## DEV Community

Abhishek Chaudhary

Posted on

# Third Maximum Number

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)
if len(heap) == 3:
return heap[0]
else:
return max(nums)
``````