Merge pull request #839 from 0xff-dev/2073

Add solution and test-cases for problem 2073

Author: 6boris

leetcode/2001-2100/2073.Time-Needed-to-Buy-Tickets/README.md

@@ -1,28 +1,35 @@
 # [2073.Time Needed to Buy Tickets][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
+There are `n` people in a line queuing to buy tickets, where the 0<sup>th</sup> person is at the **front** of the line and the (n - 1)<sup>th</sup> person is at the **back** of the line.
+
+You are given a **0-indexed** integer array `tickets` of length `n` where the number of tickets that the i<sup>th</sup> person would like to buy is `tickets[i]`.
+
+Each person takes **exactly 1 second** to buy a ticket. A person can only buy **1 ticket at a time** and has to go back to **the end** of the line (which happens **instantaneously**) in order to buy more tickets. If a person does not have any tickets left to buy, the person will **leave** the line.
+
+Return the **time taken** for the person at position k (**0-indexed**) to finish buying tickets.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: tickets = [2,3,2], k = 2
+Output: 6
+Explanation: 
+- In the first pass, everyone in the line buys a ticket and the line becomes [1, 2, 1].
+- In the second pass, everyone in the line buys a ticket and the line becomes [0, 1, 0].
+The person at position 2 has successfully bought 2 tickets and it took 3 + 3 = 6 seconds.
 ```
 
-## 题意
-> ...
+**Example 2:**
 
-## 题解
-
-### 思路1
-> ...
-Time Needed to Buy Tickets
-```go
 ```
-
+Input: tickets = [5,1,1,1], k = 0
+Output: 8
+Explanation:
+- In the first pass, everyone in the line buys a ticket and the line becomes [4, 0, 0, 0].
+- In the next 4 passes, only the person in position 0 is buying tickets.
+The person at position 0 has successfully bought 5 tickets and it took 4 + 1 + 1 + 1 + 1 = 8 seconds.
+```
 
 ## 结语
 

leetcode/2001-2100/2073.Time-Needed-to-Buy-Tickets/Solution.go

@@ -1,5 +1,16 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(tickets []int, k int) int {
+	// 每个人1s的轮下来
+	// 对于左侧的人，和右侧的人有差异，差异点在于：最后一轮到k后，右侧就不需要再走了
+	// 所以最后从整体看就是，左侧:min(tickets[k], ticket[i]), 右侧就是min(tickets[k]-1,tickets[i])
+	time := 0
+	for i := 0; i < len(tickets); i++ {
+		if i <= k {
+			time += min(tickets[i], tickets[k])
+		} else {
+			time += min(tickets[k]-1, tickets[i])
+		}
+	}
+	return time
 }

leetcode/2001-2100/2073.Time-Needed-to-Buy-Tickets/Solution_test.go

@@ -10,30 +10,30 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		inputs []int
+		k      int
+		expect int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", []int{2, 3, 2}, 2, 6},
+		{"TestCase2", []int{5, 1, 1, 1}, 0, 8},
 	}
 
 	//	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.inputs, 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 5v",
+					c.expect, got, c.inputs, 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() {
 }