Merge pull request #1343 from 0xff-dev/3350

Add solution and test-cases for problem 3350

Author: 6boris

leetcode/3301-3400/3350.Adjacent-Increasing-Subarrays-Detection-II/README.md

@@ -1,28 +1,42 @@
 # [3350.Adjacent Increasing Subarrays Detection II][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
+Given an array `nums` of `n` integers, your task is to find the **maximum** value of `k` for which there exist **two** adjacent subarrays of length `k` each, such that both subarrays are **strictly increasing**. Specifically, check if there are **two** subarrays of length `k` starting at indices `a` and `b` (`a < b`), where:
+
+- Both subarrays `nums[a..a + k - 1]` and `nums[b..b + k - 1]` are **strictly increasing**.
+- The subarrays must be **adjacent**, meaning `b = a + k`.
+
+Return the **maximum** possible value of `k`.
+
+A **subarray** is a contiguous **non-empty** sequence of elements within an array.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
-```
+Input: nums = [2,5,7,8,9,2,3,4,3,1]
 
-## 题意
-> ...
+Output: 3
 
-## 题解
+Explanation:
 
-### 思路1
-> ...
-Adjacent Increasing Subarrays Detection II
-```go
+The subarray starting at index 2 is [7, 8, 9], which is strictly increasing.
+The subarray starting at index 5 is [2, 3, 4], which is also strictly increasing.
+These two subarrays are adjacent, and 3 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.
 ```
 
+**Example 2:**
+
+```
+Input: nums = [1,2,3,4,4,4,4,5,6,7]
+
+Output: 2
+
+Explanation:
+
+The subarray starting at index 0 is [1, 2], which is strictly increasing.
+The subarray starting at index 2 is [3, 4], which is also strictly increasing.
+These two subarrays are adjacent, and 2 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.
+```
 
 ## 结语
 

leetcode/3301-3400/3350.Adjacent-Increasing-Subarrays-Detection-II/Solution.go

@@ -1,5 +1,42 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+import "sort"
+
+func ok(nums []int, k int) bool {
+	// 存储所有ok的字数组的下标
+	// 然后找index的间隔是否存在不想等就可以了
+	indies := []int{}
+	start, end := 0, 0
+	curLen := 0
+	pre := -1001
+	// 6, 13, -17, -20, 2
+	for ; end < len(nums); end++ {
+		if nums[end] <= pre {
+			start, curLen = end, 1
+		} else {
+			curLen++
+		}
+		pre = nums[end]
+		if curLen == k {
+			indies = append(indies, start)
+			start++
+			curLen--
+		}
+	}
+	keys := make(map[int]struct{})
+	for _, index := range indies {
+		keys[index] = struct{}{}
+	}
+	for i := 0; i < len(indies)-1; i++ {
+		if _, ok := keys[indies[i]+k]; ok {
+			return true
+		}
+	}
+	return false
+}
+func Solution(nums []int) int {
+	index := sort.Search(len(nums)/2, func(i int) bool {
+		return !ok(nums, i+1)
+	})
+	return index
 }

leetcode/3301-3400/3350.Adjacent-Increasing-Subarrays-Detection-II/Solution_test.go

@@ -10,12 +10,11 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		inputs []int
+		expect int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", []int{2, 5, 7, 8, 9, 2, 3, 4, 3, 1}, 3},
+		{"TestCase2", []int{1, 2, 3, 4, 4, 4, 4, 5, 6, 7}, 2},
 	}
 
 	//	开始测试
@@ -30,10 +29,10 @@ func TestSolution(t *testing.T) {
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }