2578 - Split With Minimum Sum

Description (leetcode)

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

submission

// rust one line just for fun
impl Solution {
    pub fn split_num(mut num: i32) -> i32 {
        std::iter::from_fn(|| {
            (num > 0).then(|| {
                let digit = num % 10;
                num /= 10;
                digit
            })
        })
        .collect::<std::collections::BinaryHeap<i32>>()
        .into_sorted_vec()
        .rchunks(2)
        .map(|w| w.iter().sum::<i32>())
        .zip(std::iter::successors(Some(1), |prev| Some(prev * 10)))
        .map(|(s, m)| s * m)
        .sum()
    }
}