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

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):
"""
: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)
}