# 1422.maximum-score-after-splitting-a-string

## Statement

「分割字符串的得分」为 子字符串中 0 的数量加上 子字符串中 1 的数量。

``````输入：s = "011101"

``````

``````输入：s = "00111"

``````

``````输入：s = "1111"

``````

• `2 <= s.length <= 500`
• 字符串 `s` 仅由字符 `'0'``'1'` 组成。

Given a string `s` of zeros and ones, return the maximum score after splitting the string into two non-empty substrings (i.e. left substring and right substring).

The score after splitting a string is the number of zeros in the left substring plus the number of ones in the right substring.

Example 1:

``````Input: s = "011101"
Output: 5
Explanation:
All possible ways of splitting s into two non-empty substrings are:
left = "0" and right = "11101", score = 1 + 4 = 5
left = "01" and right = "1101", score = 1 + 3 = 4
left = "011" and right = "101", score = 1 + 2 = 3
left = "0111" and right = "01", score = 1 + 1 = 2
left = "01110" and right = "1", score = 2 + 1 = 3
``````

Example 2:

``````Input: s = "00111"
Output: 5
Explanation: When left = "00" and right = "111", we get the maximum score = 2 + 3 = 5
``````

Example 3:

``````Input: s = "1111"
Output: 3
``````

Constraints:

• `2 <= s.length <= 500`
• The string `s` consists of characters `'0'` and `'1'` only.

## Solution

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif

class Solution {
public:
int maxScore(string s) {
int n = s.length();
auto pre = vector<int>(n + 5, 0);
auto suffix = vector<int>(n + 5, 0);

for (int i = 1; i <= n; i++) {
pre[i] = pre[i - 1] + (s[i - 1] == '0');
}

for (int i = n; i >= 1; i--) {
suffix[i] = suffix[i + 1] + (s[i - 1] == '1');
}

int res = 0;
for (int i = 1; i < n; i++) {
res = max(res, pre[i] + suffix[i + 1]);
}

return res;
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````