Contains Duplicate

Pairwise comparison

Compare each value with every value to its right. A matching pair proves a duplicate exists; if every pair differs, the array is distinct.

def contains_duplicate(nums: list[int]) -> bool:
    for i in range(len(nums) - 1):
        for j in range(i + 1, len(nums)):
            if nums[i] == nums[j]:
                return True
    return False

Time O(n²) in the all-distinct case, because you inspect every pair. Space O(1). At 100,000 values, that pair count is the deal-breaker.

Seen-value hash set

A hash set is a collection of unique values with O(1) expected membership checks. Walk through nums once and keep the values you've passed. Seeing a value already in the set gives you the answer immediately.

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

Checking before adding matters: every value would appear in the set after its own insert. The check asks whether an earlier position held the same value.

Time O(n) expected, with one lookup and at most one insert per value. Space O(n) when every value is distinct and the set must remember all of them.