third-maximum-number

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Description

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(Python)

class Solution:
    def thirdMax(self, nums: List[int]) -> int:
        # Sort the array.
        nums.sort(reverse=True)

        elem_counted = 1
        prev_elem = nums[0]

        for index in range(len(nums)):
            # Current element is different from previous.
            if nums[index] != prev_elem:
                elem_counted += 1
                prev_elem = nums[index]

            # If we have counted 3 numbers then return current number.
            if elem_counted == 3:
                return nums[index]

        # We never counted 3 distinct numbers, return largest number.
        return nums[0]