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+# 3445. Maximum Difference Between Even and Odd Frequency II
+
+"""
+you are given a string 's' and an integer 'k'. your task is to find the
+maximum difference between the frequency of two characters 'freq[a], -
+freq[b]', in a substring 'subs' of 's' such that 'subs' has a size of at
+least 'k', character 'a' has an odd frequency in 'subs', character 'b' has an
+even frequency in 'subs'. return the maximum difference. note that 'subs' can
+contain more than 2 distinct characters.
+"""
+
+
+class fenwick_tree:
+ inf = float("inf")
+
+ def __init__(self, n):
+ self.n = n
+ self.tree = [self.inf] * (n + 1)
+
+ def update(self, i, val):
+ pos = i + 1
+ while pos <= self.n:
+ self.tree[pos] = min(self.tree[pos], val)
+ pos += pos & -pos
+
+ def query(self, i):
+ res, pos = self.inf, i + 1
+ while pos > 0:
+ res = min(res, self.tree[pos])
+ pos -= pos & -pos
+ return res
+
+
+class Solution(object):
+ def maxDifference(self, s, k):
+ """
+ :type s: str
+ :type k: int
+ :rtype: int
+ """
+ n, ans = len(s), float("-inf")
+ for a in range(5):
+ for b in range(5):
+ if a == b:
+ continue
+ d = [0] * (n + 1)
+ pa, pb = [0] * (n + 1), [0] * (n + 1)
+ cnt_b = [0] * (n + 1)
+ for i in range(1, n + 1):
+ digit = int(s[i - 1])
+ d[i] = (
+ d[i - 1] + (1 if digit == a else 0) - (1 if digit == b else 0)
+ )
+ pa[i] = (pa[i - 1] + (1 if digit == a else 0)) & 1
+ pb[i] = (pb[i - 1] + (1 if digit == b else 0)) & 1
+ cnt_b[i] = cnt_b[i - 1] + (1 if digit == b else 0)
+ trees = [[fenwick_tree(n + 1) for _ in range(2)] for _ in range(2)]
+ for j in range(n + 1):
+ if j >= k:
+ idx = j - k
+ trees[pa[idx]][pb[idx]].update(cnt_b[idx], d[idx])
+ if j > 0:
+ need_p, need_q = (
+ 1 - pa[j],
+ pb[j],
+ )
+ if cnt_b[j] > 0:
+ best = trees[need_p][need_q].query(cnt_b[j] - 1)
+ if best != fenwick_tree.inf:
+ ans = max(ans, d[j] - best)
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.maxDifference(s="12233", k=4))
+ print(obj.maxDifference(s="1122211", k=3))
+ print(obj.maxDifference(s="110", k=3))