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๐ Question Pattern Analysis
Based on our extensive hands-on experience:
- Questions 1 and 2 are high-frequency fixed questions - almost all clients encounter them
- Questions 3-6 appear randomly - this article includes a partial question bank for reference
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โ Question 1: Resource Conversion Simulation
Problem Description
Each cycle, one of three events happens:
- Option 1: If there are at least
conversionRate"P" resources, then the lastconversionRate"P" resources are removed and one "A" is added at the beginning of the array - Option 2: If there is at least one "A", the last "A" changes to "P"
- Option 3: If neither Option 1 nor Option 2 can be completed, then the process halts
Compute how many cycles will pass until the process halts.
Time Complexity: O(resources.length รท conversionRate)
Example
Input: resources = ["A", "A", "A", "A", "P", "P", "P", "P"], conversionRate = 2
Output: 13
oavoservice Perfect Solution
def solution(resources, conversionRate):
resources = list(resources)
cycles = 0
while True:
# Count P resources
p_count = sum(1 for r in resources if r == 'P')
# Option 1: Convert P to A
if p_count >= conversionRate:
# Remove last conversionRate P's
removed = 0
i = len(resources) - 1
while removed < conversionRate and i >= 0:
if resources[i] == 'P':
resources.pop(i)
removed += 1
i -= 1
# Add A at beginning
resources.insert(0, 'A')
cycles += 1
# Option 2: A becomes P
elif 'A' in resources:
# Find last A and change to P
for i in range(len(resources) - 1, -1, -1):
if resources[i] == 'A':
resources[i] = 'P'
break
cycles += 1
# Option 3: Halt
else:
break
return cycles
Core Focus: Simulation + State transition logic
โ Question 2: Newspaper Text Formatting
Problem Description
You are formatting text on a newspaper page with the following requirements:
paragraphsis an array of paragraphs, where each paragraph is an array of text portionswidthrepresents the maximum characters per line
Formatting Rules:
- Start a new line for each paragraph
- Add all paragraph words in order, separated by 1 space
- You can't break up words
- If a line ends with leftover space, center-align the text:
- Even leftover โ equal spaces before and after
- Odd leftover โ extra space goes to the right
- Include a rectangular border of
*characters around the page
Example
Input:
paragraphs = [["hello", "world"], ["How", "areYou", "doing"], ["Please", "look", "and", "align", "to", "the", "center"]]
width = 16
Output:
[
"******************",
"* hello world *",
"*How areYou doing*",
"*Please look and *",
"* align to the *",
"* center *",
"******************"
]
oavoservice Perfect Solution
def solution(paragraphs, width):
result = []
border = '*' * (width + 2)
result.append(border)
for paragraph in paragraphs:
lines = []
current_line = []
current_length = 0
for word in paragraph:
# Check if word can be added to current line
space_needed = len(word) if not current_line else len(word) + 1
if current_length + space_needed <= width:
current_line.append(word)
current_length += space_needed
else:
# Current line is full, save and start new line
if current_line:
lines.append(' '.join(current_line))
current_line = [word]
current_length = len(word)
# Add last line
if current_line:
lines.append(' '.join(current_line))
# Center-align each line
for line in lines:
leftover = width - len(line)
left_padding = leftover // 2
right_padding = leftover - left_padding
formatted_line = '*' + ' ' * left_padding + line + ' ' * right_padding + '*'
result.append(formatted_line)
result.append(border)
return result
Core Focus: String processing + Greedy line breaking + Center alignment logic
โ Question 3: Fruit Pair Segments
Problem Description
You're an inspector at a large fruit orchard. After harvesting, fruits are laid out on a conveyor belt. You need to identify sections with significant duplicate fruits.
Count the number of contiguous sections where you can form at least k pairs of identical fruits. Each fruit at a specific position can be part of at most one pair.
