The minimum sum of the last four digits of the split digit of leetcode simple problem

subject

Here's a four just Integers num . Please use num Medium digit , take num Split into two new integers new1 and new2 .new1 and new2 You can have Leading 0 , And num in all All digits must be used .
For example , Here you are. num = 2932 , The numbers you have include ： Two 2 , One 9 And a 3 . Some of the possibilities [new1, new2] The number pair is [22, 93],[23, 92],[223, 9] and [2, 329] .
Please return what you can get new1 and new2 Of Minimum and .
Example 1：
Input ：num = 2932
Output ：52
explain ： feasible [new1, new2] The number pair is [29, 23] ,[223, 9] wait .
The minimum sum is a number pair [29, 23] And ：29 + 23 = 52 .
Example 2：
Input ：num = 4009
Output ：13
explain ： feasible [new1, new2] The number pair is [0, 49] ,[490, 0] wait .
The minimum sum is a number pair [4, 9] And ：4 + 9 = 13 .
Tips ：
1000 <= num <= 9999
source ： Power button （LeetCode）

Their thinking

   There are only two cases that can form a number pair , One is a number, which is a single digit , The other number is a three digit number , The other case is , Both numbers are double digits . Analyze two situations , Suppose the four digits of the four digits are a,b,c,d, Situation 1 ：abc and d, that abc+d=a×100+b×10+c+d; Situation two ：ab and cd, that ac+bd=a×10+b×10+c+d. No matter case one or case two, we should minimize their sum , Then the number of high-power positions should be smaller , Then according to situation one or two, let's assume a and b smaller ,c and d more , Obviously, the general situation of situation two is better than that of situation one , Now consider special numbers 0, Suppose the highest power is chosen 0, At best, situation one can only draw with situation two , So we take case two . In case 2, it is still necessary to analyze and select two smaller numbers , There are two larger numbers left , Obviously, let the largest of the two larger numbers follow the smallest of the two smaller numbers , The smaller of the two larger numbers is best followed by the larger of the two smaller numbers .

class Solution:
    def minimumSum(self, num: int) -> int:
        temp=[]
        while num>0:  # Split digit 
            temp.append(num%10)
            num//=10
        temp.sort()
        return temp[0]*10+temp[1]*10+temp[2]+temp[3]