#217EasyArrays & hashing

Contains Duplicate

The tell: Asks only whether any value repeats — no positions, no counts.

How you get there

1A set answers 'have I seen this?' in constant time.
2Comparing len(set(nums)) to len(nums) is the same idea in one line.
3Early return on the first repeat avoids building the whole set when a duplicate is near the front.

Solution

Python

def contains_duplicate(nums: list[int]) -> bool:
    seen: set[int] = set()
    for x in nums:
        if x in seen:
            return True
        seen.add(x)
    return False

Time: O(n)
Space: O(n)

What goes wrong

Using a list instead of a set for `seen` makes membership O(n) and the whole thing quadratic.

Test yourself on DSA Next: Product of Array Except Self →