482. License Key Formatting

Problem

You are given a license key represented as a string s that consists of only alphanumeric characters and dashes. The string is separated into n + 1 groups by n dashes. You are also given an integer k.

We want to reformat the string s 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 you should convert all lowercase letters to uppercase.

Return the reformatted license key.

  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.

  Constraints:

• 1 <= s.length <= 10^5

• s consists of English letters, digits, and dashes '-'.

• 1 <= k <= 10^4

Solution (Java)

class Solution {
    public String licenseKeyFormatting(String s, int k) {
        StringBuilder sb = new StringBuilder();
        int cnt = 0;
        int occ = 0;
        for (char c : s.toCharArray()) {
            if (c == '-') {
                continue;
            }
            cnt++;
        }
        int l = cnt % k;
        for (char c : s.toCharArray()) {
            if (c == '-') {
                continue;
            }
            if (occ == k) {
                occ = 0;
                sb.append('-');
            } else if (occ == l && l != 0) {
                l = 0;
                occ = 0;
                sb.append('-');
            }
            sb.append(Character.toUpperCase(c));
            occ++;
        }
        return sb.toString();
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).