permutation-in-string py3

0424_longest-repeating-character-replacement/README.md

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+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

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

0567_permutation-in-string/README.md

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+Given two strings `s1` and `s2`, return `true` _if_ `s2` _contains a permutation of_ `s1`_, or_ `false` _otherwise_.
+
+In other words, return `true` if one of `s1`'s permutations is the substring of `s2`.
+
+**Example 1:**
+
+    Input: s1 = "ab", s2 = "eidbaooo"
+    Output: true
+    Explanation: s2 contains one permutation of s1 ("ba").
+    
+
+**Example 2:**
+
+    Input: s1 = "ab", s2 = "eidboaoo"
+    Output: false
+    
+
+**Constraints:**
+
+*   `1 <= s1.length, s2.length <= 104`
+*   `s1` and `s2` consist of lowercase English letters.
+
+https://leetcode.com/problems/permutation-in-string

0567_permutation-in-string/python3/hashmaps_in_loop.py

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+# Time: O(S1 · S2) ; S2 is len(s1) and S2 is len(S2)
+# Space: O(S2 · 26)
+
+from collections import Counter
+
+class Solution:
+    def checkInclusion(self, s1: str, s2: str) -> bool:
+        if len(s1) > len(s2): return False
+        
+        if len(s1) == len(s2):
+            return Counter(s1) == Counter(s2)
+        
+        s1_counter = Counter(s1)
+        
+        target_length = len(s1)
+        end = target_length - 1
+        
+        while end < len(s2):
+            start = end - target_length + 1
+            curr_substring = s2[start:end + 1]
+            
+            # Check if counters match for the given substring
+            # indicated by start:end with a length of target_length
+            if s1_counter == Counter(curr_substring):
+                return True
+            
+            end += 1
+        
+        return False
+        
+        

0567_permutation-in-string/python3/solution.py

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+class Solution:
+    def checkInclusion(self, s1: str, s2: str) -> bool:
+        '''
+        Sliding window, better space utilization
+        '''
+        
+        if len(s1) > len(s2): return False
+        
+        def get_letter_index(letter):
+            '''
+            Problem only concerns with lowercase letters
+            '''
+            return ord(letter) - ord('a')
+        
+        
+        s1_count, s2_count = [0] * 26, [0] * 26
+        for i in range(len(s1)):
+            s1_count[get_letter_index(s1[i])] += 1
+            s2_count[get_letter_index(s2[i])] += 1
+        
+        # Keep a `count_matches` variable that tells us how many
+        # counts of a-z from s1 match a-z in the window of s2
+        count_matches = 0
+        
+        for i in range(26):
+            if s1_count[i] == s2_count[i]:
+                count_matches += 1
+        
+        # It's possible that we get a perfect match right after the
+        # above initial computation. If yes, yay!
+        if count_matches == 26:
+            return True
+        
+        
+        # Start from the very next character after figuring out the
+        # initial `count_matches` value which would be at len(s1)
+        l = 0
+        for r in range(len(s1), len(s2)):
+            #
+            # Adding rightmost letter
+            #
+            li = get_letter_index(s2[r])
+            s2_count[li] += 1
+            
+            if s1_count[li] == s2_count[li]:
+                count_matches += 1
+            # If after adding new right letter, count increased by 1 for
+            # the letter index, then that means total matches also reduced
+            # by 1
+            elif s1_count[li] + 1 == s2_count[li]:
+                count_matches -= 1
+            
+            #
+            # Removing leftmost letter
+            #
+            li = get_letter_index(s2[l])
+            s2_count[li] -= 1
+            
+            if s1_count[li] == s2_count[li]:
+                count_matches += 1
+            # If after removing leftmost letter, count decreased by 1, then
+            # our total matches also reduced by 1
+            elif s1_count[li] - 1 == s2_count[li]:
+                count_matches -= 1
+            
+            if count_matches == 26: return True
+            
+            l += 1
+        
+        return count_matches == 26