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Commit 5aa0cea

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feat: add solutions to lc problem: No.3023 (doocs#2293)
No.3023.Find Pattern in Infinite Stream I
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‎solution/2200-2299/2207.Maximize Number of Subsequences in a String/README.md‎

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# [2207. 字符串中最多数目的子字符串](https://leetcode.cn/problems/maximize-number-of-subsequences-in-a-string)
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# [2207. 字符串中最多数目的子序列](https://leetcode.cn/problems/maximize-number-of-subsequences-in-a-string)
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[English Version](/solution/2200-2299/2207.Maximize%20Number%20of%20Subsequences%20in%20a%20String/README_EN.md)
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# [3023. Find Pattern in Infinite Stream I](https://leetcode.cn/problems/find-pattern-in-infinite-stream-i)
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[English Version](/solution/3000-3099/3023.Find%20Pattern%20in%20Infinite%20Stream%20I/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>You are given a binary array <code>pattern</code> and an object <code>stream</code> of class <code>InfiniteStream</code> representing a <strong>0-indexed</strong> infinite stream of bits.</p>
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<p>The class <code>InfiniteStream</code> contains the following function:</p>
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<ul>
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<li><code>int next()</code>: Reads a <strong>single</strong> bit (which is either <code>0</code> or <code>1</code>) from the stream and returns it.</li>
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</ul>
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<p>Return <em>the <strong>first starting</strong> index where the pattern matches the bits read from the stream</em>. For example, if the pattern is <code>[1, 0]</code>, the first match is the highlighted part in the stream <code>[0, <strong><u>1, 0</u></strong>, 1, ...]</code>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> stream = [1,1,1,0,1,1,1,...], pattern = [0,1]
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> The first occurrence of the pattern [0,1] is highlighted in the stream [1,1,1,<strong><u>0,1</u></strong>,...], which starts at index 3.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> stream = [0,0,0,0,...], pattern = [0]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> The first occurrence of the pattern [0] is highlighted in the stream [<strong><u>0</u></strong>,...], which starts at index 0.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> stream = [1,0,1,1,0,1,1,0,1,...], pattern = [1,1,0,1]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong> The first occurrence of the pattern [1,1,0,1] is highlighted in the stream [1,0,<strong><u>1,1,0,1</u></strong>,...], which starts at index 2.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= pattern.length &lt;= 100</code></li>
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<li><code>pattern</code> consists only of <code>0</code> and <code>1</code>.</li>
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<li><code>stream</code> consists only of <code>0</code> and <code>1</code>.</li>
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<li>The input is generated such that the pattern&#39;s start index exists in the first <code>10<sup>5</sup></code> bits of the stream.</li>
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</ul>
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## 解法
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### 方法一:位运算 + 滑动窗口
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我们注意到,数组 $pattern$ 的长度不超过 100ドル,ドル因此,我们可以用两个 64ドル$ 位的整数 $a$ 和 $b$ 来表示 $pattern$ 左右两半的二进制数。
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接下来,我们遍历数据流,同样维护两个 64ドル$ 位的整数 $x$ 和 $y$ 表示当前 $pattern$ 长度的窗口的二进制数。如果当前达到了窗口的长度,我们比较 $a$ 和 $x$ 是否相等,以及 $b$ 和 $y$ 是否相等,如果是,我们返回当前数据流的索引即可。
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时间复杂度 $O(n + m),ドル其中 $n$ 和 $m$ 分别是数据流和 $pattern$ 的元素个数。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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# Definition for an infinite stream.
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# class InfiniteStream:
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# def next(self) -> int:
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# pass
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class Solution:
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def findPattern(
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self, stream: Optional["InfiniteStream"], pattern: List[int]
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) -> int:
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a = b = 0
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m = len(pattern)
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half = m >> 1
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mask1 = (1 << half) - 1
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mask2 = (1 << (m - half)) - 1
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for i in range(half):
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a |= pattern[i] << (half - 1 - i)
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for i in range(half, m):
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b |= pattern[i] << (m - 1 - i)
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x = y = 0
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for i in count(1):
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v = stream.next()
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y = y << 1 | v
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v = y >> (m - half) & 1
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y &= mask2
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x = x << 1 | v
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x &= mask1
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if i >= m and a == x and b == y:
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return i - m
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```
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```java
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/**
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* Definition for an infinite stream.
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* class InfiniteStream {
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* public InfiniteStream(int[] bits);
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* public int next();
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* }
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*/
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class Solution {
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public int findPattern(InfiniteStream infiniteStream, int[] pattern) {
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long a = 0, b = 0;
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int m = pattern.length;
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int half = m >> 1;
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long mask1 = (1L << half) - 1;
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long mask2 = (1L << (m - half)) - 1;
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for (int i = 0; i < half; ++i) {
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a |= (long) pattern[i] << (half - 1 - i);
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}
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for (int i = half; i < m; ++i) {
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b |= (long) pattern[i] << (m - 1 - i);
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}
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long x = 0, y = 0;
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for (int i = 1;; ++i) {
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int v = infiniteStream.next();
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y = y << 1 | v;
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v = (int) ((y >> (m - half)) & 1);
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y &= mask2;
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x = x << 1 | v;
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x &= mask1;
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if (i >= m && a == x && b == y) {
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return i - m;
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}
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}
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}
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}
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```
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```cpp
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/**
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* Definition for an infinite stream.
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* class InfiniteStream {
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* public:
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* InfiniteStream(vector<int> bits);
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* int next();
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* };
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*/
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class Solution {
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public:
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int findPattern(InfiniteStream* stream, vector<int>& pattern) {
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long long a = 0, b = 0;
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int m = pattern.size();
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int half = m >> 1;
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long long mask1 = (1LL << half) - 1;
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long long mask2 = (1LL << (m - half)) - 1;
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for (int i = 0; i < half; ++i) {
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a |= (long long) pattern[i] << (half - 1 - i);
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}
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for (int i = half; i < m; ++i) {
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b |= (long long) pattern[i] << (m - 1 - i);
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}
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long x = 0, y = 0;
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for (int i = 1;; ++i) {
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int v = stream->next();
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y = y << 1 | v;
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v = (int) ((y >> (m - half)) & 1);
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y &= mask2;
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x = x << 1 | v;
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x &= mask1;
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if (i >= m && a == x && b == y) {
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return i - m;
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}
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}
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}
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};
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```
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```go
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/**
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* Definition for an infinite stream.
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* type InfiniteStream interface {
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* Next() int
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* }
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*/
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func findPattern(stream InfiniteStream, pattern []int) int {
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a, b := 0, 0
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m := len(pattern)
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half := m >> 1
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mask1 := (1 << half) - 1
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mask2 := (1 << (m - half)) - 1
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for i := 0; i < half; i++ {
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a |= pattern[i] << (half - 1 - i)
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}
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for i := half; i < m; i++ {
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b |= pattern[i] << (m - 1 - i)
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}
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x, y := 0, 0
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for i := 1; ; i++ {
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v := stream.Next()
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y = y<<1 | v
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v = (y >> (m - half)) & 1
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y &= mask2
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x = x<<1 | v
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x &= mask1
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if i >= m && a == x && b == y {
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return i - m
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}
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}
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}
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```
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<!-- tabs:end -->
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<!-- end -->

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