[String] add Solution to One Edit Distance

String/OneEditDistance.swift

@@ -0,0 +1,61 @@
+/**
+ * Question Link: https://leetcode.com/problems/one-edit-distance/
+ * Primary idea: Two pointers to determine two strings' mutation
+ * Time Complexity: O(n), Space Complexity: O(n)
+ */
+
+class OneEditDistance {
+    func isOneEditDistance(s: String, _ t: String) -> Bool {
+        let sLen = s.characters.count
+        let tLen = t.characters.count
+
+        guard abs(sLen - tLen) <= 1 else {
+            return false
+        }
+
+        if sLen > tLen {
+            return _isAddChar(t, s)
+        } else if sLen == tLen {
+            return _isReplaceChar(s, t)
+        } else {
+            return _isAddChar(s, t)
+        }
+    }
+
+    private func _isAddChar(s: String, _ t: String) -> Bool {
+        let sChars = [Character](s.characters)
+        let tChars = [Character](t.characters)
+
+        var tIndex = 0
+        var sIndex = 0
+
+        while sIndex < sChars.count && tIndex < tChars.count {
+            if sChars[sIndex] != tChars[tIndex]  {
+                tIndex += 1
+            } else {
+                tIndex += 1
+                sIndex += 1
+            }
+        }
+
+        return tIndex - sIndex <= 1
+    }
+
+    private func _isReplaceChar(s: String, _ t: String) -> Bool {
+        let sChars = [Character](s.characters)
+        let tChars = [Character](t.characters)
+
+        var tIndex = 0
+        var diff = 0
+
+        for sIndex in 0 ..< sChars.count {
+            if sChars[sIndex] != tChars[tIndex] {
+                diff += 1
+            }
+
+            tIndex += 1
+        }
+
+        return diff == 1
+    }
+}