License Key Formatting - String - Easy - LeetCode
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License Key Formatting - String - Easy - LeetCode

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  • 1The license key string S is reformatted into groups of K characters, with the first group potentially shorter but containing at least one character.
  • 2All lowercase letters in the string are converted to uppercase, and unnecessary dashes are removed during formatting.
  • 3The provided solution has a time complexity of O(n) and a space complexity of O(n), ensuring efficient processing of the string.

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"The license key string S is reformatted into groups of K characters, with the first group potentially shorter but containing at least one character."

License Key Formatting - String - Easy - LeetCode

You are given a license key represented as a string S which consists of 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.

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. 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.

public class Solution {
 public string LicenseKeyFormatting(string S, int K) {
 var sb = new StringBuilder();
 for(int i=S.Length-1,j=0;i>=0;i--){
 if(S[i]!='-'){
 if(j!=0 && j%K==0){
 sb.Append("-");
 }
 
 sb.Append(S[i]); 
 j++;
 } 
 
 } 
 return new string(sb.ToString().ToUpper().Reverse().ToArray());
 }
}

Time Complexity: O(n)

Space Complexity: O(n)

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sumitc91

Published on 17 October 2020 · 1 min read · 254 words

Part of AskGif Blog · coding

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