From f06b00ea3cdcb1c5f8d75926d18f64d60439f4df Mon Sep 17 00:00:00 2001
From: 0xff-dev <stevenshuang521@gmail.com>
Date: Wed, 2 Apr 2025 09:29:16 +0800
Subject: [PATCH] Add solution and test-cases for problem 2873

---
 .../README.md                                 | 41 +++++++++++++++++++
 .../Solution.go                               | 17 ++++++++
 .../Solution_test.go                          | 39 ++++++++++++++++++
 3 files changed, 97 insertions(+)
 create mode 100644 leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/README.md
 create mode 100644 leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/Solution.go
 create mode 100644 leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/Solution_test.go

diff --git a/leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/README.md b/leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/README.md
new file mode 100644
index 000000000..e27dbe607
--- /dev/null
+++ b/leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/README.md
@@ -0,0 +1,41 @@
+# [2873.Maximum Value of an Ordered Triplet I][title]
+
+## Description
+You are given a **0-indexed** integer array `nums`.
+
+Return **the maximum value over all triplets of indices** `(i, j, k)` such that `i < j < k`. If all such triplets have a negative value, return `0`.
+
+The **value of a triplet of indices** `(i, j, k)` is equal to `(nums[i] - nums[j]) * nums[k]`.
+
+**Example 1:**
+
+```
+Input: nums = [12,6,1,2,7]
+Output: 77
+Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
+It can be shown that there are no ordered triplets of indices with a value greater than 77.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,10,3,4,19]
+Output: 133
+Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
+It can be shown that there are no ordered triplets of indices with a value greater than 133.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,2,3]
+Output: 0
+Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.
+```
+
+## 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/maximum-value-of-an-ordered-triplet-i
+[me]: https://github.com/kylesliu/awesome-golang-algorithm
diff --git a/leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/Solution.go b/leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/Solution.go
new file mode 100644
index 000000000..e63298bf1
--- /dev/null
+++ b/leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/Solution.go
@@ -0,0 +1,17 @@
+package Solution
+
+func Solution(nums []int) int64 {
+	var ans int64
+	cut := make([]int, len(nums))
+	_max := nums[0]
+	for i := 1; i < len(nums); i++ {
+		cut[i] = _max - nums[i]
+		_max = max(_max, nums[i])
+	}
+	_max = cut[1]
+	for i := 2; i < len(nums); i++ {
+		ans = max(ans, int64(nums[i])*int64(_max))
+		_max = max(_max, cut[i])
+	}
+	return ans
+}
diff --git a/leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/Solution_test.go b/leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/Solution_test.go
new file mode 100644
index 000000000..96f1704d1
--- /dev/null
+++ b/leetcode/2801-2900/2873.Maximum-Value-of-an-Ordered-Triplet-I/Solution_test.go
@@ -0,0 +1,39 @@
+package Solution
+
+import (
+	"reflect"
+	"strconv"
+	"testing"
+)
+
+func TestSolution(t *testing.T) {
+	//	测试用例
+	cases := []struct {
+		name   string
+		inputs []int
+		expect int64
+	}{
+		{"TestCase1", []int{12, 6, 1, 2, 7}, 77},
+		{"TestCase2", []int{1, 10, 3, 4, 19}, 133},
+		{"TestCase3", []int{1, 2, 3}, 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)
+			if !reflect.DeepEqual(got, c.expect) {
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
+					c.expect, got, c.inputs)
+			}
+		})
+	}
+}
+
+// 压力测试
+func BenchmarkSolution(b *testing.B) {
+}
+
+// 使用案列
+func ExampleSolution() {
+}