Add solution and test-cases for problem 2947

leetcode/2901-3000/2947.Count-Beautiful-Substrings-I/README.md

@@ -1,28 +1,55 @@
 # [2947.Count Beautiful Substrings I][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 a positive integer `k`.
+
+Let `vowels` and `consonants` be the number of vowels and consonants in a string.
+
+A string is **beautiful** if:
+
+- `vowels == consonants`.
+- `(vowels * consonants) % k == 0`, in other terms the multiplication of `vowesl` and `consonants` is divisible by `k`.
+
+Return the number of **non-empty beautiful substrings** in the given string `s`.
+
+A **substrings** is a contiguous sequence of characters in a string.
+
+**Vowel letters** in English are `'a'`, `'e'`, `'i'`, `'o'`, and `'u'`.
+
+**Consonant letters** in English are every letter except vowels.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: s = "baeyh", k = 2
+Output: 2
+Explanation: There are 2 beautiful substrings in the given string.
+- Substring "baeyh", vowels = 2 (["a",e"]), consonants = 2 (["y","h"]).
+You can see that string "aeyh" is beautiful as vowels == consonants and vowels * consonants % k == 0.
+- Substring "baeyh", vowels = 2 (["a",e"]), consonants = 2 (["b","y"]). 
+You can see that string "baey" is beautiful as vowels == consonants and vowels * consonants % k == 0.
+It can be shown that there are only 2 beautiful substrings in the given string.
 ```
 
-## 题意
-> ...
-
-## 题解
+**Example 2:**
 
-### 思路1
-> ...
-Count Beautiful Substrings I
-```go
+```
+Input: s = "abba", k = 1
+Output: 3
+Explanation: There are 3 beautiful substrings in the given string.
+- Substring "abba", vowels = 1 (["a"]), consonants = 1 (["b"]). 
+- Substring "abba", vowels = 1 (["a"]), consonants = 1 (["b"]).
+- Substring "abba", vowels = 2 (["a","a"]), consonants = 2 (["b","b"]).
+It can be shown that there are only 3 beautiful substrings in the given string.
 ```
 
+**Example 3:**
+
+```
+Input: s = "bcdf", k = 1
+Output: 0
+Explanation: There are no beautiful substrings in the given string.
+```
 
 ## 结语
 

leetcode/2901-3000/2947.Count-Beautiful-Substrings-I/Solution.go

@@ -1,5 +1,25 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(s string, k int) int {
+	var isVowel func(byte) bool
+	isVowel = func(b byte) bool {
+		return b == 'a' || b == 'e' || b == 'i' || b == 'o' || b == 'u'
+	}
+	ans := 0
+	vc := make([][2]int, len(s)+1)
+	for i := 1; i <= len(s); i++ {
+		vc[i] = [2]int{vc[i-1][0], vc[i-1][1]}
+		if isVowel(s[i-1]) {
+			vc[i][0]++
+		} else {
+			vc[i][1]++
+		}
+		for j := i - 1; j > 0; j -= 2 {
+			a, b := vc[i][0]-vc[j-1][0], vc[i][1]-vc[j-1][1]
+			if a == b && (a*b)%k == 0 {
+				ans++
+			}
+		}
+	}
+	return ans
 }

leetcode/2901-3000/2947.Count-Beautiful-Substrings-I/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", "baeyh", 2, 2},
+		{"TestCase2", "abba", 1, 3},
+		{"TestCase3", "bcdf", 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.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() {
 }