Gowtham Potureddi

Posted on May 2

Figma Data Engineering Interview Questions

#python #dataengineering #sql #interview

Figma data engineering interview questions lean on a narrow, high-fluency stack: SQL that ranks and dedupes per entity with ROW_NUMBER() OVER (PARTITION BY creator_id ORDER BY collab_count DESC, last_collab_at DESC), aggregation joins that pull "first event per entity" with MIN(shared_at) plus LEFT JOIN so creators with zero shares survive, and vanilla Python that splits and validates a structured string with str.split('.'), str.isdigit(), int(), and a leading-zero guard — no re, no ipaddress, no pandas. The schema you reason over feels like Figma's own product (creators, files, shares, collaborators), and the bar is fluency with window-function tie-breaks, NULL-safe LEFT JOIN aggregation, and structural-then-per-element validation.

This guide walks through the four topic clusters Figma actually tests, each with a detailed topic explanation, per-sub-topic explanation with a worked example and its solution, and an interview-style problem with a full solution that explains why it works. The mix matches the curated 3-problem Figma set (2 easy, 1 medium, no hard) plus one process-and-prep section — a SQL-plus-vanilla-Python loop where dictation of the invariant out loud is half the score and the other half is typing the right primitive on the first try.

#	Topic (sections 1–4)	Why it shows up at Figma
1	The Figma data engineering interview process	Five-stage funnel — recruiter → HM → tech screen (Python + SQL) → recruiter prep → onsite panel; staff candidates add an executive screen.
2	SQL window functions on collaboration events	Closest Collaborator per Creator (MEDIUM) — `ROW_NUMBER() OVER (PARTITION BY creator_id ORDER BY collab_count DESC, last_collab_at DESC)` then `WHERE rn = 1`.
3	Aggregation and joins for first-event logic	First Share Time per Creator (EASY) — `MIN(shared_at)` over `creators LEFT JOIN shares` so creators with no shares survive as `NULL`.
4	String parsing and validation in vanilla Python	IPv4 Address Validator (EASY) — `str.split('.')`, `len == 4`, `isdigit()`, leading-zero guard, `0 ≤ int(p) ≤ 255` — no `re`, no `ipaddress`.

1. The Figma Data Engineering Interview Process

The Figma DE interview funnel from recruiter call to onsite panel

The Figma data engineer interview process is a five-stage funnel that takes about three to four weeks end to end: a recruiter call, a hiring-manager conversation, a single technical screen that combines vanilla Python and LeetCode-medium SQL, an explicit recruiter-prep call, and a four-round virtual onsite panel. Staff candidates pick up an additional executive screen with a director.

The two stages most candidates misread are the technical screen (over-prepared on contest algorithms, under-prepared on window functions) and the data modeling round (treating it as a whiteboard schema dump when it is actually a tradeoff conversation). Both are addressed below, with the framing the interviewer is reading for.

Recruiter and hiring-manager calls

Detailed explanation. The recruiter call is a 30-minute fit-and-routing chat: background, motivation, why Figma, where you are with other companies. Hold your salary expectation and competing-offer status until the offer stage; naming a number first usually leaves money on the table. The hiring-manager call is 45 minutes and behavioral — past projects, problem-solving approach, how you collaborate cross-functionally. Recruiters often block the technical phone screen back-to-back with the HM call, so treat them as a single 105-minute commitment.

Worked example. When the recruiter asks "what salary range are you looking for?", a clean redirect is "I'd love to understand the band Figma has set for this role first — what range did the team budget?" That moves the anchor to them.

Rule of thumb: never name a number first. Restate the question back as "what's the band?" until the recruiter commits.

Technical screen — Python plus LeetCode-medium SQL

Detailed explanation. Roughly an hour in a CoderPad-style live editor. The flavor that recent interview reports converge on: Python with no libraries (list, dict, str, conditionals — no re, no pandas, no ipaddress) plus SQL similar to LeetCode-medium / DataLemur with no contest-style algorithms. Some teams use a Byteboard online assessment instead — three sections (SQL, Python, short-essay), 120 minutes total, with the SQL section alone giving you about 45 minutes for four questions. Either way, the signal Figma is reading is decomposition + edge-case awareness, not library trivia.

Worked example. A typical SQL prompt: "Given a shares(creator_id, file_id, shared_at) table, return each creator's first-share time and the file they first shared." The naive SELECT MIN(shared_at), file_id FROM shares GROUP BY creator_id is wrong — the file_id will be ambiguous. The correct decomposition is "first event per entity," covered in section 3.

Rule of thumb: if the SQL prompt mentions "per X" or "first / latest / closest per X," reach for either MIN/MAX + GROUP BY + a join or ROW_NUMBER() OVER (PARTITION BY X ORDER BY …) before anything else.

