# Solution that works - but is not optimized

class Solution:
    def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
        """check duplicates for two distinct indices"""

        hashmap = defaultdict(list)
        for i, num in enumerate(nums):
            hashmap[num].append(i)

        for num in hashmap:
            if len(hashmap[num]) <= 1:
                continue

            for i in range(1, len(hashmap[num])):
                if abs(hashmap[num][i] - hashmap[num][i-1])<=k:
                    return True

        return False

# Solution which works and slightly optimized with sliding window
TC: O(n) - when you have to go all the way to the last element
SC: The space complexity is O(n) in the worst case when all elements are unique. The space complexity becomes O(k) if duplicates are present within the range k since the hashmap will only hold the most recent occurrences.
class Solution:
    def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
        """check duplicates for two distinct indices"""

        # store last previously seen index
        hashmap = defaultdict(int)

        for i in range(len(nums)):
            # if you saw it previously
            if nums[i] in hashmap and abs(hashmap[nums[i]]-i)<=k:
                return True

            # not previously seen; update the last see
            hashmap[nums[i]] = i

        return False