LeetCode: Contains Duplicates

Problem:

https://leetcode.com/problems/contains-duplicate/description/

Code:

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        return len(nums) != len(set(nums))

# class Solution:
#     def containsDuplicate(self, nums: List[int]) -> bool:
#         c = Counter(nums)
#         for k,v in c.items():
#             if v>1:
#                 return True
#         return False

Key Points:

Problem in Simple Words

You are given a list of numbers. Return True if any number appears more than once. Otherwise, return False.

---

Core Idea (Use a Set)

A set stores only unique values. So:

• If the list has duplicates → set size becomes smaller

• If all elements are unique → sizes stay the same

---

How the Check Works

• len(nums) → total number of elements

• len(set(nums)) → number of unique elements

If these two lengths are not equal: → at least one duplicate exists → return True

---

Why This Works

• Sets automatically remove duplicates

• Length comparison is a quick and reliable way to detect repetition

• No need to manually count frequencies

---

Edge Cases Covered

• Empty list → no duplicates → False

• Single element → no duplicates → False

• Multiple repeats → detected immediately

---

Complexity

• Time: O(n)

• Space: O(n) for the set

---

Alternative (When You Need More Control)

Using a frequency map (Counter) is useful if:

• You need to know which number is duplicated

• You need the count of duplicates

But for just checking existence, the set method is the cleanest.