Onsite panel and the relational data-modeling round

Detailed explanation. The onsite is a four-round virtual panel: a coding round (CoderPad), one or two system-design rounds (often conducted in Figma the product), a behavioral round, and a project deep-dive. The data-modeling round is the one most candidates misread. Reports from recent Figma DE onsites describe it as "relational data, conversational, tradeoff-focused" — closer to a design-review whiteboard chat than a "draw the schema" exercise. Interviewers want you to talk through how the data will be used, what queries it supports, and what trades you are making (denormalize for read speed, normalize for write integrity, partition for scale).

Worked example. "Design the data model for Figma comments threading." A strong opener names the grain (one row per comment), the identity (comment_id), the parent pointer (parent_comment_id for replies, nullable), and the scoping keys (file_id, creator_id). Then trade off: a closure table for arbitrary-depth threads vs adjacency list for shallow threads with one extra join per level. State the call out loud — interviewers grade the conversation, not the diagram.

Rule of thumb: every column you add deserves one sentence — who reads it, who writes it, and at what grain.

Project deep dive and behavioral

Detailed explanation. The project deep-dive is one hour with one interviewer. Bring a real pipeline you owned end-to-end: ingestion, transformation, storage, the consumers, and the failure modes you handled. Interviewers will drill on scale (rows per day, peak QPS), tradeoffs (why Snowflake over BigQuery, why dbt over hand-rolled SQL), and what you would do differently. Behavioral rounds use STAR — Situation, Task, Action, Result — and lean on real friction (tight deadlines, conflicting stakeholders, post-incident retros).

Worked example. When asked "what would you do differently?", a strong answer names a specific tradeoff you made under time pressure and one or two lines on the alternative you would re-evaluate today — interviewers reward retrospective honesty over post-hoc heroism.

Common beginner mistakes

Naming a salary number on the recruiter call (anchors low; let them name the band first).
Treating the technical screen as a contest-algorithm round (Figma's screen is decomposition-grade, not LeetCode-hard).
Sketching a schema silently in the data-modeling round (the round is a tradeoff conversation; talk through grain, identity, and reads/writes out loud).
Bringing a half-owned project to the deep-dive (interviewers drill on what you actually wrote — pick one you owned end to end).

Practice: drill the Figma DE panel before the live screen

COMPANY
Figma — all DE problems
Figma data engineering practice set

Practice →

SQL
Figma — SQL only
Figma SQL practice

Practice →

2. SQL Window Functions on Collaboration Events

Window functions and per-entity ranking in SQL for data engineering

Figma's collaboration data is naturally event-shaped — a row each time someone opens a file, edits a frame, leaves a comment, or shares with a collaborator. The headline interview problem on the Figma practice set, Closest Collaborator per Creator, is a textbook "pick one row per entity" task on exactly this shape, and the canonical answer is a window function: ROW_NUMBER() OVER (PARTITION BY creator_id ORDER BY collab_count DESC, last_collab_at DESC) ranks each candidate collaborator inside each creator's partition, and a final WHERE rn = 1 picks the top one.

Pro tip: Window functions evaluate after WHERE and before ORDER BY in the SQL pipeline. That is why you cannot filter WHERE rn = 1 in the same SELECT — wrap the window in a CTE first, then filter in the outer query. This is the single most common Figma SQL screen mistake.

`ROW_NUMBER` vs `RANK` vs `DENSE_RANK`

The window-function invariant: all three assign positions inside a PARTITION BY block ordered by the ORDER BY clause; they only differ on ties.

ROW_NUMBER() — every row gets a distinct integer; ties resolved arbitrarily. Default pick.
RANK() — ties share a number, then the next rank skips. 1, 1, 3.
DENSE_RANK() — ties share a number, no skip. 1, 1, 2.

Worked example. Three collaborators for one creator, ordered by collab_count DESC.

collaborator	collab_count	row_number	rank	dense_rank
alice	12	1	1	1
bob	12	2	1	1
carol	8	3	3	2

Worked-example solution.

SELECT
  creator_id,
  collaborator_id,
  ROW_NUMBER() OVER (
    PARTITION BY creator_id
    ORDER BY collab_count DESC
  ) AS rn
FROM creator_collab_counts;

Rule of thumb: default to ROW_NUMBER(). Reach for RANK() only when the prompt says "include ties." DENSE_RANK() shows up almost never — it's a leaderboard primitive where the rank label matters to humans.

`PARTITION BY` for "one row per entity"

The partition invariant: PARTITION BY is the window-function equivalent of GROUP BY, except it does not collapse rows — it keeps every row and runs the window fresh inside each block.

