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Commit 12c3285

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Merge pull request #1272 from 0xff-dev/2411
Add solution and test-cases for problem 2411
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‎leetcode/2401-2500/2411.Smallest-Subarrays-With-Maximum-Bitwise-OR/README.md‎

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# [2411.Smallest Subarrays With Maximum Bitwise OR][title]
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> [!WARNING|style:flat]
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> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
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## Description
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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**.
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- 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`.
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The bitwise OR of an array is the bitwise OR of all the numbers in it.
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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.
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A **subarray** is a contiguous non-empty sequence of elements within an array.
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**Example 1:**
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```
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Input: a = "11", b = "1"
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Output: "100"
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Input: nums = [1,0,2,1,3]
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Output: [3,3,2,2,1]
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Explanation:
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The maximum possible bitwise OR starting at any index is 3.
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- Starting at index 0, the shortest subarray that yields it is [1,0,2].
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- Starting at index 1, the shortest subarray that yields the maximum bitwise OR is [0,2,1].
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- Starting at index 2, the shortest subarray that yields the maximum bitwise OR is [2,1].
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- Starting at index 3, the shortest subarray that yields the maximum bitwise OR is [1,3].
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- Starting at index 4, the shortest subarray that yields the maximum bitwise OR is [3].
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Therefore, we return [3,3,2,2,1].
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```
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## 题意
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> ...
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## 题解
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**Example 2:**
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### 思路1
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> ...
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Smallest Subarrays With Maximum Bitwise OR
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```go
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```
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Input: nums = [1,2]
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Output: [2,1]
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Explanation:
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Starting at index 0, the shortest subarray that yields the maximum bitwise OR is of length 2.
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Starting at index 1, the shortest subarray that yields the maximum bitwise OR is of length 1.
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Therefore, we return [2,1].
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```
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## 结语
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package Solution
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func Solution(x bool) bool {
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return x
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import "sort"
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func oneArray(n int) [32]int {
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res := [32]int{}
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index := 31
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for ; n > 0; index-- {
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if n&1 == 1 {
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res[index]++
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}
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n >>= 1
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}
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return res
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}
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func Solution(nums []int) []int {
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l := len(nums)
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ans := make([]int, l)
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ans[l-1] = 1
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or := make([]int, l)
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or[l-1] = nums[l-1]
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for i := l - 2; i >= 0; i-- {
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or[i] = nums[i] | or[i+1]
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}
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oneCount := make([][32]int, l)
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oneCount[0] = oneArray(nums[0])
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for i := 1; i < l; i++ {
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cur := oneArray(nums[i])
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for j := 0; j < 32; j++ {
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oneCount[i][j] = oneCount[i-1][j] + cur[j]
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}
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}
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var ok func(int, int, int) bool
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ok = func(start, end, target int) bool {
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tmp := oneCount[end]
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if start != 0 {
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for i := 0; i < 32; i++ {
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tmp[i] -= oneCount[start-1][i]
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}
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}
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base := 1
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num := 0
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for i := 31; i >= 0; i-- {
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x := 1
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if tmp[i] == 0 {
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x = 0
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}
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num += x * base
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base <<= 1
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}
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return num == target
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}
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for i := l - 2; i >= 0; i-- {
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// i = 3
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index := sort.Search(l-i, func(j int) bool {
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return ok(i, i+j, or[i])
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})
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ans[i] = index + 1
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}
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return ans
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}

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

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// 测试用例
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cases := []struct {
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name string
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inputs bool
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expect bool
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inputs []int
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expect []int
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}{
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{"TestCase", true, true},
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{"TestCase", true, true},
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{"TestCase", false, false},
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{"TestCase1", []int{1, 0, 2, 1, 3}, []int{3, 3, 2, 2, 1}},
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{"TestCase2", []int{1, 2}, []int{2, 1}},
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}
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// 开始测试
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}
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}
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//压力测试
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//压力测试
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func BenchmarkSolution(b *testing.B) {
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}
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//使用案列
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//使用案列
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func ExampleSolution() {
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}

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