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feat: add solutions to lc problem: No.1717 #3263

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215 changes: 120 additions & 95 deletions solution/1700-1799/1717.Maximum Score From Removing Substrings/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -74,7 +74,25 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：贪心

我们不妨假设子字符串 "ab" 的得分总是不低于子字符串 "ba" 的得分，如果不是，我们可以交换 "a" 和 "b"，同时交换 $x$ 和 $y$。

接下来，我们只需要考虑字符串中只包含 "a" 和 "b" 的情况。如果字符串中包含其他字符，我们可以将其视为一个分割点，将字符串分割成若干个只包含 "a" 和 "b" 的子字符串，然后分别计算每个子字符串的得分。

我们观察发现，对于一个只包含 "a" 和 "b" 的子字符串，无论采取什么样的操作，最后一定只剩下一种字符，或者空串。由于每次操作都会同时删除一个 "a" 和一个 "b"，因此总的操作次数一定是固定的。我们可以贪心地先删除 "ab"，再删除 "ba"，这样可以保证得分最大。

因此，我们可以使用两个变量 $\textit{cnt1}$ 和 $\textit{cnt2}$ 分别记录 "a" 和 "b" 的数量，然后遍历字符串，根据当前字符的不同情况更新 $\textit{cnt1}$ 和 $\textit{cnt2}$，并计算得分。

对于当前遍历到的字符 $c$：

- 如果 $c$ 是 "a"，由于要先删除 "ab"，因此此时我们不消除该字符，只增加 $\textit{cnt1}$；
- 如果 $c$ 是 "b"，如果此时 $\textit{cnt1} > 0$，我们可以消除一个 "ab"，并增加 $x$ 分，否则我们只能增加 $\textit{cnt2}$；
- 如果 $c$ 是其他字符，那么对于该子字符串，我们剩下了一个 $\textit{cnt2}$ 个 "b" 和 $\textit{cnt1}$ 个 "a"，我们可以消除 $\min(\textit{cnt1}, \textit{cnt2})$ 个 "ab"，并增加 $y$ 分。

遍历结束后，我们还需要额外处理一下剩余的 "ab"，增加若干个 $y$ 分。

时间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -83,29 +101,24 @@ tags:
```python
class Solution:
def maximumGain(self, s: str, x: int, y: int) -> int:
a, b = "a", "b"
if x < y:
return self.maximumGain(s[::-1], y, x)
ans = 0
stk1, stk2 = [], []
x, y = y, x
a, b = b, a
ans = cnt1 = cnt2 = 0
for c in s:
if c != 'b':
stk1.append(c)
else:
if stk1 and stk1[-1] == 'a':
stk1.pop()
if c == a:
cnt1 += 1
elif c == b:
if cnt1:
ans += x
cnt1 -= 1
else:
stk1.append(c)
while stk1:
c = stk1.pop()
if c != 'b':
stk2.append(c)
cnt2 += 1
else:
if stk2 and stk2[-1] == 'a':
stk2.pop()
ans += y
else:
stk2.append(c)
ans += min(cnt1, cnt2) * y
cnt1 = cnt2 = 0
ans += min(cnt1, cnt2) * y
return ans
```