Without PARTITION BY — one global ranking; useless for per-creator answers.
PARTITION BY creator_id — one ranking per creator; rn = 1 survives the filter for each creator.

Worked example. A collaborators table with two creators.

creator_id	collaborator_id	collab_count
1	alice	12
1	bob	8
2	carol	5
2	dave	3

After ROW_NUMBER() OVER (PARTITION BY creator_id ORDER BY collab_count DESC), alice and carol each get rn = 1 — two rows survive WHERE rn = 1, one per creator.

Worked-example solution.

WITH ranked AS (
  SELECT
    creator_id, collaborator_id, collab_count,
    ROW_NUMBER() OVER (
      PARTITION BY creator_id
      ORDER BY collab_count DESC
    ) AS rn
  FROM collaborators
)
SELECT creator_id, collaborator_id
FROM ranked
WHERE rn = 1;

Rule of thumb: if the prompt says "per X, return one row," the PARTITION BY X is non-negotiable. Forgetting it is the #1 Figma SQL screen failure.

Tie-breaking with a secondary `ORDER BY` key

The tie-break invariant: two rows with identical primary-sort keys create a tie; the secondary key in ORDER BY decides who wins. ROW_NUMBER() will pick one arbitrarily otherwise — same query, different answer across runs.

Primary key only — ORDER BY collab_count DESC — ties resolve nondeterministically.
With secondary — ORDER BY collab_count DESC, last_collab_at DESC — most-recent wins among ties.

Worked example. Bob and alice both have 12 collaborations with creator 1. Alice's most-recent collab was 2026-04-01; Bob's was 2026-03-15. Alice wins the tie.

Worked-example solution.

ROW_NUMBER() OVER (
  PARTITION BY creator_id
  ORDER BY collab_count DESC, last_collab_at DESC
) AS rn

Rule of thumb: state the tie-break out loud — "if X ties on Y, I'll pick the more recent" — before you start typing. Interviewers reward the dictation.

Common beginner mistakes

Filtering WHERE rn = 1 in the same SELECT as the window (windows evaluate after WHERE; wrap in a CTE).
Forgetting PARTITION BY and getting one global ranking.
Using RANK() when the prompt asks for exactly one row per entity (RANK keeps ties; you want ROW_NUMBER).
Skipping the tie-break — same query returns different rows across runs.

PipeCode's SQL for data engineering interviews course drills these three primitives across thirty-plus window-function problems, including the tie-breaking variants Figma reaches for.

SQL Interview Question on Closest Collaborator per Creator

Tables creators(creator_id INT, name VARCHAR) and collaborations(creator_id INT, collaborator_id INT, collab_at TIMESTAMP) track every collaboration event. For each creator, return the single closest collaborator — the one with the most collaborations, breaking ties by most-recent collaboration. Output creator_id, collaborator_id, collab_count.

Solution Using ROW_NUMBER over Aggregated Collaborations

WITH counts AS (
  SELECT
    creator_id,
    collaborator_id,
    COUNT(*)            AS collab_count,
    MAX(collab_at)      AS last_collab_at
  FROM collaborations
  GROUP BY creator_id, collaborator_id
),
ranked AS (
  SELECT
    creator_id, collaborator_id, collab_count,
    ROW_NUMBER() OVER (
      PARTITION BY creator_id
      ORDER BY collab_count DESC, last_collab_at DESC
    ) AS rn
  FROM counts
)
SELECT creator_id, collaborator_id, collab_count
FROM ranked
WHERE rn = 1
ORDER BY creator_id;

Step-by-step trace (input collaborations):

creator_id	collaborator_id	collab_at
1	7	2026-04-01
1	7	2026-04-02
1	9	2026-03-10
1	9	2026-03-11
2	5	2026-04-15

counts CTE — group by (creator_id, collaborator_id) and count rows. Row (1, 7) → collab_count = 2, last_collab_at = 2026-04-02. Row (1, 9) → collab_count = 2, last_collab_at = 2026-03-11. Row (2, 5) → collab_count = 1, last_collab_at = 2026-04-15.
ranked CTE — partition by creator_id — creator 1 has two rows (7 and 9), creator 2 has one row (5).
ORDER BY collab_count DESC, last_collab_at DESC — inside creator 1, both candidates tie on count 2. The secondary key last_collab_at decides: 2026-04-02 (collab 7) beats 2026-03-11 (collab 9). Collaborator 7 gets rn = 1, collaborator 9 gets rn = 2.
WHERE rn = 1 — keep one row per creator. Creator 1 keeps collaborator 7; creator 2 keeps collaborator 5.
Order final output — ORDER BY creator_id for stable output.

