 2 aside. Balanced → heavy is in the pair; weigh them. Tilted → take the heavy trio, weigh 1 v 1: tilt names it, balance means the third.

The answer

Two weighings: 3v3 first, then 1v1 inside the guilty group. Ternary search, not binary.

The trap

4v4 first — it wastes the balanced outcome (impossible with 8 split evenly) and forces a third weighing.

The puzzle, precisely

Burn rate varies wildly along each rope; only the total (60 min) is guaranteed. Using fire alone, mark a 45-minute interval.

How to think it through — out loud

You can't use length — the burn is uneven. But lighting a rope at both ends burns it in exactly 30 minutes regardless of unevenness: the flames must meet when the remaining time is spent from both sides. So: light rope A at both ends AND rope B at one end. A finishes at minute 30; at that instant B has 30 minutes left — light B's other end, halving that to 15. B dies at minute 45.

The answer

A both ends + B one end; when A dies (30 min), light B's other end — B dies at 45.

The trap

Any answer using 'half the length' — the uneven burn is stated precisely to kill that path.

The puzzle, precisely

No markings on either jug. Fill, empty and pour between them to end with exactly 4 litres in the 5-litre jug.

How to think it through — out loud

Track states as (small, large). Fill the 5 → (0,5). Pour into the 3 → (3,2). Empty the 3 → (0,2). Pour across → (2,0). Fill the 5 → (2,5). Top up the 3 — it takes exactly 1 → (3,4). The move that matters is creating the 1-litre remainder by topping up a partly-full jug.

The answer

Fill 5, pour to 3, empty 3, transfer the 2, refill 5, top up 3 — the 5-jug holds 4.

The trap

Narrating pours without tracking state — say the (x, y) numbers aloud; losing count mid-answer is the common fail.

The puzzle, precisely

Only 5 horses race at a time; you learn order, not times. Find the top 3 overall in the fewest races.

How to think it through — out loud

Five group races seed everything (5). Race the five group-winners (6): the winner is #1 overall. Who can still be #2 or #3? Only: the 2nd and 3rd from race six, the runner-up and third from the champion's group, and the runner-up from the second-place winner's group — five horses. Everyone else has three provably-faster horses above them. Race those five (7): its top two are #2 and #3.

The answer

7 races: five heats, the winners' final, then a five-horse decider for places 2–3.

The trap

Answering 6 by forgetting that the #2 group's runner-up can outrank the #3 group's winner — walk the elimination logic aloud.

The puzzle, precisely

The torch must accompany every crossing, so someone always walks it back. Pairs move at the slower member's speed. Beat 19 minutes — hit 17.

How to think it through — out loud

Greedy uses the 1-minute person as a shuttle for everyone: 10+1+7+1+2 = 21… optimised, 19. The insight: 7 and 10 should cross together, so the 7 hides inside the 10. But then someone fast must already be on the far side to bring the torch back. So: 1+2 cross (2), 1 returns (3), 7+10 cross (13), 2 returns (15), 1+2 cross (17).

The answer

1&2 over, 1 back, 7&10 over, 2 back, 1&2 over — 2+1+10+2+2 = 17 minutes.

The trap

Escorting everyone with the fastest walker — pairing the slow pair is the whole puzzle, and 19 is the wrong-but-confident answer.

The puzzle, precisely

A bar of 7 equal units, cuts at any positions, two cuts total. The worker must hold exactly n units at the end of day n.

How to think it through — out loud

Two cuts give three pieces — make them 1, 2 and 4. Day 1: give 1. Day 2: give 2, take back 1. Day 3: give 1 (holds 1+2). Day 4: give 4, take back 1 and 2. Days 5–7 repeat the pattern. Pieces 1-2-4 are binary weights: every value 1–7 is a sum of them, so every day's total is payable if change is allowed.

The answer

Cut into 1, 2 and 4 units; pay with exchanges. Binary representation does the rest.

The trap

Assuming pieces once given are gone — the unlock is asking 'can I take change back?', a requirements question. Ask it aloud.

The puzzle, precisely

Every kilometre walked — loaded, empty, forwards or back — costs one banana eaten. Bananas can be cached anywhere. Maximise what reaches the far side.

How to think it through — out loud

With 3,000 bananas the camel shuttles: moving the pile 1 km takes 5 legs (3 forward, 2 back) = 5 bananas/km. That rate holds until the pile drops to 2,000 — after 200 km. Then 3 legs = 3 bananas/km until the pile hits 1,000 — another 333⅓ km. From there it's one straight walk: 1,000 − 200 − 333⅓ = 466⅔ km remaining, eating 466⅔ bananas.

The answer

1,000 − 466⅔ = 533 bananas delivered (533⅓, rounded down to whole bananas).

The trap

Walking straight with the first 1,000 — delivering zero. The phased shuttle IS the answer; narrate the 5-per-km → 3-per-km → 1-per-km stages.

The puzzle, precisely

The host always opens a no-prize door you didn't pick, then offers a switch. Does switching help?

How to think it through — out loud

Your original door was right with probability 1/3 — nothing the host does changes that. The remaining 2/3 sat on the other two doors; the host, constrained to open an empty one, compresses that entire 2/3 onto the one unopened door. Sanity-check with 100 doors: you pick one, the host opens 98 empties — staying feels obviously wrong there. Same logic, three doors.

