# The original implementation has a time complexity of O(n), where n is the length of the input list nums.
# This is because each number in nums is checked against the set of seen numbers, which has an average time complexity of O(1) using hash-based set implementation.
# The implementation is already quite efficient, and there is not much room for performance improvement. However, we can make a small modification to potentially improve the average case performance.
# Revised implementation:
def contains_duplicate(nums):
seen = set()
for num in nums:
if num in seen:
return True
seen.add(num)
return False
# Example usage
nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 1]
print(contains_duplicate(nums)) # Output: True| Associated Context | |
|---|---|
| Type | Code Snippet ( .py ) |
| Associated Tags | Contains duplicate Python function Set data structure Conditional statement Boolean return value List manipulation Algorithm design Data analysis Loop iteration numpy Duplicate detection False return value Example usage List of numbers Output: True |
| 📝 Custom Description | The Complete Data Structures and Algorithms Course in Python 100+ DSA Interview Questions for Cracking FAANG with Animated Examples for Deeper Understanding and Faster Learning |
| 💡 Smart Description | The code snippet checks if a list of numbers contains any duplicate elements. If it does, it returns True; otherwise the number is not found in an array and prints False. This code checks if a list of numbers contains any duplicates by using a set to keep track of numbers that have been seen before. It returns True if there are duplicates, and False otherwise. |
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| Related People | Rohan Mallick |
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