From 2a7ce3a29f2e2cbea7e3a29e12344e6c6c18e928 Mon Sep 17 00:00:00 2001
From: Sangeeth Sudheer <sangeeth96@users.noreply.github.com>
Date: Sun, 24 Apr 2022 16:50:53 +0530
Subject: [PATCH] longest-repeating-character-replacement py3

---
 .../README.md                                 | 26 +++++++++++
 .../python3/sliding_window_1.py               | 42 ++++++++++++++++++
 .../python3/solution.py                       | 43 +++++++++++++++++++
 3 files changed, 111 insertions(+)
 create mode 100644 0424_longest-repeating-character-replacement/README.md
 create mode 100644 0424_longest-repeating-character-replacement/python3/sliding_window_1.py
 create mode 100644 0424_longest-repeating-character-replacement/python3/solution.py

diff --git a/0424_longest-repeating-character-replacement/README.md b/0424_longest-repeating-character-replacement/README.md
new file mode 100644
index 0000000..07e1c61
--- /dev/null
+++ b/0424_longest-repeating-character-replacement/README.md
@@ -0,0 +1,26 @@
+You are given a string `s` and an integer `k`. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k` times.
+
+Return _the length of the longest substring containing the same letter you can get after performing the above operations_.
+
+**Example 1:**
+
+    Input: s = "ABAB", k = 2
+    Output: 4
+    Explanation: Replace the two 'A's with two 'B's or vice versa.
+    
+
+**Example 2:**
+
+    Input: s = "AABABBA", k = 1
+    Output: 4
+    Explanation: Replace the one 'A' in the middle with 'B' and form "AABBBBA".
+    The substring "BBBB" has the longest repeating letters, which is 4.
+    
+
+**Constraints:**
+
+*   `1 <= s.length <= 105`
+*   `s` consists of only uppercase English letters.
+*   `0 <= k <= s.length`
+
+https://leetcode.com/problems/longest-repeating-character-replacement
\ No newline at end of file
diff --git a/0424_longest-repeating-character-replacement/python3/sliding_window_1.py b/0424_longest-repeating-character-replacement/python3/sliding_window_1.py
new file mode 100644
index 0000000..e2ed096
--- /dev/null
+++ b/0424_longest-repeating-character-replacement/python3/sliding_window_1.py
@@ -0,0 +1,42 @@
+# 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
diff --git a/0424_longest-repeating-character-replacement/python3/solution.py b/0424_longest-repeating-character-replacement/python3/solution.py
new file mode 100644
index 0000000..4f71b68
--- /dev/null
+++ b/0424_longest-repeating-character-replacement/python3/solution.py
@@ -0,0 +1,43 @@
+# 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