A fresh TikTok debrief—let's look at this real flow. The interview had three rounds and was hardcore overall. The first two were pure coding, two algorithm problems each, with tight timing, testing clarity of approach and code completeness. The third was an HM round, first deep-diving the resume, then coding.
1. TikTok SDE NG Flow Overview
| Round | Format | Focus |
|---|---|---|
| Round 1 | Pure coding (2 problems) | Mid-to-high-frequency algorithms, complexity + boundary sensitivity |
| Round 2 | Pure coding (2 problems) | Data-structure design + optimization follow-ups |
| Round 3 | HM round | Resume infra deep dive + behavioral + DP closer |
The question types are mostly mid-to-high-frequency algorithms; you need to be sensitive to complexity and boundary cases.
2. Round 1 Coding
Coding 1: one-edit sorted check
You may modify at most one element of the array—can it become non-decreasing?
def checkPossibility(nums):
modified = False
for i in range(1, len(nums)):
if nums[i] < nums[i - 1]:
if modified:
return False # already edited once, another violation fails
modified = True
# prefer lowering nums[i-1]; if not possible (nums[i] < nums[i-2]), raise nums[i]
if i < 2 or nums[i] >= nums[i - 2]:
nums[i - 1] = nums[i]
else:
nums[i] = nums[i - 1]
return True
Approach: scan for the first non-decreasing violation, greedily decide whether to edit nums[i-1] or nums[i]. Complexity: time O(n), space O(1).
Coding 2: tree node with minimum average distance
Find the node with the minimum average distance to all other nodes (each edge = 1), in O(n).
Approach: this is a "minimum height tree / tree centroid" problem. Two DFS passes—first bottom-up for subtree sizes and distance sums; second top-down rerooting, deriving children from the parent's result. Rerooting formula: dist[child] = dist[parent] + (n - 2*size[child]). Complexity: time O(n).
3. Round 2 Coding
The interviewer chatted for 2 minutes ("how did you optimize database query performance") then went into coding.
Coding 1: nested list iterator (LeetCode 341)
Given a nested integer list, implement an iterator traversing all integers in order; implement
next()andhasNext()yourself.
class NestedIterator:
def __init__(self, nestedList):
# push in reverse; the stack top is always the next element to process
self.stack = nestedList[::-1]
def next(self):
# hasNext() guarantees the top is an integer, so just pop
return self.stack.pop()
def hasNext(self):
while self.stack:
top = self.stack[-1]
if isinstance(top, int):
return True
# top is a list: pop and push its elements back in reverse
self.stack.extend(self.stack.pop()[::-1])
return False
Approach: push in reverse, and in hasNext() unfold lists until the top is an integer. The interviewer asked "why amortized O(1)"—each element is pushed/popped at most once.
Coding 2: group anagrams (better than O(nk log k))
Given a string array, group anagrams. Required better than the sorting O(nk log k).
from collections import defaultdict
def groupAnagrams(strs):
groups = defaultdict(list)
for s in strs:
# count array as key (26 letters), O(k) per string, avoiding O(k log k) sorting
count = [0] * 26
for ch in s:
count[ord(ch) - ord('a')] += 1
groups[tuple(count)].append(s)
return list(groups.values())
Approach: use a 26-dim count array as a tuple key, O(k) per string. The interviewer hinted "can you use a prime product"—the answer is that a prime product can overflow, and a count array is safer. Complexity: time O(nk).
4. Round 3: HM Round
First a resume deep dive, focusing on infra background: system design approach, technology choices, real delivery. Behavioral centered on one core project, with follow-ups on the difficulties faced and how you solved them. The final coding was a DP-leaning problem—not easy, but consistent with TikTok's real hiring style.
5. Prep Points
| Dimension | Tip |
|---|---|
| Coding | Mid-to-high-frequency algorithms, boundary + complexity sensitive, be fast |
| Data-structure design | Master nested iterator and stack-based designs; explain amortized analysis |
| Optimization follow-ups | Proactively raise count-key vs sort-key optimizations |
| HM round | Be ready to deep-dive the infra part of your resume; prepare 1-2 core projects |
FAQ
Q1: How many rounds is TikTok NG, and how hard?
Usually three: two pure coding (2 problems each) + one HM. Hardcore overall with tight timing. Mid-to-high-frequency algorithms + data-structure design, with a DP closer in the third round.
Q2: Why not use sorting for group anagrams?
A sort key is O(k log k), and the interviewer explicitly required optimization. A 26-dim count-array key drops it to O(k) per string. A prime product also works but risks overflow; the count array is safer.
Q3: How do you do the tree min-average-distance node in O(n)?
Rerooting DP: two DFS passes—first computing subtree sizes and distance sums, second top-down rerooting to derive each node's total distance. BFS from every node is O(n^2) and TLEs.
Q4: How do I prepare for the HM infra deep dive?
If infra is on your resume, be ready to explain the system design, technology choices, and delivery results. Prepare 1-2 core projects you can dig three layers into, paired with a behavioral difficulty + resolution story.
Q5: The timing is tight—is there real-time practice?
Yes. Back-to-back pure coding with two timed problems per round makes pacing easy to lose. We offer VO assistance / VO live support: frequent-problem prediction + timed mocks + direction when you stall + pacing help to lock in what you know.
Preparing for TikTok SDE NG?
TikTok NG is hardcore, fast-paced, and the third HM round also deep-dives infra. If you want timed practice on the first two rounds' frequent problems, focused work on nested iterator/anagrams/rerooting DP, or real-time VO assistance / VO live support, reach out—send the role's JD and we will predict the question types first, then plan a practice schedule.
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