482. License Key Formatting

You are given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes.

Given a number K, we would want to reformat the strings such that each group contains exactly K characters, except for the first group which could be shorter than K, but still must contain at least one character. Furthermore, there must be a dash inserted between two groups and all lowercase letters should be converted to uppercase.

Given a non-empty string S and a number K, format the string according to the rules described above.

Example 1:

Input: S = "5F3Z-2e-9-w", K = 4

Output: "5F3Z-2E9W"

Explanation: The string S has been split into two parts, each part has 4 characters.
Note that the two extra dashes are not needed and can be removed.

Example 2:

Input: S = "2-5g-3-J", K = 2

Output: "2-5G-3J"

Explanation: The string S has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above.


  1. The length of string S will not exceed 12,000, and K is a positive integer.
  2. String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
  3. String S is non-empty.

482. License Key Formatting
struct Solution;

impl Solution {
    fn license_key_formatting(s: String, k: i32) -> String {
        let mut res: Vec<char> = vec![];
        let mut i = 0;
        for c in s.chars().rev() {
            if c != '-' {
                i += 1;
                if i == k {
                    i = 0;
        if let Some(&'-') = res.last() {

fn test() {
    let s = "5F3Z-2e-9-w".to_string();
    let k = 4;
    let o = "5F3Z-2E9W".to_string();
    assert_eq!(Solution::license_key_formatting(s, k), o);

    let s = "2-5g-3-J".to_string();
    let k = 2;
    let o = "2-5G-3J".to_string();
    assert_eq!(Solution::license_key_formatting(s, k), o);