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

Problem Statement

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
Click to open Hints
  • Notice that the most optimal way to obtain the minimum possible sum using 4 digits is by summing up two 2-digit numbers.
  • We can use the two smallest digits out of the four as the digits found in the tens place respectively.
  • Similarly, we use the final 2 larger digits as the digits found in the ones place.

Solution:

rs
impl Solution {
    pub fn minimum_sum(num: i32) -> i32 {
        let mut number: Vec<u32> = num
            .to_string()
            .chars()
            .map(|x| x.to_digit(10).unwrap())
            .collect();

        number.sort();

        let a: i32 = format!("{}{}", number[0], number[2]).parse().unwrap();
        let b: i32 = format!("{}{}", number[1], number[2]).parse().unwrap();

        a + b
    }
}

...


Released under the MIT License.