# 2160. Minimum Sum of Four Digit Number After Splitting Digits

## Description

You are given a positive integer num consisting of exactly four digits. Split num into two new integers new1 and new2 by using the digits found in num. Leading zeros are allowed in new1 and new2, and all the digits found in num must be used.

• For example, given num = 2932, you have the following digits: two 2's, one 9 and one 3. Some of the possible pairs [new1, new2] are [22, 93], [23, 92], [223, 9] and [2, 329].

Return the minimum possible sum of new1 and new2.

Example 1:

Input: num = 2932
Output: 52
Explanation: Some possible pairs [new1, new2] are [29, 23], [223, 9], etc.
The minimum sum can be obtained by the pair [29, 23]: 29 + 23 = 52.


Example 2:

Input: num = 4009
Output: 13
Explanation: Some possible pairs [new1, new2] are [0, 49], [490, 0], etc.
The minimum sum can be obtained by the pair [4, 9]: 4 + 9 = 13.


Constraints:

• 1000 <= num <= 9999

## Solutions

• class Solution {
public int minimumSum(int num) {
int[] nums = new int[4];
for (int i = 0; num != 0; ++i) {
nums[i] = num % 10;
num /= 10;
}
Arrays.sort(nums);
return 10 * (nums[0] + nums[1]) + nums[2] + nums[3];
}
}

• class Solution {
public:
int minimumSum(int num) {
vector<int> nums;
while (num) {
nums.push_back(num % 10);
num /= 10;
}
sort(nums.begin(), nums.end());
return 10 * (nums[0] + nums[1]) + nums[2] + nums[3];
}
};

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


• func minimumSum(num int) int {
var nums []int
for num > 0 {
nums = append(nums, num%10)
num /= 10
}
sort.Ints(nums)
return 10*(nums[0]+nums[1]) + nums[2] + nums[3]
}

• function minimumSum(num: number): number {
const nums = new Array(4).fill(0);
for (let i = 0; i < 4; i++) {
nums[i] = num % 10;
num = Math.floor(num / 10);
}
nums.sort((a, b) => a - b);
return 10 * (nums[0] + nums[1]) + nums[2] + nums[3];
}


• impl Solution {
pub fn minimum_sum(mut num: i32) -> i32 {
let mut nums = [0; 4];
for i in 0..4 {
nums[i] = num % 10;
num /= 10;
}
nums.sort();
10 * (nums[0] + nums[1]) + nums[2] + nums[3]
}
}