From afef753f6a065f1081fdce7a36786247768d142f Mon Sep 17 00:00:00 2001
From: 0xff-dev <stevenshuang521@gmail.com>
Date: Thu, 8 May 2025 09:03:59 +0800
Subject: [PATCH] Add solution and test-cases for problem 2895

---
 .../2895.Minimum-Processing-Time/README.md    | 45 +++++++++++++++++++
 .../2895.Minimum-Processing-Time/Solution.go  | 18 ++++++++
 .../Solution_test.go                          | 38 ++++++++++++++++
 3 files changed, 101 insertions(+)
 create mode 100644 leetcode/2801-2900/2895.Minimum-Processing-Time/README.md
 create mode 100755 leetcode/2801-2900/2895.Minimum-Processing-Time/Solution.go
 create mode 100755 leetcode/2801-2900/2895.Minimum-Processing-Time/Solution_test.go

diff --git a/leetcode/2801-2900/2895.Minimum-Processing-Time/README.md b/leetcode/2801-2900/2895.Minimum-Processing-Time/README.md
new file mode 100644
index 00000000..5ce1b315
--- /dev/null
+++ b/leetcode/2801-2900/2895.Minimum-Processing-Time/README.md
@@ -0,0 +1,45 @@
+# [2895.Minimum Processing Time][title]
+
+## Description
+You have a certain number of processors, each having 4 cores. The number of tasks to be executed is four times the number of processors. Each task must be assigned to a unique core, and each core can only be used once.
+
+You are given an array `processorTime` representing the time each processor becomes available and an array `tasks` representing how long each task takes to complete. Return the minimum time needed to complete all tasks.
+
+**Example 1:**
+
+```
+Input: processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]
+
+Output: 16
+
+Explanation:
+
+Assign the tasks at indices 4, 5, 6, 7 to the first processor which becomes available at time = 8, and the tasks at indices 0, 1, 2, 3 to the second processor which becomes available at time = 10. 
+
+The time taken by the first processor to finish the execution of all tasks is max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16.
+
+The time taken by the second processor to finish the execution of all tasks is max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13.
+```
+
+**Example 2:**
+
+```
+Input: processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]
+
+Output: 23
+
+Explanation:
+
+Assign the tasks at indices 1, 4, 5, 6 to the first processor and the others to the second processor.
+
+The time taken by the first processor to finish the execution of all tasks is max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18.
+
+The time taken by the second processor to finish the execution of all tasks is max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23.
+```
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-golang-algorithm][me]
+
+[title]: https://leetcode.com/problems/minimum-processing-time
+[me]: https://github.com/kylesliu/awesome-golang-algorithm
diff --git a/leetcode/2801-2900/2895.Minimum-Processing-Time/Solution.go b/leetcode/2801-2900/2895.Minimum-Processing-Time/Solution.go
new file mode 100755
index 00000000..ac1b8174
--- /dev/null
+++ b/leetcode/2801-2900/2895.Minimum-Processing-Time/Solution.go
@@ -0,0 +1,18 @@
+package Solution
+
+import "sort"
+
+func Solution(processorTime []int, tasks []int) int {
+	sort.Slice(tasks, func(i, j int) bool {
+		return tasks[i] > tasks[j]
+	})
+	sort.Ints(processorTime)
+	ans := 0
+	index := 0
+	for _, n := range processorTime {
+		mm := max(tasks[index], tasks[index+1], tasks[index+2], tasks[index+3])
+		index += 4
+		ans = max(ans, n+mm)
+	}
+	return ans
+}
diff --git a/leetcode/2801-2900/2895.Minimum-Processing-Time/Solution_test.go b/leetcode/2801-2900/2895.Minimum-Processing-Time/Solution_test.go
new file mode 100755
index 00000000..ff18943b
--- /dev/null
+++ b/leetcode/2801-2900/2895.Minimum-Processing-Time/Solution_test.go
@@ -0,0 +1,38 @@
+package Solution
+
+import (
+	"reflect"
+	"strconv"
+	"testing"
+)
+
+func TestSolution(t *testing.T) {
+	//	测试用例
+	cases := []struct {
+		name                 string
+		processorTime, tasks []int
+		expect               int
+	}{
+		{"TestCase1", []int{8, 10}, []int{2, 2, 3, 1, 8, 7, 4, 5}, 16},
+		{"TestCase2", []int{10, 20}, []int{2, 3, 1, 2, 5, 8, 4, 3}, 23},
+	}
+
+	//	开始测试
+	for i, c := range cases {
+		t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
+			got := Solution(c.processorTime, c.tasks)
+			if !reflect.DeepEqual(got, c.expect) {
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v %v",
+					c.expect, got, c.processorTime, c.tasks)
+			}
+		})
+	}
+}
+
+// 压力测试
+func BenchmarkSolution(b *testing.B) {
+}
+
+// 使用案列
+func ExampleSolution() {
+}