Algorithm
Problem Name: 482. License Key Formatting
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 <= 105sconsists of English letters, digits, and dashes'-'.1 <= k <= 104
Code Examples
#1 Code Example with C Programming
Code -
C Programming
char* licenseKeyFormatting(char* S, int K) {
char *p;
int l, n, m;
char *head, c;
int i;
p = S;
l = 0;
while (*p) {
if (*p != '-') l ++; // get length all valid characters
p ++;
}
if (l == 0) return "";
n = l / K; // total number of groups having full of K characters
m = K - (l % K); // start number of first group
p = malloc(l + n + 1);
//assert(p);
head = p;
while (c = *(S ++)) {
if (c == '-') continue;
*p = (c >= 'a' && c < = 'z') ? c - 'a' + 'A': c;
p ++;
m = (m + 1) % K;
if (m == 0) {
*p = '-';
p ++;
}
}
if (*(p - 1) == '-') p --;
*p = 0;
return head;
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
string licenseKeyFormatting(string S, int K) {
stringstream ss(S);
string s = "", tmp = "", res = "";
while(getline(ss, tmp, '-')) s += tmp;
for(auto& c: s) c = toupper(c);
int i = 0, step = (s.size() % K == 0) ? K : s.size() % K;
while(i < s.size()){
res += s.substr(i, step) + '-';
i += step;
step = K;
}
res.pop_back();
return res;
}
};
Input
Output
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public String licenseKeyFormatting(String S, int K) {
StringBuilder sb = new StringBuilder();
int idx = S.length() - 1;
int count = 0;
while (idx >= 0) {
if (S.charAt(idx) != '-') {
if (Character.isLetter(S.charAt(idx))) {
sb.append(Character.toUpperCase(S.charAt(idx)));
}
else {
sb.append(S.charAt(idx));
}
count++;
}
idx--;
if (count == K && idx >= 0) {
sb.append('-');
count = 0;
}
}
if (sb.length() == 0) {
return "";
}
String ans = sb.reverse().toString();
return ans.charAt(0) == '-' ? ans.substring(1) : ans;
}
}
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Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const licenseKeyFormatting = function(S, K) {
if (S == null || S === "") return "";
const newStr = S.replace(/-/g, "").toUpperCase();
const arr = newStr.split("");
for (let i = arr.length - 1 - K; i >= 0; i -= K) {
arr[i] = arr[i] + "-";
}
return arr.join("");
};
Input
Output
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def licenseKeyFormatting(self, S, K):
S = S.replace("-", "").upper()[::-1]
return '-'.join([S[i:i+K] for i in range(0, len(S), K)])[::-1]
Input
Output
#6 Code Example with C# Programming
Code -
C# Programming
using System.Text;
namespace LeetCode
{
public class _0482_LicenseKeyFormatting
{
public string LicenseKeyFormatting(string S, int K)
{
S = S.ToUpper().Replace("-", "");
var sb = new StringBuilder();
var groupSize = S.Length % K;
if (groupSize == 0) groupSize = K;
for (int i = 0; i < S.Length; i++)
{
if (groupSize == 0)
{
sb.Append("-");
groupSize = K;
}
sb.Append(S[i]);
groupSize--;
}
return sb.ToString();
}
}
}
Input
Output