Output:

creator_id	collaborator_id	collab_count
1	7	2
2	5	1

Why this works — concept by concept:

Aggregate-then-rank — collapsing the raw event log into (creator, collaborator, count, last_at) first means the window function ranks aggregated rows, not events; one creator with 1,000 events still produces only as many rows as distinct collaborators.
PARTITION BY creator_id — resets ROW_NUMBER() inside each creator's block so every creator independently gets a rn = 1 row.
Composite ORDER BY — the secondary last_collab_at DESC is the tie-break the prompt specified; without it, ties resolve nondeterministically and the same query returns different answers across runs.
WHERE rn = 1 outside the window — you cannot filter a window in the same SELECT because window functions evaluate after WHERE; the CTE gives you a stable layer to filter against.
ROW_NUMBER not RANK — RANK would keep both tied collaborators; the prompt asks for exactly one per creator.
Cost — O(n log n) for the sort inside each partition (n = aggregated (creator, collaborator) rows), plus O(events) for the initial GROUP BY; both pieces parallelize across creators.

SQL
Figma — window functions
Closest Collaborator per Creator (Figma)

Practice →

SQL
Topic — window functions
Window function problems (all companies)

Practice →

3. Aggregation and Joins for First-Event Logic

`GROUP BY` aggregation, `LEFT JOIN`, and first-event-per-entity SQL

Not every "per-creator" SQL question deserves a window function. The second canonical Figma pattern is first-event-per-entity, and when the only column you need beyond the entity is the timestamp itself, plain MIN() plus GROUP BY is cheaper, simpler, and the answer the interviewer is reading for. The headline problem here is First Share Time per Creator — return each creator's earliest share timestamp, including creators who never shared (those need a LEFT JOIN with NULL survival).

Pro tip: MIN() is NULL-safe — aggregates ignore NULL inputs and a group of all-NULL returns NULL. That is exactly the behavior you want for "creators with no shares" — the result column carries NULL through as "no share yet" instead of needing a separate union.

`GROUP BY` + `MIN` for the earliest event

The aggregation invariant: SELECT entity, MIN(ts) FROM events GROUP BY entity is one pass — one aggregate per group, no sort inside the group.

Cheap path — needs only the entity and the minimum timestamp; no other columns from the matching row.
Trap — SELECT entity, MIN(ts), other_col is illegal in standard SQL because other_col is not aggregated; engines either error or return an arbitrary row.

Worked example. A shares(creator_id, file_id, shared_at) table.

creator_id	file_id	shared_at
1	100	2026-04-01
1	101	2026-03-10
2	200	2026-04-15

MIN(shared_at) per creator_id returns creator 1 → 2026-03-10, creator 2 → 2026-04-15.

Worked-example solution.

SELECT creator_id, MIN(shared_at) AS first_share_at
FROM shares
GROUP BY creator_id;

Rule of thumb: if the prompt asks for "the earliest / latest timestamp," MIN/MAX is the answer. If it asks for "the earliest / latest row (with other columns)," skip to ROW_NUMBER().

`LEFT JOIN` to keep entities with no events

The LEFT JOIN invariant: every row from the left side survives; unmatched right-side columns come through as NULL.

INNER JOIN — drops left rows with no right match (silent loss).
LEFT JOIN — keeps left rows; right columns are NULL when no match.

Worked example. Adding creator 3 with zero shares.

creator_id	name
1	alice
2	bob
3	carol

creators LEFT JOIN shares returns (1, 2026-03-10), (2, 2026-04-15), (3, NULL). An INNER JOIN would silently drop creator 3.

Worked-example solution.

SELECT
  c.creator_id, c.name,
  MIN(s.shared_at) AS first_share_at
FROM creators c
LEFT JOIN shares s ON s.creator_id = c.creator_id
GROUP BY c.creator_id, c.name;

Rule of thumb: if the prompt says "include creators who never shared" or "every creator," you need LEFT JOIN. If it says "creators who shared," INNER JOIN is fine.

Aggregation vs window functions: which to pick

The selection invariant: need only the timestamp? MIN() + GROUP BY. Need the full row at the timestamp? ROW_NUMBER() + PARTITION BY. They produce identical answers in the simple case but diverge the moment the prompt asks for "what file was first shared" instead of "when was the first share."

MIN plan — one pass over the join, single aggregate per group.
ROW_NUMBER plan — sort inside each partition (O(n log n) per partition), but every column of the chosen row survives.

Worked example. "For each creator, return the file they first shared." MIN(shared_at) gives you the timestamp but not the matching file_id. The clean answer is ROW_NUMBER() OVER (PARTITION BY creator_id ORDER BY shared_at ASC) filtered to rn = 1, which carries the full row.

Worked-example solution.

