408. Valid Word Abbreviation

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Problem

A string can be abbreviated by replacing any number of non-adjacent, non-empty substrings with their lengths. The lengths should not have leading zeros.

For example, a string such as “substitution” could be abbreviated as (but not limited to):

• “s10n” (“s ubstitutio n”)
• “sub4u4” (“sub stit u tion”)
• “12” (“substitution”)
• “su3i1u2on” (“su bst i t u ti on”)

• “substitution” (no substrings replaced) The following are not valid abbreviations:

• “s55n” (“s ubsti tutio n”, the replaced substrings are adjacent)
• “s010n” (has leading zeros)
• “s0ubstitution” (replaces an empty substring) Given a string word and an abbreviation abbr, return whether the string matches the given abbreviation.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: word = "internationalization", abbr = "i12iz4n"
Output: true
Explanation: The word "internationalization" can be abbreviated as "i12iz4n" ("i nternational iz atio n").

Example 2:

Input: word = "apple", abbr = "a2e"
Output: false
Explanation: The word "apple" cannot be abbreviated as "a2e".

Code

class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        int i = 0;
        int j = 0;

        while(i < word.length() && j < abbr.length()){
            char c = abbr.charAt(j);

            if(Character.isDigit(c)){
                int num =  c - '0';

                if(num == 0) return false;

                while(j + 1 < abbr.length() && Character.isDigit(abbr.charAt(j + 1))){
                    num = num * 10 + (abbr.charAt(j + 1) - '0');
                    j++;
                }

                i += num;
                j++;
            } else {
                if(word.charAt(i) != abbr.charAt(j)) return false;

                i++;
                j++;
            }
        }

        return i == word.length() && j == abbr.length();
    }
}