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Answer a real interview question, get graded on it, earn the rank.

Jane StreetHardBrain Teaser

The 20-sided die

You are playing a game at a casino with a fair 20-sided die for 100 rounds.

At each round the die is rolled. You have two choices: take the sum of money represented by the number facing up, or roll again, in which case you are not guaranteed to win more than the original.

How are you playing this game?

Work backwards from the last round.

With one round left you must accept whatever shows, so it is worth 10.5.

With two left, rolling on is worth 10.5, so accept only v > 10.5. That round is worth E[max(v, 10.5)] = 13.025.

In general V(k) = (1/20) * sum of max(v, V(k-1)).

The rule never changes: accept v when it beats the value of continuing. The threshold climbs 10.5, 13.025, 14.72 and converges to 20.

So over 100 rounds you hold out for 19 or 20 early, and drop your threshold as rounds run out. The answer is a stopping rule, not a number.

No test cases. The argument is the answer.

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Configurable Quant Mock Interviews.

Practice spoken quant interviews with an interactive agent that listens, interrupts, and digs into your reasoning. Configure your 15 to 60-minute live verbal or coding sessions across critical topics like brain teasers, probability, option pricing, stochastic calculus, and linear algebra.

Target your practice to mimic the interview styles of elite proprietary trading firms and hedge funds.

Configure your first interview

You

I'll calculate the odds of someone matching a specific date, meaning we'd need 253 people to hit a 50% chance.

Interviewer

Are you sure? What about the edge case where two other people in the group share a birthday with each other, rather than matching that specific date?

You

Ah, right! Since every single person can form a matching pair with anyone else, the actual answer is only 23 people.

Mic ready

Birthday Problem Class Size

EasyGeneric

Determine the smallest number of people required in a group so that the probability of at least two individuals sharing the same birthday exceeds 50%. Assume a non-leap year with 365 days.

Coding Interviews.

A real code judge, an interactive compiler, and an interviewer reading your code as you write it.

Start a coding interview

Interviewer

How do you plan on solving this?

You

One left to right pass. Skip the spaces, read an optional sign, then fold each digit into an accumulator and clamp as I go.

Interviewer

And the time complexity of doing that?

You

Linear in the length of the string, constant extra space. Every character is read once.

Interviewer

Good. Proceed with the implementation please.

Interviewer

63 of 63. Now, that clamp only holds because Python integers never overflow. In C++ the accumulator would overflow before your check ever ran. How would you rewrite the guard?

Signed Integer Parser

MediumStringSimulation

Read a signed 32-bit integer off the front of a string and return it.

Walk the string once, left to right, in four stages:

Spaces. Step past any run of leading spaces.
Sign. A single + or - may follow. Anything else means the number is positive.
Digits. Keep consuming characters while they are digits. Stop at the first character that is not one, or at the end of the string. If you never read a digit, the answer is 0.
Clamp. The result has to land inside the signed 32-bit range, -2^31 through 2^31 - 1. Anything under the floor returns the floor; anything over the ceiling returns the ceiling.

Whatever follows the run of digits is ignored, whatever it happens to be.

Examples

Input: " -042"
Output: -42

Input: "1337 is a number"
Output: 1337

Input: "words then 7"
Output: 0

Input: "99999999999"
Output: 2147483647

Constraints

0 <= len(s) <= 200
s holds English letters, digits, spaces, +, - and .

class Solution:
    def myAtoi(self, s: str) -> int:
        MAX_INT = 2**31 - 1
        MIN_INT = -(2**31)
        n = len(s)
        i = 0
        while i < n and s[i] == " ":
            i += 1
        sign = 1
        if i < n and s[i] in "+-":
            if s[i] == "-":
                sign = -1
            i += 1
        value = 0
        while i < n and s[i].isdigit():
            digit = ord(s[i]) - ord("0")
            value = value * 10 + digit
            if sign * value >= MAX_INT:
                return MAX_INT
            if sign * value <= MIN_INT:
                return MIN_INT
            i += 1
        return sign * value

Accepted63 / 63 testcases passed

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Practice across 58 topics.

Practice across 58 comprehensive topics covering probability, brain teasers, quantitative finance, machine learning, deep learning and much more.

Filter every problem by topic and difficulty level, and receive instant, structured grading with deep written feedback on every submission.

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QuestionCore ideaSubmissions

One Pass (Optimal)

Core idea

Strategy: one-pass greedy tracking with a running minimum. When a problem asks you to maximize profit from a single buy-sell pair while scanning a sequence of prices, reach for tracking the running minimum price, because the best buy for any sell day is always the smallest price seen so far. Checking every buy-sell pair takes O(n²) time; the running minimum brings it down to O(n). Next time you see a problem about finding extrema that depend on earlier values, the same reflex applies: keep a running minimum or maximum instead of recomputing from scratch.

Track the lowest price seen so far and the profit from selling today. Keep the best profit found.

class Solution:
    def maxProfit(self, prices: list[int]) -> int:
        lowest = prices[0]
        best = 0
        for price in prices[1:]:
            lowest = min(lowest, price)
            best = max(best, price - lowest)
        return best

Time: O(n) · Space: O(1)

Data Structures & Algorithms Problems.

Master your technical rounds with complete solutions and proven coding frameworks for every DSA problem.

Instead of memorizing code, you learn tactical problem-solving techniques and algorithmic frameworks that you can save and apply to instantly tackle new, unseen variations with the exact same core strategies.

Solve your first problem

Targeted Quant Interview Study Plans.

Endless scrolling through a question bank is not prep. Instead, fast-track your progress with curated study guides and specific problem-solving methodologies. Preparing for an elite firm? Access company-specific study roadmaps covering the exact question categories you will face.

Create custom prep schedules, share roadmaps with peers to earn platform rewards, and eliminate decision fatigue with clear daily goals that build unstoppable momentum.

Pick your track

Brainteaser Essentials

Essentials

45 questions

Stochastic Calculus & Core Maths

Essentials

60 questions

Probability & Statistics Essentials

Essentials

80 questions

Quant Developer Track

Role-Specific

120 questions

Quant Researcher Track

Role-Specific

95 questions

Citadel Interview Prep

Firm-Specific

70 questions

Interactive Quant Interview Skill Trees.

Visualize your preparation with interconnected learning paths that map exactly how complex quant topics build on core fundamentals. Master foundational concepts like Combinatorics, Probability Events, and Linear Algebra before advancing to advanced nodes like Markov Chains and Linear Regression.

Quant interviews test a continuous chain of knowledge. Make sure your technical prerequisites are flawless before facing elite trading panels.

Explore the roadmap

Combinatorics Calculus Events & Probability Expected Value Linear Algebra Conditional Probability Discrete Random Variables Conditional Expectation Continuous Random Variables Covariance Markov Chains Linear Regression & Stats

Global Quant Leaderboards & Candidate Networks.

Accelerate your preparation with gamified interview metrics and performance tracking. Every graded solution rewards you with 200 to 2,500 XP, scaled by your specific topical coverage to favor well-rounded candidates. Climb eleven competitive ranks from Intern to Immortal, fighting for a spot on our global top-100 leaderboard.

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Company specific questions

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