WITH ranked AS (
  SELECT creator_id, file_id, shared_at,
    ROW_NUMBER() OVER (
      PARTITION BY creator_id ORDER BY shared_at ASC
    ) AS rn
  FROM shares
)
SELECT creator_id, file_id, shared_at AS first_share_at
FROM ranked
WHERE rn = 1;

Rule of thumb: the prompt's verbs decide. "When was X?" → MIN. "What was X?" → ROW_NUMBER.

Common beginner mistakes

INNER JOIN instead of LEFT JOIN for "every creator" — silently drops zero-event creators.
Selecting non-aggregated columns alongside MIN() — error in Postgres, undefined in MySQL.
Defaulting to ROW_NUMBER() when MIN() is enough — extra sort cost for no benefit.
Treating the NULL first-share as a real timestamp downstream — wrap in COALESCE or filter at consumption.

SQL Interview Question on First Share Time per Creator

Tables creators(creator_id INT, name VARCHAR) and shares(creator_id INT, file_id INT, shared_at TIMESTAMP). Return every creator and their first-share timestamp, including creators who have never shared. Output creator_id, name, first_share_at (nullable).

Solution Using GROUP BY MIN with a LEFT JOIN

SELECT
  c.creator_id,
  c.name,
  MIN(s.shared_at) AS first_share_at
FROM creators c
LEFT JOIN shares s
  ON s.creator_id = c.creator_id
GROUP BY c.creator_id, c.name
ORDER BY c.creator_id;

Step-by-step trace (inputs):

creators:

creator_id	name
1	alice
2	bob
3	carol

shares:

creator_id	file_id	shared_at
1	100	2026-04-01
1	101	2026-03-10
2	200	2026-04-15

LEFT JOIN — match every creators row to its shares rows. Creator 1 gets two share rows; creator 2 gets one share row; creator 3 gets a single row with NULL on every shares column.
GROUP BY (creator_id, name) — collapse to one row per creator. name is included so it survives the aggregation.
MIN(s.shared_at) per creator — creator 1 has timestamps 2026-04-01 and 2026-03-10; minimum is 2026-03-10. Creator 2 has one timestamp, so MIN returns it. Creator 3 has only NULL on s.shared_at; MIN over all-NULL returns NULL.
Order final output — ORDER BY creator_id for stability.

Output:

creator_id	name	first_share_at
1	alice	2026-03-10
2	bob	2026-04-15
3	carol	NULL

Why this works — concept by concept:

LEFT JOIN preserves left side — every creators row survives even when the right side has no match; the result for an unmatched creator is one row with NULL on every shares column.
MIN is NULL-safe — aggregates ignore NULL inputs; a group of all-NULL returns NULL, which correctly carries through as "no share yet."
GROUP BY includes name — non-aggregated columns must appear in GROUP BY (standard SQL); skipping it raises an ambiguity error in Postgres and ANSI engines.
No window function needed — the prompt asks only for the timestamp, not the file; MIN is the right tool, one pass over the join.
ORDER BY outside the aggregate — final ordering is for human readability; MIN already produced the per-group minimum, no second sort needed inside the aggregate.
Cost — O(creators + shares) for the hash-join, plus one aggregation pass; cheaper than ROW_NUMBER() + WHERE rn = 1 because there is no per-partition sort.

SQL
Figma — aggregation
First Share Time per Creator (Figma)

Practice →

COMPANY
Figma — aggregation
Figma aggregation problems

Practice →

4. String Parsing and Validation in Vanilla Python

Structured-string validation in Python for data engineering

Half of the Figma technical screen is Python, and the explicit constraint reported across multiple recent onsites is no libraries — no re, no ipaddress, no pandas. The thinking the interviewer wants to see is decomposition with str.split('.'), str.isdigit(), int(), and conditionals — primitives that prove you understand the data shape rather than the API surface. The headline problem on the Figma practice set is IPv4 Address Validator — given a string, return True if it is a valid IPv4 dotted-quad and False otherwise.

Pro tip: the leading-zero rule catches most candidates. "192.168.0.1" is valid; "192.168.01.1" is not (a leading zero on a multi-digit octet is universally rejected by RFC 791). The check is one line: if len(part) > 1 and part[0] == '0': return False. Forgetting it passes seven test cases and fails the eighth.

Splitting and counting parts

The structural invariant: a valid IPv4 dotted-quad has exactly four parts separated by single dots.

str.split('.') — returns the parts as a list.
Too many — "1.2.3.4.5".split('.') → length 5, reject.
Too few — "1.2.3".split('.') → length 3, reject.
Empty parts — "1..3.4".split('.') → ['1', '', '3', '4']; the empty string fails isdigit() later.

