Add solution and test-cases for problem 2116

leetcode/2101-2200/2116.Check-if-a-Parentheses-String-Can-Be-Valid/README.md

@@ -1,28 +1,46 @@
 # [2116.Check if a Parentheses String Can Be Valid][title]
 
-> [!WARNING|style:flat]
-> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
-
 ## Description
+A parentheses string is a **non-empty** string consisting only of `'('` and `')'`. It is valid if **any** of the following conditions is **true**:
+
+- It is `()`.
+- It can be written as `AB` (`A` concatenated with `B`), where `A` and `B` are valid parentheses strings.
+- It can be written as `(A)`, where `A` is a valid parentheses string.
+
+You are given a parentheses string `s` and a string `locked`, both of length `n`. `locked` is a binary string consisting only of `'0'`s and `'1'`s. For **each** index `i` of `locked`,
+
+- If `locked[i]` is `'1'`, you **cannot** change `s[i]`.
+- But if `locked[i]` is `'0'`, you **can** change `s[i]` to either `'('` or `')'`.
+
+Return `true` if you can make `s` a valid parentheses string. Otherwise, return `false`.
+
+**Example 1:**  
 
-**Example 1:**
+![1](./1.png)
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: s = "))()))", locked = "010100"
+Output: true
+Explanation: locked[1] == '1' and locked[3] == '1', so we cannot change s[1] or s[3].
+We change s[0] and s[4] to '(' while leaving s[2] and s[5] unchanged to make s valid.
 ```
 
-## 题意
-> ...
+**Example 2:**
 
-## 题解
-
-### 思路1
-> ...
-Check if a Parentheses String Can Be Valid
-```go
 ```
+Input: s = "()()", locked = "0000"
+Output: true
+Explanation: We do not need to make any changes because s is already valid.
+```
+
+**Example 3:**
 
+```
+Input: s = ")", locked = "0"
+Output: false
+Explanation: locked permits us to change s[0]. 
+Changing s[0] to either '(' or ')' will not make s valid.
+```
 
 ## 结语
 

leetcode/2101-2200/2116.Check-if-a-Parentheses-String-Can-Be-Valid/Solution.go

@@ -1,5 +1,38 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(s string, locked string) bool {
+	l := len(s)
+	if l&1 == 1 {
+		return false
+	}
+	stack1, stack2 := make([]int, l), make([]int, l)
+	i1, i2 := -1, -1
+	for i, b := range s {
+		if locked[i] == '0' {
+			i1++
+			stack1[i1] = i
+			continue
+		}
+		if b == '(' {
+			i2++
+			stack2[i2] = i
+			continue
+		}
+		if i2 != -1 {
+			i2--
+			continue
+		}
+		if i1 != -1 {
+			i1--
+			continue
+		}
+		return false
+	}
+	for i2 >= 0 && i1 >= 0 {
+		if stack2[i2] > stack1[i1] {
+			break
+		}
+		i2, i1 = i2-1, i1-1
+	}
+	return i2 == -1
 }

leetcode/2101-2200/2116.Check-if-a-Parentheses-String-Can-Be-Valid/Solution_test.go

@@ -10,30 +10,31 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
+		inputs string
+		locked string
 		expect bool
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", "))()))", "010100", true},
+		{"TestCase2", "()()", "0000", true},
+		{"TestCase3", ")", "0", false},
 	}
 
 	//	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.inputs, c.locked)
 			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",
+					c.expect, got, c.inputs, c.locked)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }