假设我们有一个长度为偶数的小写字符串 s。我们必须找到需要更新的最小字符数,以便所有 i 都满足以下三个条件之一,其中 0 ≤ i < n/2 和 j, n/2 ≤ j < n -
s[i] > s[j]
s[i] < s[j]
s[i] == s[j]
所以,如果输入像 s = "pppxxp",那么输出将是 1 因为如果我们把最后一个 "p" 改为 "x",那么这可以满足条件 s[i] < s[j]
让我们看看以下实现以获得更好的理解 -
from collections import Counter from string import ascii_lowercase def solve(s): n = len(s) left = Counter(s[: n >> 1]) right = Counter(s[n >> 1 :]) ans = n for pivot in ascii_lowercase: ans = min(ans, n - left[pivot] - right[pivot]) good = sum(left[c] for c in left if c <= pivot) good += sum(right[c] for c in right if c > pivot) ans = min(ans, n - good) good = sum(left[c] for c in left if c > pivot) good += sum(right[c] for c in right if c <= pivot) ans = min(ans, n - good) return ans s = "pppxxp" print(solve(s))
"pppxxp"输出结果
1