Worked example. Three quick splits.

input	split	length	verdict
`"192.168.0.1"`	`['192','168','0','1']`	4	proceed
`"1.2.3.4.5"`	`['1','2','3','4','5']`	5	reject
`"1..3.4"`	`['1','','3','4']`	4	proceed (per-element will reject)

Worked-example solution.

def split_into_octets(s: str) -> list[str] | None:
    parts = s.split('.')
    if len(parts) != 4:
        return None
    return parts

Rule of thumb: check structural invariants (length, presence) before per-element invariants (range, format). Cheap checks first.

Validating each octet — `isdigit` plus range

The per-element invariant: each octet must be all decimal digits (isdigit()) and convert to an integer in [0, 255].

isdigit() — True only for non-empty strings of '0'..'9'. Rejects '', '-1', 'abc' in one call.
int() — safe to call after isdigit() passes.
Range bound — 0 <= n <= 255.

Worked example. Five octet candidates.

octet	isdigit	int(p)	in [0,255]	verdict
`'255'`	✓	255	✓	pass
`'256'`	✓	256	✗	reject
`'-1'`	✗	—	—	reject
`'abc'`	✗	—	—	reject
`''`	✗	—	—	reject

Worked-example solution.

def is_valid_octet_basic(part: str) -> bool:
    if not part.isdigit():
        return False
    n = int(part)
    return 0 <= n <= 255

Rule of thumb: s.isdigit() is the cheap gate that covers empty, signed, and non-numeric in one call. Always run it before int().

Leading-zero edge cases

The leading-zero invariant: a multi-digit octet cannot start with '0'. '01' is invalid; '0' alone is valid; '00' is invalid; '001' is invalid.

Length-1 octet — '0' is fine; the guard skips because len(part) > 1 is False.
Length-2+ octet starting with '0' — reject.
Why this rule — RFC 791 forbids leading zeros to avoid octal-literal ambiguity in legacy parsers.

Worked example. Four leading-zero edge cases.

input	first octet	len > 1	starts with '0'	verdict
`"192.168.0.1"`	`'192'`	✓	✗	pass
`"192.168.01.1"`	`'01'` (third)	✓	✓	reject
`"0.0.0.0"`	`'0'`	✗	— (skipped)	pass
`"00.0.0.0"`	`'00'`	✓	✓	reject

Worked-example solution.

def is_valid_octet(part: str) -> bool:
    if not part.isdigit():
        return False
    if len(part) > 1 and part[0] == '0':
        return False
    n = int(part)
    return 0 <= n <= 255

Rule of thumb: the length guard len(part) > 1 is the half of the leading-zero check most candidates miss — without it, '0' would be wrongly rejected.

Common beginner mistakes

Skipping the structural length check and crashing later on parts[3] when the input has 3 parts.
Calling int(part) before isdigit() and getting a ValueError on 'abc'.
Writing if part[0] == '0' without the len(part) > 1 guard — wrongly rejects the literal '0' octet in "0.0.0.0".
Reaching for re or ipaddress despite the "no libraries" rule — interviewers note it as a signal you did not internalize the constraint.

Python Interview Question on IPv4 Address Validator

Write a function is_valid_ipv4(s: str) -> bool that returns True if s is a valid IPv4 dotted-quad address (four octets separated by single dots, each octet is a non-leading-zero decimal in [0, 255]) and False otherwise. Do not import re, ipaddress, or any other module.

Solution Using Split-and-Validate Without Regex

def is_valid_ipv4(s: str) -> bool:
    parts = s.split('.')
    if len(parts) != 4:
        return False
    for p in parts:
        if not p.isdigit():
            return False
        if len(p) > 1 and p[0] == '0':
            return False
        n = int(p)
        if n < 0 or n > 255:
            return False
    return True

Step-by-step trace (input s = "192.168.0.1"):

step	value	check
split	`['192', '168', '0', '1']`	len == 4 ✓
octet `'192'`	`isdigit` ✓, len 3 / no leading zero ✓, `int → 192` in `[0,255]` ✓	pass
octet `'168'`	`isdigit` ✓, no leading zero ✓, `168` in range ✓	pass
octet `'0'`	`isdigit` ✓, len 1 / leading-zero check skipped ✓, `0` in range ✓	pass
octet `'1'`	`isdigit` ✓, no leading zero ✓, `1` in range ✓	pass

Split on '.' — produces ['192', '168', '0', '1'], length 4. The structural check passes, so move to per-octet validation.
Validate '192' — '192'.isdigit() is True. Length is 3, but the first character is '1', not '0', so the leading-zero check is irrelevant. int('192') == 192, which is in [0, 255]. Pass.
Validate '168' — same path, passes.
Validate '0' — '0'.isdigit() is True. Length is 1, so the leading-zero check (len(p) > 1) short-circuits to skip. int('0') == 0, which is in [0, 255]. Pass. This is the case that catches most candidates if they wrote if p[0] == '0' without the length guard.
Validate '1' — same path as '0', passes.
Return True — every octet passed; the input is a valid IPv4 address.

