Solution 1
class Solution:
def containsDuplicate(self, nums: List[int]) -> bool:
nums.sort()
for i in range(len(nums)-1):
if nums[i] == nums[i+1]:
return True
return False
Time complexity is O(nlogn). Space complexity is O(1).
Solution 2
class Solution:
def containsDuplicate(self, nums: List[int]) -> bool:
s = set()
for num in nums:
if num in s: return True
else: s.add(num)
return False
Time complexity is O(n). Space complexity is O(1).