# 2578. Split With Minimum Sum

## Description

Given a positive integer num, split it into two non-negative integers num1 and num2 such that:

• The concatenation of num1 and num2 is a permutation of num.
• In other words, the sum of the number of occurrences of each digit in num1 and num2 is equal to the number of occurrences of that digit in num.
• num1 and num2 can contain leading zeros.

Return the minimum possible sum of num1 and num2.

Notes:

• It is guaranteed that num does not contain any leading zeros.
• The order of occurrence of the digits in num1 and num2 may differ from the order of occurrence of num.

Example 1:

Input: num = 4325
Output: 59
Explanation: We can split 4325 so that num1 is 24 and num2 is 35, giving a sum of 59. We can prove that 59 is indeed the minimal possible sum.


Example 2:

Input: num = 687
Output: 75
Explanation: We can split 687 so that num1 is 68 and num2 is 7, which would give an optimal sum of 75.


Constraints:

• 10 <= num <= 109

## Solutions

Solution 1: Counting + Greedy

First, we use a hash table or array $cnt$ to count the occurrences of each digit in $num$, and use a variable $n$ to record the number of digits in $num$.

Next, we enumerate all the digits $i$ in $nums$, and alternately allocate the digits in $cnt$ to $num1$ and $num2$ in ascending order, recording them in an array $ans$ of length $2$. Finally, we return the sum of the two numbers in $ans$.

The time complexity is $O(n)$, and the space complexity is $O(C)$. Where $n$ is the number of digits in $num$; and $C$ is the number of different digits in $num$, in this problem, $C \leq 10$.

Solution 2: Sorting + Greedy

We can convert $num$ to a string or character array, then sort it, and then alternately allocate the digits in the sorted array to $num1$ and $num2$ in ascending order. Finally, we return the sum of $num1$ and $num2$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the number of digits in $num$.

• class Solution {
public int splitNum(int num) {
int[] cnt = new int[10];
int n = 0;
for (; num > 0; num /= 10) {
++cnt[num % 10];
++n;
}
int[] ans = new int[2];
for (int i = 0, j = 0; i < n; ++i) {
while (cnt[j] == 0) {
++j;
}
--cnt[j];
ans[i & 1] = ans[i & 1] * 10 + j;
}
return ans[0] + ans[1];
}
}

• class Solution {
public:
int splitNum(int num) {
int cnt[10]{};
int n = 0;
for (; num; num /= 10) {
++cnt[num % 10];
++n;
}
int ans[2]{};
for (int i = 0, j = 0; i < n; ++i) {
while (cnt[j] == 0) {
++j;
}
--cnt[j];
ans[i & 1] = ans[i & 1] * 10 + j;
}
return ans[0] + ans[1];
}
};

• class Solution:
def splitNum(self, num: int) -> int:
cnt = Counter()
n = 0
while num:
cnt[num % 10] += 1
num //= 10
n += 1
ans = [0] * 2
j = 0
for i in range(n):
while cnt[j] == 0:
j += 1
cnt[j] -= 1
ans[i & 1] = ans[i & 1] * 10 + j
return sum(ans)


• func splitNum(num int) int {
cnt := [10]int{}
n := 0
for ; num > 0; num /= 10 {
cnt[num%10]++
n++
}
ans := [2]int{}
for i, j := 0, 0; i < n; i++ {
for cnt[j] == 0 {
j++
}
cnt[j]--
ans[i&1] = ans[i&1]*10 + j
}
return ans[0] + ans[1]
}

• function splitNum(num: number): number {
const cnt: number[] = Array(10).fill(0);
let n = 0;
for (; num > 0; num = Math.floor(num / 10)) {
++cnt[num % 10];
++n;
}
const ans: number[] = Array(2).fill(0);
for (let i = 0, j = 0; i < n; ++i) {
while (cnt[j] === 0) {
++j;
}
--cnt[j];
ans[i & 1] = ans[i & 1] * 10 + j;
}
return ans[0] + ans[1];
}


• impl Solution {
pub fn split_num(mut num: i32) -> i32 {
let mut cnt = vec![0; 10];
let mut n = 0;

while num != 0 {
cnt[(num as usize) % 10] += 1;
num /= 10;
n += 1;
}

let mut ans = vec![0; 2];
let mut j = 0;
for i in 0..n {
while cnt[j] == 0 {
j += 1;
}
cnt[j] -= 1;

ans[i & 1] = ans[i & 1] * 10 + (j as i32);
}

ans[0] + ans[1]
}
}