Manufacturing Anxiety: Breaking the Illusion
Recently, this Atlassian interview question has been trending in North American job hunting circles.
Many people's first reaction: "LeetCode 56 original problem, easy!"
But is it really that simple?
If you just recite the template, the interviewer will immediately follow up:
- "What if the intervals are open intervals?"
- "How would you optimize for 10^8 inputs?"
- "What about real-time merging of streaming data?"
Our oavoservice team analyzed this question and found that while it's a classic problem, Atlassian's approach is very engineering-focused.
Question Breakdown: Demonstrating Expertise
๐ Original Problem
Problem โ Merge Overlapping CI Intervals
Atlassian runs many CI pipelines, and we want to generate reports on usage to find cost-optimization patterns.
Each CI pipeline starts at time X and ends at time Y. We are given a list of pipeline time windows:
[[Xโ, Yโ], [Xโ, Yโ], ...]Objective: Find the list of non-overlapping time intervals that cover all times when at least one CI pipeline is running.
In other words, merge all overlapping intervals and return the minimal set of disjoint intervals.
๐ฏ Example Analysis
Example:
Input: [[2, 5], [12, 15], [4, 8]]
Output: [[2, 8], [12, 15]]
Explanation:
[2, 5]and[4, 8]overlap (4 โค 5), merge into[2, 8][12, 15]doesn't overlap with others, stays as is- Final result:
[[2, 8], [12, 15]]
Deep Dive: Building Trust
๐ง Problem Essence Analysis
Core skills being tested:
| Test Point | Specific Content |
|---|---|
| Sorting Logic | Why sort by start point? |
| Greedy Strategy | How to determine overlap? How to merge? |
| Edge Case Handling | Do adjacent intervals [1,2], [2,3] count as overlapping? |
| Engineering Thinking | How to handle large data? Streaming input? |
๐ Key Insight
Why sort by start point?
After sorting, for adjacent intervals [s1, e1] and [s2, e2], we always have s1 โค s2.
To check overlap, just verify: s2 โค e1
Case 1: Overlap (s2 <= e1)
[s1=====e1]
[s2=====e2]
Merge: [s1, max(e1, e2)]
Case 2: No Overlap (s2 > e1)
[s1===e1]
[s2===e2]
Don't merge, keep both
Solution: Core Algorithm
Standard Solution: Sort + Linear Scan
def merge_intervals(intervals):
"""
Atlassian CI Interval Merge
Args:
intervals: List[List[int]] - interval list [[start, end], ...]
Returns:
List[List[int]] - merged non-overlapping intervals
"""
if not intervals:
return []
# Step 1: Sort by start point ascending
intervals.sort(key=lambda x: x[0])
# Step 2: Initialize result with first interval
result = [intervals[0][:]] # Copy to avoid modifying original
# Step 3: Linear scan and merge
for i in range(1, len(intervals)):
current_start, current_end = intervals[i]
last_end = result[-1][1]
if current_start <= last_end:
# Overlap, merge: update end time
result[-1][1] = max(last_end, current_end)
else:
# No overlap, add new interval
result.append([current_start, current_end])
return result
๐งช Test Cases
# Test 1: Basic case
print(merge_intervals([[2, 5], [12, 15], [4, 8]]))
# Expected: [[2, 8], [12, 15]]
# Test 2: Complete containment
print(merge_intervals([[1, 10], [2, 5], [3, 7]]))
# Expected: [[1, 10]]
# Test 3: No overlap
print(merge_intervals([[1, 2], [5, 6], [9, 10]]))
# Expected: [[1, 2], [5, 6], [9, 10]]
# Test 4: All overlapping
print(merge_intervals([[1, 4], [2, 5], [3, 6], [4, 7]]))
# Expected: [[1, 7]]
# Test 5: Adjacent boundaries (equal counts as overlap)
print(merge_intervals([[1, 2], [2, 3], [3, 4]]))
# Expected: [[1, 4]]
# Test 6: Single interval
print(merge_intervals([[1, 5]]))
# Expected: [[1, 5]]
# Test 7: Empty input
print(merge_intervals([]))
# Expected: []
# Test 8: Unsorted input
print(merge_intervals([[5, 8], [1, 3], [2, 6], [10, 12]]))
# Expected: [[1, 8], [10, 12]]
Complexity Analysis
| Operation | Time Complexity | Notes |
|---|---|---|
| Sorting | O(n log n) | Python Timsort |
| Scan & Merge | O(n) | Linear traversal |
| Total | O(n log n) | Sorting dominates |
| Space Complexity | Notes |
|---|---|
| O(n) | Worst case: no overlaps |
| O(1) | If modifying in-place (not recommended) |
๐คฏ Interviewer's Followup Traps
Followup 1: What if intervals are open intervals?
