Add solution and test-cases for problem 2327

Author: 0xff-dev

leetcode/2301-2400/2327.Number-of-People-Aware-of-a-Secret/README.md

@@ -1,28 +1,37 @@
 # [2327.Number of People Aware of a Secret][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
+On day `1`, one person discovers a secret.
+
+You are given an integer `delay`, which means that each person will **share** the secret with a new person **every day**, starting from `delay` days after discovering the secret. You are also given an integer `forget`, which means that each person will **forget** the secret `forget` days after discovering it. A person **cannot** share the secret on the same day they forgot it, or on any day afterwards.
+
+Given an integer `n`, return the number of people who know the secret at the end of day `n`. Since the answer may be very large, return it **modulo** `10^9 + 7`.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: n = 6, delay = 2, forget = 4
+Output: 5
+Explanation:
+Day 1: Suppose the first person is named A. (1 person)
+Day 2: A is the only person who knows the secret. (1 person)
+Day 3: A shares the secret with a new person, B. (2 people)
+Day 4: A shares the secret with a new person, C. (3 people)
+Day 5: A forgets the secret, and B shares the secret with a new person, D. (3 people)
+Day 6: B shares the secret with E, and C shares the secret with F. (5 people)
 ```
 
-## 题意
-> ...
+**Example 2:**
 
-## 题解
-
-### 思路1
-> ...
-Number of People Aware of a Secret
-```go
 ```
-
+Input: n = 4, delay = 1, forget = 3
+Output: 6
+Explanation:
+Day 1: The first person is named A. (1 person)
+Day 2: A shares the secret with B. (2 people)
+Day 3: A and B share the secret with 2 new people, C and D. (4 people)
+Day 4: A forgets the secret. B, C, and D share the secret with 3 new people. (6 people)
+```
 
 ## 结语
 

leetcode/2301-2400/2327.Number-of-People-Aware-of-a-Secret/Solution.go

@@ -1,5 +1,28 @@
 package Solution
 
-func Solution(x bool) bool {
-return x
+func Solution(n int, delay int, forget int) int {
+MOD := 1_000_000_007
+dp := make([]int, n+forget+1)
+inc := make([]int, n+forget+1)
+
+// init
+dp[1] = 1
+dp[1+forget] = -1
+
+for j := 1 + delay; j < 1+forget; j++ {
+inc[j] = 1
+}
+
+for i := 2; i <= n; i++ {
+dp[i] = (dp[i] + dp[i-1] + inc[i]) % MOD
+dp[i+forget] = (dp[i+forget] - inc[i]) % MOD
+for j := i + delay; j < i+forget; j++ {
+inc[j] = (inc[j] + inc[i]) % MOD
+}
+}
+result := dp[n] % MOD
+if result < 0 {
+result += MOD
+}
+return result
 }

leetcode/2301-2400/2327.Number-of-People-Aware-of-a-Secret/Solution_test.go

@@ -9,31 +9,30 @@ import (
 func TestSolution(t *testing.T) {
 //	测试用例
 cases := []struct {
-name string
-inputs bool
-expect bool
+name  string
+n, delay, forget int
+expect  int
 }{
-{"TestCase", true, true},
-{"TestCase", true, true},
-{"TestCase", false, false},
+{"TestCase1", 6, 2, 4, 5},
+{"TestCase2", 4, 1, 3, 6},
 }
 
 //	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.n, c.delay, c.forget)
 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.n, c.delay, c.forget)
 }
 })
 }
 }
 
-//压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//使用案列
+// 使用案列
 func ExampleSolution() {
 }