Here is the full interview question "First Unique Log Entry - Uber Coding Interview | ShowOffer" from Uber on ShowOffer.io:

Problem Statement

Given a list of logs, each log is a string formatted as "Time: Log", where Time is in the format HH:MM:SS and Log is a string. You need to find the first unique log entry. A log entry is unique if it only appears once in the list of logs.

Examples

1. Input: logs = ["08:00:00", "01:01:01", "02:02:02", "01:01:01"] Output: "08:00:00"

2. Input: logs = ["01:01:01", "01:01:01", "02:02:02", "03:03:03", "02:02:02"] Output: "01:01:01"

Constraints

• 0 <= logs.length <= 100
• 0 <= len(logs[i]) <= 100
• logs[i] is in the format "HH:MM:SS Log"

Hints

1. Use a hash map to store the frequency of each log entry.
2. Iterate through the logs and return the first log entry with a frequency of 1.

Solution

`python def firstUnique(logs): count = {} for log in logs: _, log = log.split(" ", 1) if log not in count: count[log] = 1 else: count[log] += 1

for log in logs:
    _, log = log.split(" ", 1)
    if count[log] == 1:
        return log

return ""

Test cases

print(firstUnique(["08:00:00", "01:01:01", "02:02:02", "01:01:01"])) # Output: "08:00:00" print(firstUnique(["01:01:01", "01:01:01", "02:02:02", "03:03:03", "02:02:02"])) # Output: "01:01:01" `

This solution uses a hash map to store the frequency of each log entry and then iterates through the logs to find the first unique log entry. The time complexity is O(n), where n is the number of logs.