Leetcode discuss: 408. Valid Word Abbreviation

 March 8, 2022

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C# | learn from failed test cases

March 6, 2022
Introduction
It is an easy algorithm, but it is so challenge since I failed so many test cases, so I continuously modified the logic in order to pass all test cases. I have two weeks to practice Leetcode in order to prepare Meta phone screen, I just choose to solve as many algorithms as I can and warmup my coding skills.

Learn from failed test cases | Write down and review later
test case 166/322
Input:
"internationalization"
"i5a11o1"
Output:
false
Expected:
true
Fix: Do not mix c with 'c', c is in the range from 0 to 9. '9' - c >= 0

11 / 322 test cases passed.
Input:
"a"
"01"
Output:
true
Expected:
false
Fix: Add edge case to check number should not start from 0 digit.

12 / 322 test cases passed.
Runtime Error Message:
Unhandled exception. System.IndexOutOfRangeException: Index was outside the bounds of the array.

Last executed input:
"hi"
"2i"
Fix: Add index-out-of-range check for word array, wIndex - check in the range

314 / 322 test cases passed.
Input:
"hi"
"1"
Output:
true
Expected:
false
Fix: make sure that return statement should check that search in word is completed

return wIndex == length;

The following C# code passes online judge.

public class Solution {
    public bool ValidWordAbbreviation(string word, string abbr) {
        if (word == null || word.Length == 0 || abbr == null || abbr.Length == 0)
                return false;

            var length = word.Length;
            var aLength = abbr.Length;
            var wIndex = 0;
            var index = 0;
            var digits = 0;
            while (index < aLength)
            {
                var c = abbr[index];
                if ((c - '0') >= 0 && ('9' - c) >= 0)
                {
                    var digit = c - '0';
                    // test case: "a", "01"
                    if(digit == 0 && (index == 0 || abbr[index - 1] >='a') && (index + 1 < aLength && (abbr[index + 1] - '0' >=0)))
                    {
                        return false;
                    }
                                                                               
                    digits = digits * 10 + (c - '0');

                    if (index == aLength - 1 || abbr[index + 1] >= 'a')
                    {
                        if (wIndex + digits > length)
                            return false;

                        wIndex += digits;
                    }
                }
                else
                {
                    digits = 0;
                    if (wIndex >= length || word[wIndex++] != abbr[index])
                    {
                        return false;
                    }                    
                }

                index++;
            }

            return wIndex == length;
    }
}