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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.

Example 1:

Input: s = "12233", k = 4

Output: -1

Explanation:

For the substring "12233", the frequency of '1' is 1 and the frequency of '3' is 2. The difference is 1 - 2 = -1.

Example 2:

Input: s = "1122211", k = 3

Output: 1

Explanation:

For the substring "11222", the frequency of '2' is 3 and the frequency of '1' is 2. The difference is 3 - 2 = 1.

Example 3:

Input: s = "110", k = 3

Output: -1

Constraints:

3 <= s.length <= 3 * 10⁴
s consists only of digits '0' to '4'.
The input is generated that at least one substring has a character with an even frequency and a character with an odd frequency.
1 <= k <= s.length

解题思路

一开始想的做法太复杂了，赛时没做出来，赛后磨了快一个小时才过。

关键是要想到，对于任意一段子串，必然会选出一个出现频率是奇数的数字 $a$ ，以及出现频率是偶数的数字 $b$ 。观察到数字范围非常小只有 $0 \sim 4$ ，意味着我们可以暴力枚举 $a$ 和 $b$ 。对于 $[l, r]$ 的子串，答案为 $sa_r - sa_{l-1} - \left(sb_r - sb_{l-1}\right)$ ，其中 $sa_i = \sum\limits_{j=1}^{i}{[s_j = a]}$ ， $sb_i = \sum\limits_{j=1}^{i}{[s_j = b]}$ 。该式子可以等价为 $sa_r - sb_r + sb_{l-1} - sa_{l-1}$ ，因此我们可以枚举右端点 $r$ ，然后快速得到满足条件的左端点 $l$ 关于 $sb_{l-1} - sa_{l-1}$ 的最大值。

另外在 $[l,r]$ 中由于 $a$ 要出现奇数次， $b$ 要出现偶数次，因此 $sa_{l-1}$ 的奇偶性要与 $sa_r$ 不同， $sb_{l-1}$ 的奇偶性要与 $sb_r$ 相同。为此定义 $g(0/1,0/1)$ 表示在满足条件的 $l$ 中， $sa_{l-1}$ 是奇/偶， $sb_{l-1}$ 是奇/偶时，关于 $sb_{l-1} - sa_{l-1}$ 的最大值。

下面考虑 $l$ 应该满足什么条件。首先由于子串的长度至少为 $k$ ，因此有 $r - l + 1 \geq k$ 。另外由于区间内必须要出现 $a$ 和 $b$ ，因此有 $sa_r - sa_{l-1} > 0$ 且 $sb_r - sb_{l-1} > 0$ 。注意到 $sa_r$ 和 $sb_r$ 随着 $r$ 递增而递增，因此满足条件的 $l$ 实际上是一个前缀，且随着 $r$ 递增而往右扩展。因此我们可以用一个指针来维护这个前缀的右端点，同时维护这个前缀中的 $g(0/1,0/1)$ 。

因此对于右端点为 $r$ 且选择 $a$ 和 $b$ 分别作为出现奇数和偶数的数字的子串中，最大的答案是 $sa_r - sb_r + g(\neg (sa_r \bmod{2}), sb_r \bmod{2})$ 。

AC 代码如下，时间复杂度为 $O(4^2 n)$ ：

class Solution {
public:
    int maxDifference(string s, int k) {
        int n = s.size();
        int ret = -0x3f3f3f3f;
        for (int a = 0; a <= 4; a++) {
            for (int b = 0; b <= 4; b++) {
                vector<int> s1(n + 1), s2(n + 1);
                for (int i = 1; i <= n; i++) {
                    s1[i] = s1[i - 1] + (s[i - 1] - '0' == a);
                    s2[i] = s2[i - 1] + (s[i - 1] - '0' == b);
                }
                vector<vector<int>> g(2, vector<int>(2, -0x3f3f3f3f));
                for (int i = k, j = 0; i <= n; i++) {
                    while (i - j >= k && s1[i] > s1[j] && s2[i] > s2[j]) {
                        g[s1[j] & 1][s2[j] & 1] = max(g[s1[j] & 1][s2[j] & 1], s2[j] - s1[j]);
                        j++;
                    }
                    ret = max(ret, s1[i] - s2[i] + g[~s1[i] & 1][s2[i] & 1]);
                }
            }
        }
        return ret;
    }
};