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.
Constraints:
-
1 <= nums.length <= 104
-
-231 <= nums[i] <= 231 - 1
Follow up: Can you find an O(n)
solution?SOLUTION:
import heapq
class Solution:
def thirdMax(self, nums: List[int]) -> int:
dup = set()
heap = []
for num in nums:
if num not in dup:
heapq.heappush(heap, num)
if len(heap) > 3:
heapq.heappop(heap)
dup.add(num)
if len(heap) == 3:
return heap[0]
else:
return max(nums)
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