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feat: add solutions to lc problem: No.2606
No.2606.Find the Substring With Maximum Cost
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‎solution/2600-2699/2606.Find the Substring With Maximum Cost/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：前缀和 + 维护前缀和的最小值**
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我们根据题目描述，遍历字符串 $s$ 的每个字符 $c$，求出其对应的价值 $v$，然后更新当前的前缀和 $tot=tot+v$，那么以 $c$ 结尾的最大开销子字符串的开销为 $tot$ 减去前缀和的最小值 $mi$，即 $tot-mi$，我们更新答案 $ans=max(ans,tot-mi)$，并维护前缀和的最小值 $mi=min(mi,tot)$。
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遍历结束后返回答案 $ans$ 即可。
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时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 为字符串 $s$ 的长度；而 $C$ 为字符集的大小，本题中 $C=26$。
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**方法二：转化为最大子数组和问题**
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我们可以将每个字符 $c$ 的价值 $v$ 看作是一个整数，那么题目实际上转化为求最大子数组和问题。
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我们用变量 $f$ 维护以当前字符 $c$ 结尾的最大开销子字符串的开销，每次遍历到一个字符 $c$，更新 $f=max(f, 0) + v$，然后更新答案 $ans=max(ans,f)$ 即可。
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时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 为字符串 $s$ 的长度；而 $C$ 为字符集的大小，本题中 $C=26$。
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<!-- tabs:start -->
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### **Python3**
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return ans
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```
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```python
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class Solution:
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def maximumCostSubstring(self, s: str, chars: str, vals: List[int]) -> int:
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d = {c: i for i, c in enumerate(chars)}
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ans = f = 0
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for c in s:
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v = vals[d[c]] if c in d else ord(c) - ord('a') + 1
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f = max(f, 0) + v
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ans = max(ans, f)
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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}
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```
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```java
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class Solution {
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public int maximumCostSubstring(String s, String chars, int[] vals) {
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int[] d = new int[26];
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Arrays.fill(d, -1);
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int m = chars.length();
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for (int i = 0; i < m; ++i) {
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d[chars.charAt(i) - 'a'] = i;
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}
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int ans = 0, f = 0;
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int n = s.length();
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for (int i = 0; i < n; ++i) {
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int j = s.charAt(i) - 'a';
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int v = d[j] == -1 ? j + 1 : vals[d[j]];
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f = Math.max(f, 0) + v;
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ans = Math.max(ans, f);
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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};
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```
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```cpp
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class Solution {
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public:
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int maximumCostSubstring(string s, string chars, vector<int>& vals) {
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vector<int> d(26, -1);
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int m = chars.size();
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for (int i = 0; i < m; ++i) {
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d[chars[i] - 'a'] = i;
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}
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int ans = 0, f = 0;
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for (char& c : s) {
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int j = c - 'a';
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int v = d[j] == -1 ? j + 1 : vals[d[j]];
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f = max(f, 0) + v;
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ans = max(ans, f);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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}
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```
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```go
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func maximumCostSubstring(s string, chars string, vals []int) (ans int) {
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d := [26]int{}
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for i := range d {
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d[i] = -1
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}
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for i, c := range chars {
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d[c-'a'] = i
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}
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f := 0
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for _, c := range s {
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j := int(c - 'a')
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v := j + 1
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if d[j] != -1 {
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v = vals[d[j]]
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}
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f = max(f, 0) + v
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ans = max(ans, f)
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**
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```

‎solution/2600-2699/2606.Find the Substring With Maximum Cost/README_EN.md

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return ans
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```
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```python
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class Solution:
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def maximumCostSubstring(self, s: str, chars: str, vals: List[int]) -> int:
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d = {c: i for i, c in enumerate(chars)}
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ans = f = 0
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for c in s:
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v = vals[d[c]] if c in d else ord(c) - ord('a') + 1
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f = max(f, 0) + v
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ans = max(ans, f)
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return ans
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```
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### **Java**
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```java
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}
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```
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```java
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class Solution {
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public int maximumCostSubstring(String s, String chars, int[] vals) {
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int[] d = new int[26];
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Arrays.fill(d, -1);
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int m = chars.length();
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for (int i = 0; i < m; ++i) {
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d[chars.charAt(i) - 'a'] = i;
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}
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int ans = 0, f = 0;
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int n = s.length();
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for (int i = 0; i < n; ++i) {
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int j = s.charAt(i) - 'a';
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int v = d[j] == -1 ? j + 1 : vals[d[j]];
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f = Math.max(f, 0) + v;
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ans = Math.max(ans, f);
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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};
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```
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```cpp
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class Solution {
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public:
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int maximumCostSubstring(string s, string chars, vector<int>& vals) {
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vector<int> d(26, -1);
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int m = chars.size();
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for (int i = 0; i < m; ++i) {
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d[chars[i] - 'a'] = i;
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}
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int ans = 0, f = 0;
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for (char& c : s) {
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int j = c - 'a';
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int v = d[j] == -1 ? j + 1 : vals[d[j]];
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f = max(f, 0) + v;
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ans = max(ans, f);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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}
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```
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```go
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func maximumCostSubstring(s string, chars string, vals []int) (ans int) {
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d := [26]int{}
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for i := range d {
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d[i] = -1
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}
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for i, c := range chars {
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d[c-'a'] = i
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}
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f := 0
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for _, c := range s {
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j := int(c - 'a')
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v := j + 1
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if d[j] != -1 {
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v = vals[d[j]]
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}
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f = max(f, 0) + v
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ans = max(ans, f)
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**
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```

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