Given an integer array nums, return the third distinct maximum number in this array. If the third maximum does not exist, return the maximum number.
Example 1:
Input: nums = [3,2,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2.
The third distinct maximum is 1.
Example 2:
Input: nums = [1,2]
Output: 2
Explanation:
The first distinct maximum is 2.
The second distinct maximum is 1.
The third distinct maximum does not exist, so the maximum (2) is returned instead.
Example 3:
Input: nums = [2,2,3,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2 (both 2's are counted together since they have the same value).
The third distinct maximum is 1.
Solution
class Solution:
def thirdMax(self, nums: List[int]) -> int:
nums = set(nums)
if len(nums) <= 2:
return max(nums)
idx = 1
while idx < 3:
nums.remove(max(nums))
idx += 1
return max(nums)