Add solution and test-cases for problem 2411

leetcode/2401-2500/2411.Smallest-Subarrays-With-Maximum-Bitwise-OR/README.md

@@ -1,28 +1,41 @@
 # [2411.Smallest Subarrays With Maximum Bitwise OR][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
+You are given a **0-indexed** array `nums` of length `n`, consisting of non-negative integers. For each index `i` from `0` to `n - 1`, you must determine the size of the **minimum sized** non-empty subarray of `nums` starting at `i` (inclusive) that has the **maximum** possible **bitwise OR**.
+
+- In other words, let `Bij` be the bitwise OR of the subarray `nums[i...j]`. You need to find the smallest subarray starting at `i`, such that bitwise OR of this subarray is equal to `max(Bik`) where `i <= k <= n - 1`.
+
+The bitwise OR of an array is the bitwise OR of all the numbers in it.
+
+Return an integer array `answer` of size `n` where `answer[i]` is the length of the **minimum** sized subarray starting at `i` with **maximum** bitwise OR.
+
+A **subarray** is a contiguous non-empty sequence of elements within an array.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: nums = [1,0,2,1,3]
+Output: [3,3,2,2,1]
+Explanation:
+The maximum possible bitwise OR starting at any index is 3. 
+- Starting at index 0, the shortest subarray that yields it is [1,0,2].
+- Starting at index 1, the shortest subarray that yields the maximum bitwise OR is [0,2,1].
+- Starting at index 2, the shortest subarray that yields the maximum bitwise OR is [2,1].
+- Starting at index 3, the shortest subarray that yields the maximum bitwise OR is [1,3].
+- Starting at index 4, the shortest subarray that yields the maximum bitwise OR is [3].
+Therefore, we return [3,3,2,2,1].
 ```
 
-## 题意
-> ...
-
-## 题解
+**Example 2:**
 
-### 思路1
-> ...
-Smallest Subarrays With Maximum Bitwise OR
-```go
 ```
-
+Input: nums = [1,2]
+Output: [2,1]
+Explanation:
+Starting at index 0, the shortest subarray that yields the maximum bitwise OR is of length 2.
+Starting at index 1, the shortest subarray that yields the maximum bitwise OR is of length 1.
+Therefore, we return [2,1].
+```
 
 ## 结语
 

leetcode/2401-2500/2411.Smallest-Subarrays-With-Maximum-Bitwise-OR/Solution.go

@@ -1,5 +1,64 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+import "sort"
+
+func oneArray(n int) [32]int {
+	res := [32]int{}
+	index := 31
+	for ; n > 0; index-- {
+		if n&1 == 1 {
+			res[index]++
+		}
+		n >>= 1
+	}
+	return res
+}
+
+func Solution(nums []int) []int {
+	l := len(nums)
+	ans := make([]int, l)
+	ans[l-1] = 1
+
+	or := make([]int, l)
+	or[l-1] = nums[l-1]
+	for i := l - 2; i >= 0; i-- {
+		or[i] = nums[i] | or[i+1]
+	}
+
+	oneCount := make([][32]int, l)
+	oneCount[0] = oneArray(nums[0])
+	for i := 1; i < l; i++ {
+		cur := oneArray(nums[i])
+		for j := 0; j < 32; j++ {
+			oneCount[i][j] = oneCount[i-1][j] + cur[j]
+		}
+	}
+	var ok func(int, int, int) bool
+	ok = func(start, end, target int) bool {
+		tmp := oneCount[end]
+		if start != 0 {
+			for i := 0; i < 32; i++ {
+				tmp[i] -= oneCount[start-1][i]
+			}
+		}
+		base := 1
+		num := 0
+		for i := 31; i >= 0; i-- {
+			x := 1
+			if tmp[i] == 0 {
+				x = 0
+			}
+			num += x * base
+			base <<= 1
+		}
+		return num == target
+	}
+	for i := l - 2; i >= 0; i-- {
+		// i = 3
+		index := sort.Search(l-i, func(j int) bool {
+			return ok(i, i+j, or[i])
+		})
+		ans[i] = index + 1
+	}
+	return ans
 }

leetcode/2401-2500/2411.Smallest-Subarrays-With-Maximum-Bitwise-OR/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{1, 0, 2, 1, 3}, []int{3, 3, 2, 2, 1}},
+		{"TestCase2", []int{1, 2}, []int{2, 1}},
 	}
 
 	//	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() {
 }