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45 changes: 45 additions & 0 deletions leetcode/2801-2900/2895.Minimum-Processing-Time/README.md
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# [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
18 changes: 18 additions & 0 deletions leetcode/2801-2900/2895.Minimum-Processing-Time/Solution.go
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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
}
38 changes: 38 additions & 0 deletions leetcode/2801-2900/2895.Minimum-Processing-Time/Solution_test.go
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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() {
}
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