From f915f6aab8e02397f0517abe6196cfc52088ea8c Mon Sep 17 00:00:00 2001
From: 0xff-dev <stevenshuang521@gmail.com>
Date: Thu, 27 Feb 2025 09:21:38 +0800
Subject: [PATCH] Add solution and test-cases for problem 2606

---
 .../README.md                                 | 43 +++++++++++++++++++
 .../Solution.go                               | 25 ++++++++++-
 .../Solution_test.go                          | 22 +++++-----
 3 files changed, 77 insertions(+), 13 deletions(-)
 create mode 100644 leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/README.md

diff --git a/leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/README.md b/leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/README.md
new file mode 100644
index 000000000..38a985a11
--- /dev/null
+++ b/leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/README.md
@@ -0,0 +1,43 @@
+# [2606.Find the Substring With Maximum Cost][title]
+
+## Description
+You are given a string `s`, a string `chars` of **distinct** characters and an integer array `vals` of the same length as `chars`.
+
+The **cost of the substring** is the sum of the values of each character in the substring. The cost of an empty string is considered `0`.
+
+The **value of the character** is defined in the following way:
+
+- If the character is not in the string `chars`, then its value is its corresponding position **(1-indexed)** in the alphabet.
+
+    - For example, the value of `'a'` is `1`, the value of `'b'` is `2`, and so on. The value of `'z'` is `26`.
+
+- Otherwise, assuming `i` is the index where the character occurs in the string `chars`, then its value is `vals[i]`.
+
+Return the maximum cost among all substrings of the string `s`.
+
+**Example 1:**
+
+```
+Input: s = "adaa", chars = "d", vals = [-1000]
+Output: 2
+Explanation: The value of the characters "a" and "d" is 1 and -1000 respectively.
+The substring with the maximum cost is "aa" and its cost is 1 + 1 = 2.
+It can be proven that 2 is the maximum cost.
+```
+
+**Example 2:**
+
+```
+Input: s = "abc", chars = "abc", vals = [-1,-1,-1]
+Output: 0
+Explanation: The value of the characters "a", "b" and "c" is -1, -1, and -1 respectively.
+The substring with the maximum cost is the empty substring "" and its cost is 0.
+It can be proven that 0 is the maximum cost.
+```
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-golang-algorithm][me]
+
+[title]: https://leetcode.com/problems/find-the-substring-with-maximum-cost/
+[me]: https://github.com/kylesliu/awesome-golang-algorithm
diff --git a/leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/Solution.go b/leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/Solution.go
index d115ccf5e..9d9f978d5 100755
--- a/leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/Solution.go
+++ b/leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/Solution.go
@@ -1,5 +1,26 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(s string, chars string, vals []int) int {
+	value := [26]int{}
+	for i := range 26 {
+		value[i] = i + 1
+	}
+	for i, c := range chars {
+		value[c-'a'] = vals[i]
+	}
+	vv := make([]int, len(s))
+	for i := range s {
+		vv[i] = value[s[i]-'a']
+	}
+
+	ans, minSum, sum := 0, 0, 0
+	for i := 0; i < len(vv); i++ {
+		ans = max(ans, vv[i])
+		sum += vv[i]
+		ans = max(ans, sum-minSum)
+		if sum < minSum {
+			minSum = sum
+		}
+	}
+	return ans
 }
diff --git a/leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/Solution_test.go b/leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/Solution_test.go
index 14ff50eb4..56e6a9a2b 100755
--- a/leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/Solution_test.go
+++ b/leetcode/2601-2700/2606.Find-the-Substring-With-Maximum-Cost/Solution_test.go
@@ -9,31 +9,31 @@ import (
 func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
-		name   string
-		inputs bool
-		expect bool
+		name     string
+		s, chars string
+		vals     []int
+		expect   int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", "adaa", "d", []int{-1000}, 2},
+		{"TestCase2", "abc", "abc", []int{-1, -1, -1}, 0},
 	}
 
 	//	开始测试
 	for i, c := range cases {
 		t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
-			got := Solution(c.inputs)
+			got := Solution(c.s, c.chars, c.vals)
 			if !reflect.DeepEqual(got, c.expect) {
-				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
-					c.expect, got, c.inputs)
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v %v %v",
+					c.expect, got, c.s, c.chars, c.vals)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }