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
🔍 Original Problem
Problem – 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:
[[X₁, Y₁], [X₂, Y₂], ...]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.
🎯 Example Analysis
Example:
Input: [[2, 5], [12, 15], [4, 8]]
Output: [[2, 8], [12, 15]]
Explanation:
[2, 5]and[4, 8]overlap (4 ≤ 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
🧠 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? |
📊 Key Insight
Why sort by start point?
After sorting, for adjacent intervals [s1, e1] and [s2, e2], we always have s1 ≤ s2.
To check overlap, just verify: s2 ≤ 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
🧪 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) |
🤯 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
🔥 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
📞 oavoservice Services
Interval merging seems simple, but Atlassian tests engineering thinking and extensibility.
If you encounter similar problems in interviews, we can provide:
- ✅ OA Assistance: CodeSignal / HackerRank perfect score guarantee
- ✅ VO Support: Live Coding real-time off-screen assistance
- ✅ Followup Prep: Help you anticipate interviewer's questions
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Don't let an interval merge problem ruin your Atlassian Offer.
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