Add Solution.py to problem 1234.Replace the Substring for Balanced String

solution/1234.Replace the Substring for Balanced String/Solution.py

@@ -0,0 +1,91 @@
+'''
+https://leetcode.com/problems/replace-the-substring-for-balanced-string/
+
+You are given a string containing only 4 kinds of characters 'Q', 'W', 'E' and 'R'.
+
+A string is said to be balanced if each of its characters appears n/4 times where n is the length of the string.
+
+Return the minimum length of the substring that can be replaced with any other string of the same length to make the original string s balanced.
+
+Return 0 if the string is already balanced.
+
+
+
+Example 1:
+
+Input: s = "QWER"
+Output: 0
+Explanation: s is already balanced.
+Example 2:
+
+Input: s = "QQWE"
+Output: 1
+Explanation: We need to replace a 'Q' to 'R', so that "RQWE" (or "QRWE") is balanced.
+Example 3:
+
+Input: s = "QQQW"
+Output: 2
+Explanation: We can replace the first "QQ" to "ER". 
+Example 4:
+
+Input: s = "QQQQ"
+Output: 3
+Explanation: We can replace the last 3 'Q' to make s = "QWER".
+
+
+Constraints:
+
+1 <= s.length <= 10^5
+s.length is a multiple of 4
+s contains only 'Q', 'W', 'E' and 'R'.
+'''
+'''
+Runtime: 172 ms, faster than 92.84% of Python3 online submissions for Replace the Substring for Balanced String.
+Memory Usage: 13.4 MB, less than 100.00% of Python3 online submissions for Replace the Substring for Balanced String.
+
+'''
+
+
+from collections import defaultdict, Counter
+class Solution:
+    def balancedString(self, s: str) -> int:
+        # count the occurence of each char
+        count_chars = Counter(s)
+
+        required = len(s) // 4
+
+        # hold the number of excessive occurences 
+        more_chars = defaultdict(int)
+        for char, count_char in count_chars.items():
+            more_chars[char] = max(0, count_char - required)
+
+        min_len = len(s)
+
+        # count the number of total replacements
+        need_replace = sum(more_chars.values())
+        if need_replace == 0:
+            return  0
+
+        # Sliding windows
+        # First, move the second cursors until satisfy the conditions
+        # Second, move the first_cursor so that it still satisfy the requirement
+
+        first_cursor, second_cursor = 0, 0
+        while second_cursor < len(s):
+            # Move second_cursor
+            if more_chars[s[second_cursor]] > 0:
+                need_replace -= 1    
+            more_chars[s[second_cursor]] -= 1
+            second_cursor += 1
+
+            # Move first_cursor
+            while first_cursor < second_cursor and need_replace == 0:
+                min_len = min(min_len, second_cursor - first_cursor)
+                if s[first_cursor] in more_chars:
+                    more_chars[s[first_cursor]] += 1
+                    if more_chars[s[first_cursor]] > 0:
+                        need_replace += 1
+                first_cursor += 1
+
+        return min_len
+