Add solution and test-cases for problem 3003

leetcode/3001-3100/3003.Maximize-the-Number-of-Partitions-After-Operations/README.md

@@ -1,28 +1,61 @@
 # [3003.Maximize the Number of Partitions After Operations][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 string `s` and an integer `k`.
+
+First, you are allowed to change **at most one** index in `s` to another lowercase English letter.
+
+After that, do the following partitioning operation until `s` is **empty**:
+
+- Choose the **longest prefix** of `s` containing at most `k` **distinct** characters.
+- **Delete** the prefix from `s` and increase the number of partitions by one. The remaining characters (if any) in `s` maintain their initial order.
+
+Return an integer denoting the **maximum** number of resulting partitions after the operations by optimally choosing at most one index to change.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: s = "accca", k = 2
+
+Output: 3
+
+Explanation:
+
+The optimal way is to change s[2] to something other than a and c, for example, b. then it becomes "acbca".
+
+Then we perform the operations:
+
+The longest prefix containing at most 2 distinct characters is "ac", we remove it and s becomes "bca".
+Now The longest prefix containing at most 2 distinct characters is "bc", so we remove it and s becomes "a".
+Finally, we remove "a" and s becomes empty, so the procedure ends.
+Doing the operations, the string is divided into 3 partitions, so the answer is 3.
+```
+
+**Example 2:**
+
 ```
+Input: s = "aabaab", k = 3
+
+Output: 1
+
+Explanation:
 
-## 题意
-> ...
+Initially s contains 2 distinct characters, so whichever character we change, it will contain at most 3 distinct characters, so the longest prefix with at most 3 distinct characters would always be all of it, therefore the answer is 1.
+```
 
-## 题解
+**Example 3:**
 
-### 思路1
-> ...
-Maximize the Number of Partitions After Operations
-```go
 ```
+Input: s = "xxyz", k = 1
+
+Output: 4
+
+Explanation:
 
+The optimal way is to change s[0] or s[1] to something other than characters in s, for example, to change s[0] to w.
+
+Then s becomes "wxyz", which consists of 4 distinct characters, so as k is 1, it will divide into 4 partitions.
+```
 
 ## 结语
 

leetcode/3001-3100/3003.Maximize-the-Number-of-Partitions-After-Operations/Solution.go

@@ -1,5 +1,63 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(s string, k int) int {
+	n := len(s)
+	left := make([][3]int, n)
+	right := make([][3]int, n)
+
+	num, mask, count := 0, 0, 0
+	for i := 0; i < n-1; i++ {
+		binary := 1 << (s[i] - 'a')
+		if mask&binary == 0 {
+			count++
+			if count <= k {
+				mask |= binary
+			} else {
+				num++
+				mask = binary
+				count = 1
+			}
+		}
+		left[i+1][0] = num
+		left[i+1][1] = mask
+		left[i+1][2] = count
+	}
+
+	num, mask, count = 0, 0, 0
+	for i := n - 1; i > 0; i-- {
+		binary := 1 << (s[i] - 'a')
+		if mask&binary == 0 {
+			count++
+			if count <= k {
+				mask |= binary
+			} else {
+				num++
+				mask = binary
+				count = 1
+			}
+		}
+		right[i-1][0] = num
+		right[i-1][1] = mask
+		right[i-1][2] = count
+	}
+
+	maxVal := 0
+	for i := 0; i < n; i++ {
+		seg := left[i][0] + right[i][0] + 2
+		totMask := left[i][1] | right[i][1]
+		totCount := 0
+		for totMask != 0 {
+			totMask = totMask & (totMask - 1)
+			totCount++
+		}
+		if left[i][2] == k && right[i][2] == k && totCount < 26 {
+			seg++
+		} else if min(totCount+1, 26) <= k {
+			seg--
+		}
+		if seg > maxVal {
+			maxVal = seg
+		}
+	}
+	return maxVal
 }

leetcode/3001-3100/3003.Maximize-the-Number-of-Partitions-After-Operations/Solution_test.go

@@ -10,30 +10,31 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		s      string
+		k      int
+		expect int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", "accca", 2, 3},
+		{"TestCase2", "aabaab", 3, 1},
+		{"TestCase3", "xxyz", 1, 4},
 	}
 
 	//	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.k)
 			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.s, c.k)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }