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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

  • 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)
}

  • function licenseKeyFormatting(s: string, k: number): string {
    const n = s.length;
    let cnt = (n - (s.match(/-/g) || []).length) % k || k;
    const ans: string[] = [];
    for (let i = 0; i < n; i++) {
        const c = s[i];
        if (c === '-') {
            continue;
        }
        ans.push(c.toUpperCase());
        if (--cnt === 0) {
            cnt = k;
            if (i !== n - 1) {
                ans.push('-');
            }
        }
    }
    while (ans.at(-1) === '-') {
        ans.pop();
    }
    return ans.join('');
}

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