Expand All @@ -114,37 +127,35 @@ class Solution:
```java
class Solution {
public int maximumGain(String s, int x, int y) {
char a = 'a', b = 'b';
if (x < y) {
return maximumGain(new StringBuilder(s).reverse().toString(), y, x);
int t = x;
x = y;
y = t;
char c = a;
a = b;
b = c;
}
int ans = 0;
Deque<Character> stk1 = new ArrayDeque<>();
Deque<Character> stk2 = new ArrayDeque<>();
for (char c : s.toCharArray()) {
if (c != 'b') {
stk1.push(c);
} else {
if (!stk1.isEmpty() && stk1.peek() == 'a') {
stk1.pop();
int ans = 0, cnt1 = 0, cnt2 = 0;
int n = s.length();
for (int i = 0; i < n; ++i) {
char c = s.charAt(i);
if (c == a) {
cnt1++;
} else if (c == b) {
if (cnt1 > 0) {
ans += x;
cnt1--;
} else {
stk1.push(c);
cnt2++;
}
}
}
while (!stk1.isEmpty()) {
char c = stk1.pop();
if (c != 'b') {
stk2.push(c);
} else {
if (!stk2.isEmpty() && stk2.peek() == 'a') {
stk2.pop();
ans += y;
} else {
stk2.push(c);
}
ans += Math.min(cnt1, cnt2) * y;
cnt1 = 0;
cnt2 = 0;
}
}
ans += Math.min(cnt1, cnt2) * y;
return ans;
}
}
Expand All @@ -156,37 +167,30 @@ class Solution {
class Solution {
public:
int maximumGain(string s, int x, int y) {
char a = 'a', b = 'b';
if (x < y) {
reverse(s.begin(), s.end());
return maximumGain(s, y, x);
swap(x, y);
swap(a, b);
}
int ans = 0;
stack<char> stk1;
stack<char> stk2;

int ans = 0, cnt1 = 0, cnt2 = 0;
for (char c : s) {
if (c != 'b')
stk1.push(c);
else {
if (!stk1.empty() && stk1.top() == 'a') {
stk1.pop();
if (c == a) {
cnt1++;
} else if (c == b) {
if (cnt1) {
ans += x;
} else
stk1.push(c);
}
}
while (!stk1.empty()) {
char c = stk1.top();
stk1.pop();
if (c != 'b')
stk2.push(c);
else {
if (!stk2.empty() && stk2.top() == 'a') {
stk2.pop();
ans += y;
} else
stk2.push(c);
cnt1--;
} else {
cnt2++;
}
} else {
ans += min(cnt1, cnt2) * y;
cnt1 = 0;
cnt2 = 0;
}
}
ans += min(cnt1, cnt2) * y;
return ans;
}
};
Expand All @@ -195,42 +199,63 @@ public:
#### Go

```go
func maximumGain(s string, x int, y int) int {
func maximumGain(s string, x int, y int) (ans int) {
a, b := 'a', 'b'
if x < y {
t := []rune(s)
for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
t[i], t[j] = t[j], t[i]
}
return maximumGain(string(t), y, x)
x, y = y, x
a, b = b, a
}
ans := 0
var stk1 []byte
var stk2 []byte
for i := range s {
if s[i] != 'b' {
stk1 = append(stk1, s[i])
} else {
if len(stk1) > 0 && stk1[len(stk1)-1] == 'a' {
stk1 = stk1[0 : len(stk1)-1]

var cnt1, cnt2 int
for _, c := range s {
if c == a {
cnt1++
} else if c == b {
if cnt1 > 0 {
ans += x
cnt1--
} else {
stk1 = append(stk1, s[i])
cnt2++
}
}
}
for _, c := range stk1 {
if c != 'a' {
stk2 = append(stk2, c)
} else {
if len(stk2) > 0 && stk2[len(stk2)-1] == 'b' {
stk2 = stk2[0 : len(stk2)-1]
ans += y
} else {
stk2 = append(stk2, c)
}
ans += min(cnt1, cnt2) * y
cnt1, cnt2 = 0, 0
}
}
return ans
ans += min(cnt1, cnt2) * y
return
}
```

#### TypeScript

```ts
function maximumGain(s: string, x: number, y: number): number {
let [a, b] = ['a', 'b'];
if (x < y) {
[x, y] = [y, x];
[a, b] = [b, a];
}

let [ans, cnt1, cnt2] = [0, 0, 0];
for (let c of s) {
if (c === a) {
cnt1++;
} else if (c === b) {
if (cnt1) {
ans += x;
cnt1--;
} else {
cnt2++;
}
} else {
ans += Math.min(cnt1, cnt2) * y;
cnt1 = 0;
cnt2 = 0;
}
}
ans += Math.min(cnt1, cnt2) * y;
return ans;
}
```

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