Add solution and test-cases for problem 1963

leetcode/1901-2000/1963.Minimum-Number-of-Swaps-to-Make-the-String-Balanced/README.md

@@ -1,28 +1,45 @@
 # [1963.Minimum Number of Swaps to Make the String Balanced][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
+You are given a **0-indexed** string `s` of **even** length `n`. The string consists of **exactly** `n / 2` opening brackets `'['` and `n / 2` closing brackets `']'`.
+
+A string is called **balanced** if and only if:
+
+- It is the empty string, or
+- It can be written as `AB`, where both `A` and `B` are **balanced** strings, or
+- It can be written as `[C]`, where `C` is a **balanced** string.
+
+You may swap the brackets at **any** two indices **any** number of times.
+
+Return the **minimum** number of swaps to make `s` **balanced**.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: s = "][]["
+Output: 1
+Explanation: You can make the string balanced by swapping index 0 with index 3.
+The resulting string is "[[]]".
 ```
 
-## 题意
-> ...
+**Example 2:**
 
-## 题解
-
-### 思路1
-> ...
-Minimum Number of Swaps to Make the String Balanced
-```go
 ```
+Input: s = "]]][[["
+Output: 2
+Explanation: You can do the following to make the string balanced:
+- Swap index 0 with index 4. s = "[]][][".
+- Swap index 1 with index 5. s = "[[][]]".
+The resulting string is "[[][]]".
+```
+
+**Example 3:**
 
+```
+Input: s = "[]"
+Output: 0
+Explanation: The string is already balanced.
+```
 
 ## 结语
 

leetcode/1901-2000/1963.Minimum-Number-of-Swaps-to-Make-the-String-Balanced/Solution.go

@@ -1,5 +1,26 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(s string) int {
+	bs := []byte(s)
+	stackIndex := -1
+	for index := 0; index < len(s); index++ {
+		if s[index] == ']' {
+			if stackIndex != -1 && bs[stackIndex] == '[' {
+				stackIndex--
+				continue
+			}
+		}
+		stackIndex++
+		bs[stackIndex] = s[index]
+	}
+	if stackIndex == -1 {
+		return 0
+	}
+	// 最终就是]]]]]][[[[ 这情况
+	count := (stackIndex + 1) / 2
+	need := count / 2
+	if count&1 == 1 {
+		need++
+	}
+	return need
 }

leetcode/1901-2000/1963.Minimum-Number-of-Swaps-to-Make-the-String-Balanced/Solution_test.go

@@ -10,12 +10,12 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		inputs string
+		expect int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", "][][", 1},
+		{"TestCase2", "]]][[[", 2},
+		{"TestCase3", "[]", 0},
 	}
 
 	//	开始测试
@@ -30,10 +30,10 @@ func TestSolution(t *testing.T) {
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }