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482. License Key Formatting

Description

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 <= 105
• s consists of English letters, digits, and dashes '-'.
• 1 <= k <= 104

Solutions

• Java
• C++
• Python
• Go

• class Solution {
      public String licenseKeyFormatting(String s, int k) {
          s = s.replace("-", "").toUpperCase();
          StringBuilder sb = new StringBuilder();
          int t = 0;
          int cnt = s.length() % k;
          if (cnt == 0) {
              cnt = k;
          }
          for (int i = 0; i < s.length(); ++i) {
              sb.append(s.charAt(i));
              ++t;
              if (t == cnt) {
                  t = 0;
                  cnt = k;
                  if (i != s.length() - 1) {
                      sb.append('-');
                  }
              }
          }
          return sb.toString();
      }
  }
  

• 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;
      }
  };
  

• 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)
  
  

• 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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