Skip to content

3023. Find Pattern in Infinite Stream I πŸ”’

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 $O(1)$.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
# 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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
/**
 * 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;
            }
        }
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
/**
 * 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;
            }
        }
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
/**
 * 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
        }
    }
}

Comments