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.
Note:
The length of string S will not exceed 12,000, and K is a positive integer.
String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
String S is non-empty.
Constraints
Approach
Links
GeeksforGeeks
ProgramCreek
YouTube
Examples
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.
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.
Solutions
/**
* Time complexity : O(N)
* Space complexity : O(N)
*/
class Solution {
public String licenseKeyFormatting(String S, int K) {
if(S == null || S.length() == 0) return S;
int len = S.length();
StringBuilder sb = new StringBuilder();
int count = 0;
for(int i = len-1; i >= 0; i--) {
char ch = S.charAt(i);
if(ch == '-') continue;
if(count == K) {
sb.append("-");
count = 0;
}
count++;
sb.append(ch);
}
return sb.reverse().toString().toUpperCase();
}
}
/**
* Time complexity : O(N)
* Space complexity : O(N)
*/
class Solution {
public String licenseKeyFormatting(String S, int K) {
if(S == null || S.length() == 0) return S;
int len = S.length();
char[] licenseKeyChars = new char[len*2];
int count = 0, index = len*2;
for(int i = len-1; i >= 0; i--) {
char ch = S.charAt(i);
if(ch == '-') continue;
if(count == K) {
licenseKeyChars[--index] = '-';
count = 0;
}
count++;
licenseKeyChars[--index] = Character.toUpperCase(ch);
}
return new String(licenseKeyChars, index, (len*2)-index);
}
}