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Formatted question description: https://leetcode.ca/all/482.html

482. License Key Formatting

Level

Easy

Description

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.

Note:

  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.

Solution

First turn all the letters in S to upper case and remove all dashes from S. Then loop over S from right to left and add a dash before every K characters. Finally, return the new string.

  • class Solution {
    public String licenseKeyFormatting(String S, int K) {
        S = S.toUpperCase();
        S = S.replaceAll("-", "");
        int length = S.length();
        StringBuffer sb = new StringBuffer(S);
        int count = 0;
        for (int i = length - 1; i > 0; i--) {
            count++;
            if (count == K) {
                sb.insert(i, '-');
                count = 0;
            }
        }
        return sb.toString();
    }
}

  • class Solution:
    def licenseKeyFormatting(self, s: str, k: int) -> str:
        s = s.replace('-', '').upper()
        res = []
        cnt = (len(s) % k) or k
        t = 0
        for i, c in enumerate(s):
            res.append(c)
            t += 1
            if t == cnt:
                t = 0
                cnt = k
                if i != len(s) - 1:
                    res.append('-')
        return ''.join(res)

############

class Solution(object):
  def licenseKeyFormatting(self, S, K):
    """
    :type S: str
    :type K: int
    :rtype: str
    """
    s = S.split("-")
    s = "".join(s)
    n = len(s)
    start = n % K
    res = []
    if start != 0:
      res.append(s[:start].upper())
    for k in range(0, (len(s) - start) / K):
      res.append(s[start:start + K].upper())
      start += K
    return "-".join(res)


  • class Solution {
public:
    string licenseKeyFormatting(string s, int k) {
        string ss = "";
        for (char c : s) {
            if (c == '-') continue;
            if ('a' <= c && c <= 'z') c += 'A' - 'a';
            ss += c;
        }
        int cnt = ss.size() % k;
        if (cnt == 0) cnt = k;
        int t = 0;
        string res = "";
        for (int i = 0; i < ss.size(); ++i) {
            res += ss[i];
            ++t;
            if (t == cnt) {
                t = 0;
                cnt = k;
                if (i != ss.size() - 1) res += '-';
            }
        }
        return res;
    }
};

  • func licenseKeyFormatting(s string, k int) string {
	s = strings.ReplaceAll(s, "-", "")
	cnt := len(s) % k
	if cnt == 0 {
		cnt = k
	}
	t := 0
	res := []byte{}
	for i, c := range s {
		res = append(res, byte(unicode.ToUpper(c)))
		t++
		if t == cnt {
			t = 0
			cnt = k
			if i != len(s)-1 {
				res = append(res, byte('-'))
			}
		}
	}
	return string(res)
}

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