Zoom does not publish a round-by-round description of its engineering interviews, so what follows is a pattern drawn from candidate reports and third-party prep sites as of October 2026. Expect a recruiter call, one or more coding rounds at roughly easy-to-medium LeetCode level, system design for mid-level and senior roles, and a behavioral conversation. Details differ by team and level, so ask your recruiter what your loop looks like.<\/p>\n
Reports from candidates describe these stages, though not every candidate sees all of them:<\/p>\n
Treat anything more specific than this (exact number of rounds, exact timing) as unconfirmed unless your recruiter tells you.<\/p>\n
In the interview you can usually pick the language you are fastest in. Pick one, and know its standard library well enough to write a hash map, a heap and a sort with a custom key without looking anything up.<\/p>\n
State the time and space complexity of every solution before the interviewer asks.<\/p>\n
For a company that runs real-time media, it is reasonable to know the difference between TCP and UDP, why real-time audio and video usually tolerate packet loss better than delay, what latency, jitter and bandwidth mean, and what a load balancer and a cache do. You do not need to be a networking specialist for a general software role, but a team working on media or networking code may go deeper.<\/p>\n
Zoom’s problems are not public, so these are practice problems in the same spirit, not leaked questions. Each one is a common pattern with a meeting-flavored setup.<\/p>\n
A warm-up to check you handle messy input. In Python, Ask whether punctuation counts as part of a word and whether to preserve the original spacing. Preserving it changes the problem.<\/p>\n Given join and leave times for participants, find the largest number present at the same time. Turn each participant into two events (+1 at join, -1 at leave), sort, and keep a running count. This is the sweep-line pattern, and it also solves “minimum number of meeting rooms”.<\/p>\n The tie-break matters. If someone leaves at minute 5 and someone else joins at minute 5, counting the join first would report 3 people when there were never more than 2. Say out loud that you made this choice and why. Complexity is O(n log n) for the sort and O(n) space.<\/p>\n “Allow at most Follow-ups interviewers like: what happens to memory for users who never come back (evict idle users), how this changes with many servers (a shared store such as Redis, or per-server limits), and the trade-off against a token bucket, which allows bursts but uses O(1) memory per user.<\/p>\n For roles above entry level, expect an open question such as designing a chat service, a notification system for millions of users, or a simplified video conferencing service. A structure that works:<\/p>\n Our list of questions to ask in a system design interview<\/a> helps with step 1.<\/p>\n Prepare four or five real stories from your work or projects: a hard bug, a disagreement, a project that slipped, something you owned end to end, something you learned from a mistake. For each, be ready to say what the situation was, what you did personally, and what happened. Specifics beat polish. Do not memorize scripts, because interviewers follow up on the details.<\/p>\nsplit()<\/code> without arguments already collapses repeated whitespace and drops leading and trailing spaces.<\/p>\ndef reverse_words(s):\n return \" \".join(reversed(s.split()))\n\nprint(reverse_words(\" Hello World from Zoom \"))\n# Zoom from World Hello<\/code><\/pre>\n2. Peak number of participants in a meeting<\/h3>\n
def peak_participants(events):\n # events: (time, +1 for join \/ -1 for leave)\n best = cur = 0\n # at the same time, process leaves (-1) before joins (+1)\n for _, delta in sorted(events, key=lambda e: (e[0], e[1])):\n cur += delta\n best = max(best, cur)\n return best\n\nevents = [(0, 1), (1, 1), (5, -1), (5, 1), (7, -1), (9, -1)]\nprint(peak_participants(events))\n# 2<\/code><\/pre>\n3. A rate limiter<\/h3>\n
limit<\/code> requests per user in any window<\/code> seconds.” The simple version keeps a queue of recent timestamps per user and drops old ones. This is a sliding window over time.<\/p>\nfrom collections import deque\n\nclass RateLimiter:\n def __init__(self, limit, window):\n self.limit, self.window = limit, window\n self.requests = {} # user -> deque of timestamps\n\n def allow(self, user, now):\n q = self.requests.setdefault(user, deque())\n while q and q[0] <= now - self.window:\n q.popleft()\n if len(q) < self.limit:\n q.append(now)\n return True\n return False\n\nr = RateLimiter(3, 10)\nprint([r.allow(\"u\", t) for t in [0, 1, 2, 3, 9, 10, 11]])\n# [True, True, True, False, False, True, True]<\/code><\/pre>\nSystem design<\/h2>\n
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Behavioral questions<\/h2>\n
Common mistakes<\/h2>\n
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A four-week plan<\/h2>\n
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