MEDIUMasked at 1 company

Bag of Tokens

A medium-tier problem at 59% community acceptance, tagged with Array, Two Pointers, Greedy. Reported in interviews at Flexport and 0 others.

Founder's read

You're in an OA and hit Bag of Tokens. It's a greedy problem wrapped in a two-pointer pattern, and the acceptance rate hovers around 59 percent, which means roughly 4 in 10 candidates miss the key insight on first attempt. Flexport asks this one. The trap is thinking you need to process tokens sequentially. You don't. The trick lives in sorting and a two-pointer sweep, and if the pattern doesn't click under pressure, StealthCoder surfaces the working solution invisible to the proctor so you move forward without burning time on a dead-end approach.

Companies asking
1
Difficulty
MEDIUM
Acceptance
59%

Companies that ask "Bag of Tokens"

If this hits your live OA

Bag of Tokens is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built by an engineer at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share.

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What this means

The problem tests whether you recognize that greedy + two pointers beats simulation. Sort the tokens first. Then use one pointer at the start (smallest tokens) and one at the end (largest tokens). Buy score with cheap tokens on your left, and when you run dry on power, sacrifice your score to burn the most expensive token on the right and refill. The trap: candidates try to simulate token spending without sorting, or they don't realize the point where you flip from buying to selling. Common fail is also not tracking the maximum score seen, only the current score. Array, Sorting, Two Pointers, and Greedy all surface here because you're not just iterating, you're making strategic trades under a constraint. If you blank on the pattern live, this is exactly the kind of problem where StealthCoder bridges the gap between panic and a clean implementation.

Pattern tags

The honest play

You know the problem. Make sure you actually pass it.

Bag of Tokens recycles across companies for a reason. It's medium-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built by an engineer at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share. Works on HackerRank, CodeSignal, CoderPad, and Karat.

Bag of Tokens interview FAQ

Why does sorting the tokens help?+

Sorting lets you access the cheapest tokens from the left and the most expensive from the right. Without sorting, you can't make optimal greedy choices. After sort, two pointers become a clean way to alternate between buying low and selling high.

What's the key thing I'm tracking?+

Three things: your current score, your current power, and the maximum score you've ever seen. You buy tokens greedily from the left when you have power. When power runs low, you sell the highest-value token from the right to regain power and bump your score down, but you remember the peak.

Why does this show up at Flexport?+

Flexport values supply-chain optimization and resource trading logic. This problem mirrors real decision-making under constraints: trade a resource now to unlock future gains. It's a small, focused version of that thinking, which is why they test it.

How do I know when to stop the two-pointer loop?+

You stop when left and right pointers collide or cross. If you have enough power to buy one more token and it's worth it, you check if left <= right before buying. Once they cross, you're done because you've exhausted meaningful trades.

What's the common mistake?+

Not realizing you can sell tokens to get power back. Candidates often treat power as a one-way expense and get stuck. Also, forgetting to track max score across the entire loop, not just the current score at the end.

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