Output:

input	result
`"192.168.0.1"`	`True`
`"192.168.01.1"`	`False` (leading zero on `'01'`)
`"1.2.3.4.5"`	`False` (5 octets)
`"256.0.0.0"`	`False` (out of range)
`"0.0.0.0"`	`True`

Why this works — concept by concept:

Structural check first — len(parts) != 4 rejects too-short and too-long inputs before any per-octet work; the per-octet logic does not have to assume four parts.
isdigit rejects empty and signed — ''.isdigit() and '-1'.isdigit() both return False, so this single call covers two failure modes.
Leading-zero guard with length check — len(p) > 1 and p[0] == '0' rejects '01' while accepting '0'; without len(p) > 1, '0' would be wrongly rejected.
int is safe after isdigit — once isdigit passed, int(p) cannot raise; the conversion is the cheap check, the range bound is the final gate.
Boolean short-circuit — every failure returns False immediately; only the all-pass path falls through to return True. No mutable state, no flags.
Cost — O(L) where L is the input length; one split, four constant-size validations.

PYTHON
Figma — string parsing
IPv4 Address Validator (Figma)

Practice →

PYTHON
Topic — string parsing
String parsing problems

Practice →

Tips to crack Figma data engineering interviews

These are habits that move the needle in real Figma DE loops — not a re-statement of the topics above.

Practice with Figma's actual data shape

Figma's interview prompts model their own product. The schema you reason over is creators, files, shares, collaborators, comments — collaboration events with timestamps. Drilling on order-line ecommerce schemas wastes prep time. Stick to event-shaped tables with a per-entity grain, and pull problems from the aggregation topic page and the joins topic page for Figma-tagged shapes.

Master `ROW_NUMBER` plus `PARTITION BY` cold

The headline #506 problem and most window-function variants live here. Type the WITH ranked AS (... ROW_NUMBER() OVER (PARTITION BY ...) AS rn ...) SELECT ... WHERE rn = 1 skeleton until it is muscle memory. Add a deterministic tie-break to every ORDER BY — interviewers reward stating "if X ties on Y, I'll pick the more recent" before you start typing. Drill it on the window-functions topic page.

Write Python without imports

Train yourself to reach for str.split, dicts, lists, and conditionals before touching re or ipaddress. PipeCode's Python for data engineering interviews course drills the vanilla-Python primitives Figma's screen rewards, and the string-parsing topic page has more validators in the same shape as the IPv4 problem.

Talk through tradeoffs in the data-modeling round

The data-modeling round is relational and conversational. Explain why a design choice — denormalize for read speed, normalize for write integrity, partition for scale — not just what the schema looks like. Every column you add deserves a sentence about who reads it, who writes it, and at what grain.

Where to practice on PipeCode

Skill lane	Practice path
Curated Figma practice set	/explore/practice/company/figma
Window functions on collaboration events	/explore/practice/topic/window-functions
Aggregation + first-event SQL (Figma-tagged)	/explore/practice/company/figma/topic/aggregation
Joins for relational data modeling (Figma-tagged)	/explore/practice/company/figma/topic/joins
String parsing in vanilla Python	/explore/practice/topic/string-parsing
Python validation primitives	/explore/practice/topic/validation
All practice topics	/explore/practice/topics
Interview courses	/explore/courses

Communication under time pressure

State assumptions before typing: "I'll assume creator_id is never NULL and that ties on collab_count resolve by most-recent." State grain: "One row per creator per collaborator after the aggregation." State edge cases: "If a creator has zero shares, my LEFT JOIN keeps them with NULL on first_share_at." Interviewers grade clear reasoning above silent-and-perfect.

Frequently asked questions

What is the Figma data engineering interview process?

The Figma data engineer interview process is a five-stage funnel: a 30-minute recruiter call, a 45-minute hiring-manager conversation, a one-hour technical screen mixing vanilla Python and LeetCode-medium SQL, an explicit recruiter-prep call, and a four-round virtual onsite panel covering coding, system design, behavioral, and a project deep-dive. Staff candidates pick up an additional executive screen with a director. Total elapsed time is typically three to four weeks. The curated Figma practice set on PipeCode mirrors the technical-screen flavor.

What programming languages does Figma test for data engineering?

Figma's data engineering interviews lean on SQL and Python — that is the entire technical screen. Python is vanilla, with explicit "no library" rules reported across recent onsites: list, dict, str, conditionals, no re, no pandas, no ipaddress. SQL is at LeetCode-medium / DataLemur grade — heavy on window functions, aggregation, and joins. The string-parsing topic page on PipeCode matches the Figma Python flavor closely.

