[String] Add a solution to Valid Word Abbreviation

README.md

@@ -4,7 +4,7 @@
 ![Leetcode](./logo.png?style=centerme)
 
 ## Progress
-[Problem Status](#problem-status) shows the latest progress to all 1000+ questions. Currently we have 320 completed solutions. Note: questions with ♥ mark means that you have to **Subscript to premium membership** of LeetCode to unlock them.
+[Problem Status](#problem-status) shows the latest progress to all 1000+ questions. Currently we have 321 completed solutions. Note: questions with ♥ mark means that you have to **Subscript to premium membership** of LeetCode to unlock them.
 
 ## Contributors
 
@@ -104,6 +104,7 @@
 [Fizz Buzz](https://leetcode.com/problems/fizz-buzz/)| [Swift](./String/FizzBuzz.swift)| Easy| O(n)| O(1)|
 [First Unique Character in a String](https://leetcode.com/problems/first-unique-character-in-a-string/)| [Swift](./String/FirstUniqueCharacterInString.swift)| Easy| O(n)| O(1)|
 [Keyboard Row](https://leetcode.com/problems/keyboard-row/)| [Swift](./String/KeyboardRow.swift)| Easy| O(nm)| O(n)|
+[Valid Word Abbreviation](https://leetcode.com/problems/valid-word-abbreviation/)| [Swift](./String/ValidWordAbbreviation.swift)| Easy| O(n)| O(n)|
 [Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)| [Swift](./String/ValidPalindrome.swift)| Easy| O(n)| O(n)|
 [Valid Palindrome II](https://leetcode.com/problems/valid-palindrome-ii/)| [Swift](./String/ValidPalindromeII.swift)| Easy| O(n)| O(n)|
 [Detect Capital](https://leetcode.com/problems/detect-capital/)| [Swift](./String/DetectCapital.swift)| Easy| O(n)| O(1)|

String/ValidWordAbbreviation.swift

@@ -0,0 +1,43 @@
+/**
+ * Question Link: https://leetcode.com/problems/valid-word-abbreviation/
+ * Primary idea: Go through both string and compare characters or skip by the number
+ * 
+ * Time Complexity: O(n), Space Complexity: O(1)
+ *
+ */
+
+class ValidWordAbbreviation {
+    func validWordAbbreviation(_ word: String, _ abbr: String) -> Bool {
+        var i = 0, j = 0
+        let word = Array(word), abbr = Array(abbr)
+
+        while i < word.count && j < abbr.count {
+            if abbr[j].isNumber {
+                // edge case: "abbc" vs. "a02c"
+                if abbr[j] == "0" {
+                    return false
+                }
+
+                let start = j
+
+                while j < abbr.count && abbr[j].isNumber {
+                    j += 1
+                }
+
+                let end = j - 1
+
+                i += Int(String(abbr[start...end]))!
+            } else {
+                if abbr[j] != word[i] {
+                    return false
+                } else {
+                    i += 1
+                    j += 1
+                }
+            }
+        }
+
+        // edge case: "hi" vs. "hi1"
+        return i == word.count && j == abbr.count
+    }
+}