longest-repeating-character-replacement py3

0424_longest-repeating-character-replacement/python3/sliding_window_1.py

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+# Time: O(N^2)
+# Space: O(N)
+
+from collections import Counter
+
+class Solution:
+    def characterReplacement(self, s: str, k: int) -> int:
+        '''
+        Sliding window approach
+        '''
+        
+        def window_size(l, r):
+            return r - l + 1
+        
+        # Need to keep track of frequency of each character in the
+        # window
+        count = Counter()
+        l = r = 0
+        result = 0
+        
+        while r < len(s):
+            # Increment the frequency of the `r`th char
+            count.update(s[r]) 
+            
+            # We can perform upto K replacements. So we prefer least num
+            # of replacements (<= K). In the current window, we can perform
+            # least replacements for the most common character in the range.
+            #
+            # e.g. AAABB K=2, in range 0 to 4, most common char is A, count=3
+            # so it just needs 5 - 3 = 2 replacements which is <= K which is ok.
+            #
+            # But, if K = 1, then we need 5 - 2 = 2 replacements for the same range
+            # but 2 > K so this window is not valid. We need to move left pointer in
+            # this case and reduce its count.
+            if window_size(l, r) - count.most_common(1)[0][1] > k:
+                count.subtract(s[l])
+                l += 1
+            
+            result = max(result, window_size(l, r))
+            r += 1
+        
+        return result

0424_longest-repeating-character-replacement/python3/solution.py

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+# Time: O(N)
+# Space: O(N)
+
+from collections import Counter
+
+class Solution:
+    def characterReplacement(self, s: str, k: int) -> int:
+        '''
+        Sliding window approach, optimized
+        '''
+        
+        def window_size(l, r):
+            return r - l + 1
+        
+        # Need to keep track of frequency of each character in the
+        # window
+        count = Counter()
+        
+        # We can optimize the prev sliding window approach by just keeping
+        # track of maximum frequency/count of an element we've seen so far.
+        maxf = 0
+        
+        l = r = 0
+        result = 0
+        
+        while r < len(s):
+            # Increment the frequency of the `r`th char
+            count.update(s[r]) 
+            
+            # If right's frequency is bigger, make it the new
+            # max frequency
+            maxf = max(maxf, count[s[r]])
+            
+            # Refer to neetcode: https://www.youtube.com/watch?v=gqXU1UyA8pk
+            # TODO: Document this approach my own way
+            if window_size(l, r) - maxf > k:
+                count.subtract(s[l])
+                l += 1
+            
+            result = max(result, window_size(l, r))
+            r += 1
+        
+        return result