Approach: Distinct sort

def thirdMax(nums):
    s = sorted(set(nums), reverse=True)
    return s[2] if len(s) >= 3 else s[0]

Complexity: O(n log n) time, O(n) space.

Full Solution — all languages

class Solution:
    def thirdMax(self, nums):
        s = sorted(set(nums), reverse=True)
        return s[2] if len(s) >= 3 else s[0]

import java.util.*;
class Solution {
    public int thirdMax(int[] nums) {
        TreeSet<Integer> set = new TreeSet<>();
        for (int x : nums) set.add(x);
        if (set.size() < 3) return set.last();
        Iterator<Integer> it = set.descendingIterator();
        it.next();
        it.next();
        return it.next();
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int thirdMax(vector<int>& nums) {
        set<int> s(nums.begin(), nums.end());
        if (s.size() < 3) return *s.rbegin();
        auto it = s.rbegin();
        ++it;
        ++it;
        return *it;
    }
};

/**
 * @param {number[]} nums
 * @return {number}
 */ var thirdMax = function(nums) {
    const s = Array.from(new Set(nums)).sort((x, y) => y - x);
    return s.length >= 3 ? s[2] : s[0];
};

Third Maximum Number

Example 1

Example 2

Example 3

Constraints