482. License Key Formatting

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.

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.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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.

Solutions

Solution 1

/**
 * 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();
    }
}

Solution 2

/**
 * 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);
    }
}

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