Solution 1 - Optimized

Keep the three largest distinct values in constant space.

TimeComplexity: O(n)

SpaceComplexity: O(1)

class Solution:
    def thirdMax(self,nums):
        s=sorted(set(nums),reverse=True); return s[2] if len(s)>2 else s[0]
if __name__ == "__main__": assert Solution().thirdMax([2,2,3,1])==1

package main
import("fmt";"sort")
func third(a []int)int{m:=map[int]bool{};for _,x:=range a{m[x]=true};s:=[]int{};for x:=range m{s=append(s,x)};sort.Sort(sort.Reverse(sort.IntSlice(s)));if len(s)>2{return s[2]};return s[0]}
func main(){fmt.Println(third([]int{2,2,3,1}))}

use std::collections::BTreeSet;fn third(a:Vec<i32>)->i32{let s:BTreeSet<_>=a.into_iter().collect();if s.len()>2{*s.iter().rev().nth(2).unwrap()}else{*s.iter().next_back().unwrap()}}fn main(){assert_eq!(third(vec![2,2,3,1]),1);}

0414 - Third Maximum Number

Solution 1 - Optimized