Refactor code style to Longest Substring with At Most K Distinct Characters and Longest Substring with At Most Two Distinct Characters

Author: soapyigu

SlidingWindow/LongestSubstringMostKDistinctCharacters.swift

@@ -13,29 +13,33 @@ class LongestSubstringMostKDistinctCharacters {
         guard k > 0 else {
             return 0
         }
-        
+
+        var charFreqMap = [Character: Int](), left = 0, res = 0
         let s = Array(s)
-        var start = 0, longest = 0, charsFreq = [Character: Int]()
-        
+
         for (i, char) in s.enumerated() {
-            if let freq = charsFreq[char] {
-                charsFreq[char] = freq + 1
+            if let freq = charFreqMap[char] {
+                charFreqMap[char] = freq + 1
             } else {
-                while charsFreq.count == k {
-                    longest = max(i - start, longest)
-                    
-                    guard let freq = charsFreq[s[start]] else {
+               
+                // update res
+                res = max(i - left, res)
+
+                // move left and window
+                while charFreqMap.count == k {
+                    if let leftFreq = charFreqMap[s[left]] {
+                        charFreqMap[s[left]] = leftFreq == 1 ? nil : leftFreq - 1
+                        left += 1
+                    } else {
                         fatalError()
                     }
-                    charsFreq[s[start]] = freq == 1 ? nil : freq - 1
-                    
-                    start += 1
                 }
-                
-                charsFreq[char] = 1
+
+                // update window for current char
+                charFreqMap[char] = 1
             }
         }
-        
-        return max(longest, s.count - start)
+
+        return max(res, s.count - left)
     }
-}
+}

SlidingWindow/LongestSubstringMostTwoDistinctCharacters.swift

@@ -1,6 +1,6 @@
 /**
  * Question Link: https://leetcode.com/problems/longest-substring-with-at-most-two-distinct-characters/
- * Primary idea: Slding window, use dictionary to check substring is valid or not, and 
+ * Primary idea: Slding window. Use char freq map to check substring is valid or not, and 
  				 note to handle the end of string edge case
  *
  * Time Complexity: O(n), Space Complexity: O(n)
@@ -9,32 +9,32 @@
 
  class LongestSubstringMostTwoDistinctCharacters {
     func lengthOfLongestSubstringTwoDistinct(_ s: String) -> Int {
-        var start = 0, longest = 0, charFreq = [Character: Int]()
-        let sChars = Array(s)
-        
-        for (i, char) in sChars.enumerated() {
-            if let freq = charFreq[char] {
-                charFreq[char] = freq + 1
+        var charFreqMap = [Character: Int](), left = 0, res = 0
+        let s = Array(s)
+
+        for (i, char) in s.enumerated() {
+            if let freq = charFreqMap[char] {
+                charFreqMap[char] = freq + 1
             } else {
-               if charFreq.count == 2 {
-                    longest = max(longest, i - start)
-                    
-                    while charFreq.count == 2 {
-                        let charStart = sChars[start]
-                        charFreq[charStart]! -= 1
-                        
-                        if charFreq[charStart] == 0 {
-                            charFreq[charStart] = nil
-                        }
-                        
-                        start += 1
-                    }   
+               
+                // update res
+                res = max(i - left, res)
+
+                // move left and window
+                while charFreqMap.count == 2 {
+                    if let leftFreq = charFreqMap[s[left]] {
+                        charFreqMap[s[left]] = leftFreq == 1 ? nil : leftFreq - 1
+                        left += 1
+                    } else {
+                        fatalError()
+                    }
                 }
-                
-                charFreq[char] = 1
+
+                // update window for current char
+                charFreqMap[char] = 1
             }
         }
-        
-        return max(longest, sChars.count - start)
+
+        return max(res, s.count - left)
     }
-}
+}