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| 1 | +''' |
| 2 | +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: |
| 3 | + |
| 4 | +subs has a size of at least k. |
| 5 | +Character a has an odd frequency in subs. |
| 6 | +Character b has an even frequency in subs. |
| 7 | +Return the maximum difference. |
| 8 | + |
| 9 | +Note that subs can contain more than 2 distinct characters. |
| 10 | +''' |
| 11 | + |
| 12 | + |
| 13 | +class Solution: |
| 14 | + def maxDifference(self, s: str, k: int) -> int: |
| 15 | + n = len(s) |
| 16 | + ans = float('-inf') |
| 17 | + |
| 18 | + # Step 1: Try all (a, b) pairs |
| 19 | + for a in range(5): |
| 20 | + for b in range(5): |
| 21 | + if a == b: continue |
| 22 | + |
| 23 | + s1 = [0] * (n + 1) |
| 24 | + s2 = [0] * (n + 1) |
| 25 | + |
| 26 | + # Step 2: Prefix counts |
| 27 | + for i in range(1, n + 1): |
| 28 | + s1[i] = s1[i - 1] + (int(s[i - 1]) == a) |
| 29 | + s2[i] = s2[i - 1] + (int(s[i - 1]) == b) |
| 30 | + |
| 31 | + # Step 3: Track best past difference for each parity |
| 32 | + g = [[float('-inf')] * 2 for _ in range(2)] |
| 33 | + j = 0 |
| 34 | + |
| 35 | + # Step 4: Two-pointer sliding window |
| 36 | + for i in range(k, n + 1): |
| 37 | + while i - j >= k and s1[i] > s1[j] and s2[i] > s2[j]: |
| 38 | + pa = s1[j] % 2 |
| 39 | + pb = s2[j] % 2 |
| 40 | + g[pa][pb] = max(g[pa][pb], s2[j] - s1[j]) |
| 41 | + j += 1 |
| 42 | + |
| 43 | + # Step 5: Check candidate answer |
| 44 | + pa = s1[i] % 2 |
| 45 | + pb = s2[i] % 2 |
| 46 | + best = g[1 - pa][pb] |
| 47 | + if best != float('-inf'): |
| 48 | + ans = max(ans, (s1[i] - s2[i]) + best) |
| 49 | + |
| 50 | + return -1 if ans == float('-inf') else ans |
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