The answer

Switch. Staying wins 1/3; switching wins 2/3.

The trap

'50-50 now' — the doors aren't symmetric, because the host's choice was constrained by where the prize is. The 100-door version is your out-loud defence.

The puzzle, precisely

Each displaced passenger picks uniformly among free seats. What's the chance the last passenger finds seat 100 free?

How to think it through — out loud

Follow the chaos and it resolves to one observation: the process ends the moment anyone picks seat 1 (everyone after sits correctly — passenger 100 wins) or seat 100 (passenger 100 loses). Every random chooser faces those two seats symmetrically — no one prefers one over the other. All other choices just postpone the same coin flip.

The answer

1/2. The whole cascade is one fair choice between seat 1 and seat 100 in disguise.

The trap

Trying to sum over 100 cases — the symmetry argument is three sentences; the summation is a whiteboard funeral.

The puzzle, precisely

Starting at 12:00, count the moments the two hands point the same way over the next 12 hours.

How to think it through — out loud

The minute hand laps the hour hand; each lap is one overlap. In 12 hours the minute hand makes 12 revolutions, the hour hand 1 — so 11 laps, 11 overlaps. That's why there's no overlap 'around 11:55': the 11-o'clock-hour one IS the 12:00 one. They're evenly spaced: every 12/11 hours ≈ 65 minutes 27 seconds.

The answer

11 times in 12 hours (22 per day), once every 65 5/11 minutes.

The trap

Answering 12 by assuming one per hour — say the relative-speed argument and name the missing 11-hour overlap.

The puzzle, precisely

Any tester can sip from any number of bottles now; in 24 hours each is either sick or fine. Identify the poisoned bottle among 1,000 with one round.

How to think it through — out loud

Ten testers produce a 10-bit result — 1,024 distinct outcomes, enough for 1,000 bottles. Number the bottles 0–999 in binary. Tester k sips every bottle whose k-th bit is 1. After 24 hours, write sick=1, fine=0 in position order: the resulting binary number IS the poisoned bottle's label.

The answer

Binary-label the bottles; each tester covers one bit position. The sickness pattern spells the bottle number. 10 testers handle up to 1,024 bottles.

The trap

Splitting 1,000 by 10 into groups of 100 — that finds the group, not the bottle. Parallel bits, not sequential groups.

The puzzle, precisely

You can't see or feel which side is up, but you may move and flip coins freely. Produce two piles with equally many heads.

How to think it through — out loud

You can't find the heads — so engineer equality instead. Take any 10 coins as pile B and flip every one of them. If k of the original 10 heads landed in pile B, then pile A has 10−k heads — and pile B, after flipping, turned its 10−k tails into heads: also 10−k. Equal, for every k, with zero information needed.

The answer

Any 10 coins into a pile, flip that whole pile. Both piles now hold 10−k heads each.

The trap

Reasoning toward 'impossible without seeing' — the flip is a move that creates the symmetry you couldn't observe.

The puzzle, precisely

Guessing your own hat colour right = freed. They may agree a strategy beforehand. Maximise guaranteed survivors.

How to think it through — out loud

The rearmost prisoner sees 99 hats and encodes one fact everyone can use: he says 'red' if he sees an odd number of red hats, 'black' if even — a parity bit; his own fate is a coin flip. Prisoner 99 counts reds ahead: if the parity he sees differs from the announced one, his own hat is red. Each subsequent prisoner tracks every 'red' spoken behind them and the count they see, and deduces their own hat exactly.

The answer

Parity strategy: 99 guaranteed correct, the first speaker 50/50 — 99.5 expected.

The trap

Strategies where each helps only their neighbour (~50 saved) — one shared parity bit serves all 99, and that leap is what's tested.

The puzzle, precisely

Priorities: survive, then maximise gold, then prefer others overboard. Pirates A (senior) to E vote on A's proposal first. What split survives?

How to think it through — out loud

Work backwards. Two left (D,E): D proposes 100-0 and wins on his tie-break — E gets nothing there. Three (C,D,E): C needs one vote — buys E with 1 coin (E's alternative is 0): 99-0-1. Four (B..E): B buys D (who'd get 0 under C) with 1: 99-0-1-0. Five: A needs two votes — buys C and E, who each get 0 in B's world, with 1 coin apiece.

The answer

A proposes 98–0–1–0–1 (A, B, C, D, E) — and it passes with A, C and E voting yes.

The trap

Splitting 'fairly' at 20 each — rationality plus the voting rule makes near-total greed stable; the backward walk is the answer they're grading.

The puzzle, precisely

The boat holds the farmer plus one passenger. Neither predator-prey pair may be left alone on either bank. Get all three across.

How to think it through — out loud

The goat is the shared threat — it can't be left with either other item. So the goat moves first, and the unlock most people miss: the goat comes back. Take goat over. Return empty. Take wolf over, bring goat back. Take cabbage over (wolf+cabbage coexist happily). Return empty. Take goat over again.

The answer

Goat over → back empty → wolf over, goat back → cabbage over → back empty → goat over. Seven crossings.

The trap

Refusing to un-ship progress — the backward move is the point, and it's a lesson interviewers like: sometimes state must temporarily regress.