ByteDance OA question bank has a certain scale, but the question types and patterns are relatively fixed. As long as you are familiar with the common routines in advance, you can complete the first two questions in ten minutes during the exam and securely win 4 questions in 40 minutes. This article has compiled four real questions from my latest ByteDance OA, with detailed problem-solving ideas and code examples. I hope it will help students who are preparing for Byte overseas positions, TikTok, domestic autumn recruitment or summer internships to familiarize themselves with the rhythm in advance and avoid pitfalls.
ByteDance OA Overview
Platform: CodeSignal
Duration: 70 minutes
Number of questions: 4
Difficulty: Lower to medium, you can complete it quickly once you are familiar with the routine
T1: User Ratings and Levels
Brief description of the question
Given an initial rating Initial And an array of changes Changes, the final score is obtained after being accumulated in sequence, and it is judged in which grade range the score falls.
Ideas
- Directly to
InitialAndChangesSum - Use
If-elseJudgment interval:< 1000→"beginner"1000–1499→"intermediate"1500–1999→"advanced"≥ 2000→"pro"
Example
Def solution(initial, changes):
total = initial + sum(changes)
if total < 1000:
return "beginner"
elif total < 1500:
return "intermediate"
elif total < 2000:
return "advanced"
else:
return "pro"T2: word transformation
Brief description of the question
For each word:
- If the first and last letters are both vowels (case sensitive,
AeiouAEIOU), keep the beginning and end, invert the middle part, capitalize the first letter of the middle part, and lowercase the rest. - Otherwise leave unchanged.
Ideas
- Define vowel set
- Judge
S[0] in vowels and s[-1] in vowels - Middle part:
S[1:-1]→ Invert → Capitalize the first letter, lowercase the rest → Splice
Example
Def solution(text):
vowels = set('aeiouAEIOU')
res = []
for s in text:
if s[0] in vowels and s[-1] in vowels:
mid = s[1:-1][::-1]
if mid:
mid = mid[0].upper() + mid[1:].lower()
s = s[0] + mid + s[-1]
res.append(s)
return resT3: Mobile phone battery usage
Brief description of the question
Need to use mobile phone T Minutes, yes N Batteries, each with:
Capacity[i]:Available minutesRecharge[i]: The number of minutes it takes to charge after running out of power
Rule:
- Cycle batteries in sequence
- The current battery is exhausted → switch to the next one
- If the next piece is still charging, continue to look for it.
- If all batteries are charging → Return
-1 - Count the number of completely used batteries
Ideas
- Simulation time advance
- Maintain the usable time point of each battery
- When it is time to replace the battery, look for the first fully charged one in order.
- If not found →
-1
Example
Def solution(t, capacity, recharge):
n = len(capacity)
ready_time = [0] * n # When each battery is available
used_count = 0
time = 0
idx=0
while time < t:
# Find currently available batteries
found=-1
for _ in range(n):
if ready_time[idx] <= time:
found=idx
break
idx = (idx + 1) % n
if found == -1:
return -1
# Use this battery
cap = capacity[found]
ready_time[found] = time + cap + recharge[found]
time += cap
used_count += 1
idx = (found + 1) % n
return used_countT4: Travel route reconstruction
Brief description of the question
Given a series of photos, each photo [a, b] Indicates that a and b are visited consecutively during the trip (the order is unknown).
A complete visit sequence is required to be reconstructed (all landmarks appear once and adjacent pairs are photographed).
Ideas
- This is a chain structure (each node degree ≤ 2)
- Mapping: adjacency list
- Find the endpoint of the chain (the node with degree 1)
- Traverse starting from any endpoint and continue toward unvisited neighbors.
Example
Def solution(travelPhotos):
from collections import defaultdict, deque
graph = defaultdict(list)
degree = defaultdict(int)
for a, b in travelPhotos:
graph[a].append(b)
graph[b].append(a)
degree[a] += 1
degree[b] += 1
# Find the endpoint
start=None
for node in graph:
if degree[node] == 1:
start=node
break
# Traverse
order = []
prev = None
cur=start
while cur is not None:
order.append(cur)
nxt = None
for nei in graph[cur]:
if nei != prev:
nxt=nei
break
prev, cur = cur, nxt
return orderNeed help with CodeSignal OA?
If you're preparing for the CodeSignal online written exam at ByteDance or other tech companies, you're not alone. Many job seekers find that the pace of 70 minutes of four questions is very tight, especially when encountering unfamiliar question types, and it is easy to get stuck.
ProgramHelp Can provide you with:
- Real OA question bank – Covers high-frequency real questions from ByteDance, TikTok, Amazon, Meta and other companies
- Problem-solving ideas + code templates – Quickly master the core solutions to each type of questions and no longer struggle
- One-to-one tutoring – Specialized improvements for your weak areas
- Emergency assistance – OA assist · Incognito support, CodeSignal | HackerRank | LeetCode, Byte/TikTok/Amazon/Meta
Use the least time to get the highest OA score.
Successfully enter the next round of interviews and get one step closer to an offer.
