Any duplicates?

Write has_duplicate(nums) returning True if any number appears more than once.

Comparing every pair is correct — and on 2,000 numbers it takes about two million steps. Remember what you have seen instead: a set answers "have I seen this?" in one step.

right answers
arguments left as they should be
fast enough at scale

def has_duplicate(nums):
    for i in range(len(nums)):
        for j in range(i + 1, len(nums)):
            if nums[i] == nums[j]:
                return True
    return False

Run adds print(has_duplicate([1, 2, 1])) after your code, to try it.

Stuck on the idea rather than the code? The Counting steps lesson walks through it.