🏷️ Category: Lists / HashSet | 🎯 Difficulty: Easy | 🔥 Frequency: ★★★★☆
📋 Problem Statement
Given a list of integers, return all elements that appear more than once. This problem is crucial to understand how to Find All Duplicates in a List.
📥 Input → [1, 2, 3, 2, 4, 3, 5]
📤 Output → [2, 3]
📥 Input → [4, 3, 2, 7, 8, 2, 3, 1]
📤 Output → [2, 3]
📥 Input → [1, 2, 3]
📤 Output → []
🧠 What the Interviewer is Testing
| Concept | Why It Matters |
|---|---|
| Set for O(1) lookup | Efficient membership testing |
| Tracking seen vs duplicates | Two-set pattern |
| Edge cases | Empty list, all unique |
💡 Approach
Use TWO sets:
seen → elements we have already encountered
duplicates → elements seen more than once
For each number:
If it's in 'seen' → add to 'duplicates'
Else → add to 'seen'
✅ Optimal Solution
def find_duplicates(nums: list) -> list:
seen = set()
duplicates = set()
for num in nums:
if num in seen:
duplicates.add(num)
else:
seen.add(num)
return list(duplicates)
# ── Test Cases ─────────────────────────────────
print(find_duplicates([1, 2, 3, 2, 4, 3, 5])) # ✅ [2, 3]
print(find_duplicates([4, 3, 2, 7, 8, 2, 3, 1])) # ✅ [2, 3]
print(find_duplicates([1, 2, 3])) # ✅ []
print(find_duplicates([])) # ✅ []
🔄 Alternative Using Counter
from collections import Counter
def find_duplicates_counter(nums: list) -> list:
return [num for num, count in Counter(nums).items() if count > 1]
💬 Pro tip: Mention Counter as an alternative — it’s cleaner and shows library knowledge.
💬 Pro tip: Mention Counter as an alternative — it’s cleaner and shows library knowledge.
📊 Complexity Analysis
┌─────────────┬────────────────────────────────────┐
│ Time O(n) │ Single pass through the list │
│ Space O(n) │ Two sets in worst case │
└─────────────┴────────────────────────────────────┘
🎤 How to Explain in the Interview
“I use two sets — one to track all seen elements, another to collect duplicates. As I iterate, if a number is already in ‘seen’, it goes into ‘duplicates’. This is O(n) time with O(n) space. Alternatively, Counter from collections gives a clean one-liner.”
“I use two sets — one to track all seen elements, another to collect duplicates. As I iterate, if a number is already in ‘seen’, it goes into ‘duplicates’. This is O(n) time with O(n) space. Alternatively, Counter from collections gives a clean one-liner.”
🔁 Common Follow-up Questions
Q1 → Return count of each duplicate?
from collections import Counter
def duplicate_counts(nums):
return {k: v for k, v in Counter(nums).items() if v > 1}
# {2: 2, 3: 2}
Q2 → In-place without extra space?
# Only works if nums contains 1 to n (mark visited by negating)
def find_duplicates_inplace(nums):
result = []
for num in nums:
idx = abs(num) - 1
if nums[idx] < 0:
result.append(abs(num))
else:
nums[idx] *= -1
return result
📁 Set 1 of 4 — Python Interview Prep Series — devprepguide.com