1+ '''
2+ https://leetcode.com/problems/replace-the-substring-for-balanced-string/
3+
4+ You are given a string containing only 4 kinds of characters 'Q', 'W', 'E' and 'R'.
5+
6+ A string is said to be balanced if each of its characters appears n/4 times where n is the length of the string.
7+
8+ 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.
9+
10+ Return 0 if the string is already balanced.
11+
12+
13+
14+ Example 1:
15+
16+ Input: s = "QWER"
17+ Output: 0
18+ Explanation: s is already balanced.
19+ Example 2:
20+
21+ Input: s = "QQWE"
22+ Output: 1
23+ Explanation: We need to replace a 'Q' to 'R', so that "RQWE" (or "QRWE") is balanced.
24+ Example 3:
25+
26+ Input: s = "QQQW"
27+ Output: 2
28+ Explanation: We can replace the first "QQ" to "ER".
29+ Example 4:
30+
31+ Input: s = "QQQQ"
32+ Output: 3
33+ Explanation: We can replace the last 3 'Q' to make s = "QWER".
34+
35+
36+ Constraints:
37+
38+ 1 <= s.length <= 10^5
39+ s.length is a multiple of 4
40+ s contains only 'Q', 'W', 'E' and 'R'.
41+ '''
42+ '''
43+ Runtime: 172 ms, faster than 92.84% of Python3 online submissions for Replace the Substring for Balanced String.
44+ Memory Usage: 13.4 MB, less than 100.00% of Python3 online submissions for Replace the Substring for Balanced String.
45+
46+ '''
47+ 48+ 49+ from collections import defaultdict , Counter
50+ class Solution :
51+ def balancedString (self , s : str ) -> int :
52+ # count the occurence of each char
53+ count_chars = Counter (s )
54+ 55+ required = len (s ) // 4
56+ 57+ # hold the number of excessive occurences
58+ more_chars = defaultdict (int )
59+ for char , count_char in count_chars .items ():
60+ more_chars [char ] = max (0 , count_char - required )
61+ 62+ min_len = len (s )
63+ 64+ # count the number of total replacements
65+ need_replace = sum (more_chars .values ())
66+ if need_replace == 0 :
67+ return 0
68+ 69+ # Sliding windows
70+ # First, move the second cursors until satisfy the conditions
71+ # Second, move the first_cursor so that it still satisfy the requirement
72+ 73+ first_cursor , second_cursor = 0 , 0
74+ while second_cursor < len (s ):
75+ # Move second_cursor
76+ if more_chars [s [second_cursor ]] > 0 :
77+ need_replace -= 1
78+ more_chars [s [second_cursor ]] -= 1
79+ second_cursor += 1
80+ 81+ # Move first_cursor
82+ while first_cursor < second_cursor and need_replace == 0 :
83+ min_len = min (min_len , second_cursor - first_cursor )
84+ if s [first_cursor ] in more_chars :
85+ more_chars [s [first_cursor ]] += 1
86+ if more_chars [s [first_cursor ]] > 0 :
87+ need_replace += 1
88+ first_cursor += 1
89+ 90+ return min_len
91+
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