Merge pull request #1322 from 0xff-dev/1817

Add solution and test-cases for problem 1817

Author: 6boris

leetcode/1801-1900/1817.Finding-the-Users-Active-Minutes/README.md

@@ -1,28 +1,38 @@
 # [1817.Finding the Users Active Minutes][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 the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with IDi performed an action at the minute `timei`.
+
+**Multiple users** can perform actions simultaneously, and a single user can perform **multiple actions** in the same minute.
+
+The **user active minutes (UAM)** for a given user is defined as the **number of unique minutes** in which the user performed an action on LeetCode. A minute can only be counted once, even if multiple actions occur during it.
+
+You are to calculate a **1-indexed** array `answer` of size `k` such that, for each `j` (`1 <= j <= k`), `answer[j]` is the **number of users** whose **UAM** equals `j`.
+
+Return the array `answer` as described above.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: logs = [[0,5],[1,2],[0,2],[0,5],[1,3]], k = 5
+Output: [0,2,0,0,0]
+Explanation:
+The user with ID=0 performed actions at minutes 5, 2, and 5 again. Hence, they have a UAM of 2 (minute 5 is only counted once).
+The user with ID=1 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.
+Since both users have a UAM of 2, answer[2] is 2, and the remaining answer[j] values are 0.
 ```
 
-## 题意
-> ...
-
-## 题解
+**Example 2:**
 
-### 思路1
-> ...
-Finding the Users Active Minutes
-```go
 ```
-
+Input: logs = [[1,1],[2,2],[2,3]], k = 4
+Output: [1,1,0,0]
+Explanation:
+The user with ID=1 performed a single action at minute 1. Hence, they have a UAM of 1.
+The user with ID=2 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.
+There is one user with a UAM of 1 and one with a UAM of 2.
+Hence, answer[1] = 1, answer[2] = 1, and the remaining values are 0.
+```
 
 ## 结语
 

leetcode/1801-1900/1817.Finding-the-Users-Active-Minutes/Solution.go

@@ -1,5 +1,18 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(logs [][]int, k int) []int {
+	ret := make([]int, k)
+	// 1:1
+	// 2:2
+	count := make(map[int]map[int]struct{})
+	for _, log := range logs {
+		if _, ok := count[log[0]]; !ok {
+			count[log[0]] = make(map[int]struct{})
+		}
+		count[log[0]][log[1]] = struct{}{}
+	}
+	for _, c := range count {
+		ret[len(c)-1]++
+	}
+	return ret
 }

leetcode/1801-1900/1817.Finding-the-Users-Active-Minutes/Solution_test.go

@@ -10,30 +10,30 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		logs   [][]int
+		k      int
+		expect []int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", [][]int{{0, 5}, {1, 2}, {0, 2}, {0, 5}, {1, 3}}, 5, []int{0, 2, 0, 0, 0}},
+		{"TestCase2", [][]int{{1, 1}, {2, 2}, {2, 3}}, 4, []int{1, 1, 0, 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.logs, 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.logs, 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() {
 }