Question

3023. Find Pattern in Infinite Stream I

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Description

You are given a binary array pattern and an object stream of class InfiniteStream representing a 0-indexed infinite stream of bits.

The class InfiniteStream contains the following function:

• int next(): Reads a single bit (which is either 0 or 1) from the stream and returns it.

Return _the first starting index where the pattern matches the bits read from the stream_. For example, if the pattern is [1, 0], the first match is the highlighted part in the stream [0, 1, 0, 1, ...].

 

Example 1:

Input: stream = [1,1,1,0,1,1,1,...], pattern = [0,1]
Output: 3
Explanation: The first occurrence of the pattern [0,1] is highlighted in the stream [1,1,1,0,1,...], which starts at index 3.

Example 2:

Input: stream = [0,0,0,0,...], pattern = [0]
Output: 0
Explanation: The first occurrence of the pattern [0] is highlighted in the stream [0,...], which starts at index 0.

Example 3:

Input: stream = [1,0,1,1,0,1,1,0,1,...], pattern = [1,1,0,1]
Output: 2
Explanation: The first occurrence of the pattern [1,1,0,1] is highlighted in the stream [1,0,1,1,0,1,...], which starts at index 2.

 

Constraints:

• 1 <= pattern.length <= 100
• pattern consists only of 0 and 1.
• stream consists only of 0 and 1.
• The input is generated such that the pattern's start index exists in the first 105 bits of the stream.

Solutions

Solution 1: Bit Manipulation + Sliding Window

We notice that the length of the array $pattern$ does not exceed $100$, therefore, we can use two $64$-bit integers $a$ and $b$ to represent the binary numbers of the left and right halves of $pattern$.

Next, we traverse the data stream, also maintaining two $64$-bit integers $x$ and $y$ to represent the binary numbers of the current window of the length of $pattern$. If the current length reaches the window length, we compare whether $a$ and $x$ are equal, and whether $b$ and $y$ are equal. If they are, we return the index of the current data stream.

The time complexity is $O(n + m)$, where $n$ and $m$ are the number of elements in the data stream and $pattern$ respectively. The space complexity is $O(1)$.

# Definition for an infinite stream.
# class InfiniteStream:
#   def next(self) -> int:
#     pass
class Solution:
  def findPattern(
    self, stream: Optional["InfiniteStream"], pattern: List[int]
  ) -> int:
    a = b = 0
    m = len(pattern)
    half = m >> 1
    mask1 = (1 << half) - 1
    mask2 = (1 << (m - half)) - 1
    for i in range(half):
      a |= pattern[i] << (half - 1 - i)
    for i in range(half, m):
      b |= pattern[i] << (m - 1 - i)
    x = y = 0
    for i in count(1):
      v = stream.next()
      y = y << 1 | v
      v = y >> (m - half) & 1
      y &= mask2
      x = x << 1 | v
      x &= mask1
      if i >= m and a == x and b == y:
        return i - m

/**
 * Definition for an infinite stream.
 * class InfiniteStream {
 *   public InfiniteStream(int[] bits);
 *   public int next();
 * }
 */
class Solution {
  public int findPattern(InfiniteStream infiniteStream, int[] pattern) {
    long a = 0, b = 0;
    int m = pattern.length;
    int half = m >> 1;
    long mask1 = (1L << half) - 1;
    long mask2 = (1L << (m - half)) - 1;
    for (int i = 0; i < half; ++i) {
      a |= (long) pattern[i] << (half - 1 - i);
    }
    for (int i = half; i < m; ++i) {
      b |= (long) pattern[i] << (m - 1 - i);
    }
    long x = 0, y = 0;
    for (int i = 1;; ++i) {
      int v = infiniteStream.next();
      y = y << 1 | v;
      v = (int) ((y >> (m - half)) & 1);
      y &= mask2;
      x = x << 1 | v;
      x &= mask1;
      if (i >= m && a == x && b == y) {
        return i - m;
      }
    }
  }
}

/**
 * Definition for an infinite stream.
 * class InfiniteStream {
 * public:
 *   InfiniteStream(vector<int> bits);
 *   int next();
 * };
 */
class Solution {
public:
  int findPattern(InfiniteStream* stream, vector<int>& pattern) {
    long long a = 0, b = 0;
    int m = pattern.size();
    int half = m >> 1;
    long long mask1 = (1LL << half) - 1;
    long long mask2 = (1LL << (m - half)) - 1;
    for (int i = 0; i < half; ++i) {
      a |= (long long) pattern[i] << (half - 1 - i);
    }
    for (int i = half; i < m; ++i) {
      b |= (long long) pattern[i] << (m - 1 - i);
    }
    long x = 0, y = 0;
    for (int i = 1;; ++i) {
      int v = stream->next();
      y = y << 1 | v;
      v = (int) ((y >> (m - half)) & 1);
      y &= mask2;
      x = x << 1 | v;
      x &= mask1;
      if (i >= m && a == x && b == y) {
        return i - m;
      }
    }
  }
};

/**
 * Definition for an infinite stream.
 * type InfiniteStream interface {
 *   Next() int
 * }
 */
 func findPattern(stream InfiniteStream, pattern []int) int {
  a, b := 0, 0
  m := len(pattern)
  half := m >> 1
  mask1 := (1 << half) - 1
  mask2 := (1 << (m - half)) - 1
  for i := 0; i < half; i++ {
    a |= pattern[i] << (half - 1 - i)
  }
  for i := half; i < m; i++ {
    b |= pattern[i] << (m - 1 - i)
  }
  x, y := 0, 0
  for i := 1; ; i++ {
    v := stream.Next()
    y = y<<1 | v
    v = (y >> (m - half)) & 1
    y &= mask2
    x = x<<1 | v
    x &= mask1
    if i >= m && a == x && b == y {
      return i - m
    }
  }
}