What SQL topics show up most in Figma data engineering interviews?

The topics are narrow and consistent: window functions (ROW_NUMBER with PARTITION BY and a deterministic ORDER BY), aggregation with GROUP BY and MIN/MAX for first-event-per-entity questions, and joins — especially LEFT JOIN to keep entities with no matching events. PipeCode's aggregation problems and joins problems tagged to Figma drill these directly.

How difficult are Figma data engineering interview questions?

Figma data engineering interview questions are calibrated at LeetCode-medium for SQL and easy-to-medium for Python — the bar is fluency, not contest difficulty. Reports describe the panel as "very high bar but creativity-encouraged" — interviewers reward candidates who think out loud and ask clarifying questions. The 3-problem PipeCode set is intentionally split 2 easy / 1 medium so you can build confidence and then stretch.

How should I prepare for a Figma data engineering interview?

Solve the 3-problem Figma practice set end to end — that maps the exact pattern coverage. Then back-fill: 20+ window-function problems for the headline #506 pattern, 10+ aggregation/join problems for first-event logic, and 10+ vanilla-Python parsing problems for the IPv4-style screen. Add product familiarity — spend 30 minutes in Figma the day before your interview so you can frame answers in product terms.

Does the Figma data engineering interview include a data modeling round?

Yes — it is part of the four-round onsite panel and is relational, conversational, and tradeoff-focused, not a whiteboard schema dump. Expect to design something close to Figma's collaboration model (comments threading, file sharing, workspace permissions) and to be graded on whether you state the grain, the identity column, the scoping keys, and the read/write tradeoffs out loud. Candidates who treat the round as a discussion tend to pass; candidates who silently sketch a schema tend to fail.

Start practicing Figma data engineering problems

Reading patterns is not the same as typing them under time pressure. PipeCode pairs company-tagged Figma problems with tests, AI feedback, and a coding environment so you can drill the exact SQL window-function and Python parsing patterns Figma asks — without the noise of generic algorithm prep.

Pipecode.ai is Leetcode for Data Engineering.

Browse Figma practice →
SQL for DE interviews course →

Top Figma data engineering interview topics

1. The Figma Data Engineering Interview Process

The Figma DE interview funnel from recruiter call to onsite panel

Recruiter and hiring-manager calls

Technical screen — Python plus LeetCode-medium SQL

Onsite panel and the relational data-modeling round

Project deep dive and behavioral

Practice: drill the Figma DE panel before the live screen

2. SQL Window Functions on Collaboration Events

Window functions and per-entity ranking in SQL for data engineering

ROW_NUMBER vs RANK vs DENSE_RANK

PARTITION BY for "one row per entity"

Tie-breaking with a secondary ORDER BY key

SQL Interview Question on Closest Collaborator per Creator

Solution Using ROW_NUMBER over Aggregated Collaborations

3. Aggregation and Joins for First-Event Logic

GROUP BY aggregation, LEFT JOIN, and first-event-per-entity SQL

GROUP BY + MIN for the earliest event

LEFT JOIN to keep entities with no events

Aggregation vs window functions: which to pick

SQL Interview Question on First Share Time per Creator

Solution Using GROUP BY MIN with a LEFT JOIN

4. String Parsing and Validation in Vanilla Python

Structured-string validation in Python for data engineering

Splitting and counting parts

Validating each octet — isdigit plus range

Leading-zero edge cases

Python Interview Question on IPv4 Address Validator

Solution Using Split-and-Validate Without Regex

Tips to crack Figma data engineering interviews

Practice with Figma's actual data shape

Master ROW_NUMBER plus PARTITION BY cold

Write Python without imports

Talk through tradeoffs in the data-modeling round

Where to practice on PipeCode

Communication under time pressure

Frequently asked questions

What is the Figma data engineering interview process?

What programming languages does Figma test for data engineering?

What SQL topics show up most in Figma data engineering interviews?

How difficult are Figma data engineering interview questions?

How should I prepare for a Figma data engineering interview?

Does the Figma data engineering interview include a data modeling round?

Start practicing Figma data engineering problems

`ROW_NUMBER` vs `RANK` vs `DENSE_RANK`

`PARTITION BY` for "one row per entity"

Tie-breaking with a secondary `ORDER BY` key

`GROUP BY` aggregation, `LEFT JOIN`, and first-event-per-entity SQL

`GROUP BY` + `MIN` for the earliest event

`LEFT JOIN` to keep entities with no events

Validating each octet — `isdigit` plus range

Master `ROW_NUMBER` plus `PARTITION BY` cold