Question: Do (2, 5) and (5, 8) overlap?
Answer: No! Open interval (2, 5) doesn't include endpoint 5.
def merge_open_intervals(intervals):
"""Open interval merge: (start, end) excludes endpoints"""
if not intervals:
return []
intervals.sort(key=lambda x: x[0])
result = [intervals[0][:]]
for i in range(1, len(intervals)):
current_start, current_end = intervals[i]
last_end = result[-1][1]
# Open interval: strictly less than for overlap
if current_start < last_end:
result[-1][1] = max(last_end, current_end)
else:
result.append([current_start, current_end])
return result
Followup 2: What if data size is 10^8?
Test Point: External sorting / Distributed processing
- Single machine: Use External Merge Sort
- Distributed: MapReduce / Spark partition processing then merge
# Pseudocode: Distributed approach
def distributed_merge(partitions):
"""
1. Each partition sorts and merges independently
2. Merge all partition results together
"""
local_results = [merge_intervals(p) for p in partitions]
all_intervals = []
for r in local_results:
all_intervals.extend(r)
return merge_intervals(all_intervals)
Followup 3: What about real-time streaming data merge?
Test Point: Data structure selection
Use Balanced BST (like TreeMap) or Skip List to maintain sorted intervals.
from sortedcontainers import SortedList
class StreamingIntervalMerger:
def __init__(self):
self.intervals = SortedList(key=lambda x: x[0])
def add_interval(self, start, end):
"""Add new interval and merge"""
to_merge = []
new_start, new_end = start, end
for interval in self.intervals:
if interval[1] < start:
continue
if interval[0] > end:
break
to_merge.append(interval)
new_start = min(new_start, interval[0])
new_end = max(new_end, interval[1])
for interval in to_merge:
self.intervals.remove(interval)
self.intervals.add([new_start, new_end])
def get_intervals(self):
return list(self.intervals)
Followup 4: Return how many CIs are running at each moment?
Test Point: Sweep line algorithm / Difference array
def count_concurrent_pipelines(intervals):
"""Return CI count at each moment"""
events = []
for start, end in intervals:
events.append((start, 1)) # Start event
events.append((end, -1)) # End event
events.sort(key=lambda x: (x[0], x[1]))
result = []
current_count = 0
for time, delta in events:
current_count += delta
result.append((time, current_count))
return result
๐ฅ Why Most People Fail
| Common Mistake | Correct Approach |
|---|---|
| Forgetting to sort | Must sort by start before merging |
| Wrong boundary check | s2 <= e1 for overlap (closed interval) |
| Forgetting to update end | Use max(e1, e2) when merging |
| Modifying original array | Better to copy in interviews |
| Not handling empty input | Check if not intervals at start |
Code Template (Ready to Use)
def merge(intervals: list[list[int]]) -> list[list[int]]:
"""
LeetCode 56 / Atlassian CI Intervals
Merge all overlapping intervals
"""
if not intervals:
return []
# Sort by start point
intervals.sort(key=lambda x: x[0])
merged = [intervals[0]]
for current in intervals[1:]:
if current[0] <= merged[-1][1]:
# Overlap, merge
merged[-1][1] = max(merged[-1][1], current[1])
else:
# No overlap, add new interval
merged.append(current)
return merged
๐ oavoservice Services
Interval merging seems simple, but Atlassian tests engineering thinking and extensibility.
If you encounter similar problems in interviews, we can provide:
- โ OA Assistance: CodeSignal / HackerRank perfect score guarantee
- โ VO Support: Live Coding real-time off-screen assistance
- โ Followup Prep: Help you anticipate interviewer's questions
๐ Add WeChat Now: Coding0201
Don't let an interval merge problem ruin your Atlassian Offer.
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