Flexport Software Engineer Interview Questions

I applied to Flexport on November 11 without refferal and received an assessment link on November 21. The assessment consisted of four DSA questions, all of LeetCode Medium/Hard difficulty. Technical Interview...

Experience: ~4 yrs HackerRank Round: 4 questions had to be solved in 90 mins. Question were medium to hard difficulty. On clearing hackerrank round, 4 coding rounds were scheduled. Round 1: Coding Round 2...

Current experience: Education: 2021 Grad, IIIT College Position: SDE1 at Fintech Location: Bengaluru, IN Interview experience: Applied Position: SDE2 at Flexport Location: Bengaluru, IN Date: April, 2024 Round 1: Technical [1 hour] Question: Consider an infinite grid. There are two...

Online Assessment # Round 1: 1. My Calendar 1: Leetcode - Solved this 2. Follow up on this: - Code not working as expected \t3. Merge events: \t\t \t \tmerge(10, 20) => [[10, 20]] \tmerge(40, 50) => [[10,...

There were 4 problems to be solved in 120 mins. 3 DSA and 1 REST API based Problem-1 Given an array of n integers, we define score for a pair of indices...

Experience: 2 years Online Assessment: Completed 4 medium-level coding questions, successfully solving all. Round 1: Coding Round Asked a problem similar to this Round 1 question. Round 2: Machine Coding Round Designed an LRU Cache, incorporating...

Part 1 Imagine you have a shipments, that can consist of following fields: weight volume crewSize and a rate for calculating shipment cost. Rate can be only applicable to the one of...

LeetCode #299: Bulls and Cows. Difficulty: Medium. Topics: Hash Table, String, Counting. Asked at Flexport in the last 6 months.

## Problem Find shortest or cheapest flight routes between airports, likely using BFS or Dijkstra on a weighted graph. ## Likely LeetCode equivalent Related to LC 787 Cheapest Flights Within K Stops. ## Tags graph, BFS, phone-screen

## Problem Simulate a flooding scenario on a grid, propagating water from source cells using BFS. ## Likely LeetCode equivalent Related to LC 542 01 Matrix / LC 994 Rotting Oranges. ## Tags matrix, BFS, graph, phone-screen

## Problem Restore or validate IP addresses from a string of digits, enumerating all valid dotted-quad forms. ## Likely LeetCode equivalent Related to LC 93 Restore IP Addresses. ## Tags coding, strings, backtracking, phone-screen

## Round 1 - Coding / OOD ## Problem Implement a token bucket rate limiter. Each bucket refills at a constant rate (tokens per second) up to a maximum capacity. A request consumes one token; if the bucket is empty the request is rejected. ```python class TokenBucket: def __init__(self, capacity: int, refill_rate: float): # capacity: max tokens; refill_rate: tokens added per second pass def allow_request(self, current_time: float) -> bool: # Returns True if a token is available (and consumes it), else False pass ``` ## Example ``` bucket = TokenBucket(capacity=5, refill_rate=2.0) # 2 tokens/sec bucket.allow_request(t=0.0) # -> True (tokens: 4) bucket.allow_request(t=0.0) # -> True (tokens: 3) bucket.allow_request(t=0.0) # -> True (tokens: 2) bucket.allow_request(t=0.0) # -> True (tokens: 1) bucket.allow_request(t=0.0) # -> True (tokens: 0) bucket.allow_request(t=0.0) # -> False (empty) bucket.allow_request(t=1.0) # -> True (refilled 2; consumed 1 -> tokens: 1) ``` ## Follow-ups 1. How do you make this thread-safe for concurrent requests? 2. How do you implement a per-user rate limiter using a dict of buckets with LRU eviction? 3. How does token bucket differ from a leaky bucket and sliding window algorithms? 4. How would you distribute this rate limiter across multiple API servers sharing state in Redis?

## Round 1 - Coding ## Problem A vehicle starts at station 0 with a full tank of `capacity` fuel units. Each station `i` has a fuel amount `fuel[i]` available and a `cost[i]` to travel to the next station. Determine the minimum starting station index from which the vehicle can complete a circular tour of all stations. If no solution exists, return -1. ```python def find_start_station( fuel: list[int], cost: list[int], capacity: int ) -> int: pass ``` ## Example ``` fuel = [1, 2, 3, 4, 5] cost = [3, 4, 5, 1, 2] capacity = 10 Output: 3 # Start at station 3: pick up 4, spend 1 -> arrive 4 # Pick up 5, spend 2 -> arrive 0 ... # Tank never drops below 0 throughout the loop ``` ## Follow-ups 1. How does the solution change if the tank cannot exceed `capacity` at any point (true fuel cap)? 2. What if some stations can be skipped but skipping costs a flat penalty? 3. How do you prove that there is at most one valid starting station when the total fuel >= total cost? 4. Extend the problem: the vehicle can refuel at any station proportionally — how does that change your greedy logic?