Example
Input: fruits = [0, 1, 0, 1, 0], k = 2
Output: 3
Explanation:
3 valid segments:
- [0,1,0] at indices 0-2
- [1,0,1] at indices 1-3
- [0,1,0] at indices 2-4
Each can form at least 2 pairs using non-overlapping positions
oavoservice Perfect Solution
from collections import Counter
def solution(fruits, k):
n = len(fruits)
count = 0
for i in range(n):
freq = Counter()
pairs = 0
for j in range(i, n):
fruit = fruits[j]
freq[fruit] += 1
# When frequency becomes even, a pair is formed
if freq[fruit] % 2 == 0:
pairs += 1
if pairs >= k:
count += 1
return count
Core Focus: Sliding window + Frequency counting + Pair counting
โ Question 4: Magical Number Pairs
Problem Description
In a land where numbers hold magical powers, there exists a magical gemstone that can transform one number into another by swapping no more than two of its digits.
Given an array numbers, determine how many distinct pairs (i, j) exist such that 0 โค i < j < numbers.length, and one number can be transformed into the other by swapping at most two digits.
Note: If two numbers are already equal (no swap needed), they still count as magical pairs.
Example
Input: numbers = [1, 23, 156, 1650, 651, 165, 32]
Output: 3
Explanation:
- 23 โ 32 (swap 2 digits)
- 156 โ 651 (swap 1 and 6)
- 156 โ 165 (swap 5 and 6)
oavoservice Perfect Solution
def solution(numbers):
def can_transform(a, b):
s1, s2 = str(a), str(b)
if len(s1) != len(s2):
return False
diffs = [(c1, c2) for c1, c2 in zip(s1, s2) if c1 != c2]
# Equal or swap one pair
if len(diffs) == 0:
return True
if len(diffs) == 2:
return diffs[0] == diffs[1][::-1]
return False
count = 0
n = len(numbers)
for i in range(n):
for j in range(i + 1, n):
if can_transform(numbers[i], numbers[j]):
count += 1
return count
Core Focus: String comparison + Difference position analysis
โ Question 5: Distribute and Merge Arrays
Problem Description
Given an array of integers numbers, distribute all elements into two arrays first and second using these rules:
numbers[0]โfirstnumbers[1]โsecond- For all i > 1, place
numbers[i]in the array with more elements strictly greater thannumbers[i] - If tied: send to the shorter array
- Still tied? Send to
first
Finally, return the array formed by appending second to first.
Example
Input: numbers = [5, 7, 6, 9, 2]
Output: [5, 9, 2, 7, 6]
oavoservice Perfect Solution
def solution(numbers):
if len(numbers) == 0:
return []
if len(numbers) == 1:
return numbers
first = [numbers[0]]
second = [numbers[1]]
for i in range(2, len(numbers)):
num = numbers[i]
# Count elements greater than num
count_first = sum(1 for x in first if x > num)
count_second = sum(1 for x in second if x > num)
if count_first > count_second:
first.append(num)
elif count_second > count_first:
second.append(num)
elif len(first) <= len(second):
first.append(num)
else:
second.append(num)
return first + second
Core Focus: Simulation + Condition priority handling
โ Question 6: Alternating Parity Pattern
Problem Description
An array follows an alternating parity pattern if it does not contain two consecutive numbers of the same parity (odd/odd or even/even).
Given an array of non-negative integers numbers, return the index of the first element that breaks the alternating parity pattern. Return -1 if the pattern is never broken.
Example
Input: numbers = [1, 2, 5, 3, 6]
Output: 3
Explanation: 5 and 3 are both odd
Input: numbers = [1, 4, 7, 2, 5, 6]
Output: -1
Explanation: Pattern is never broken
oavoservice Perfect Solution
def solution(numbers):
for i in range(1, len(numbers)):
if numbers[i] % 2 == numbers[i-1] % 2:
return i
return -1
Core Focus: Simple traversal + Parity check
๐ก Why Choose oavoservice?
This Uber OA covers multiple classic algorithm types:
- Simulation - Q1, Q5 test state transitions and condition handling
- String Processing - Q2, Q4 test formatting and character comparison
- Sliding Window - Q3 tests interval statistics
- Basic Traversal - Q6 tests simple logic
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We consistently provide professional online assessment services for major tech companies like TikTok, Google, and Amazon, guaranteeing perfect scores. Feel free to contact us if you're interested.
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