Algorithms to Detect Pattern of Length M Repeated K or More Times in a String

Given an array of positive integers arr, find a pattern of length m that is repeated k or more times.

A pattern is a subarray (consecutive sub-sequence) that consists of one or more values, repeated multiple times consecutively without overlapping. A pattern is defined by its length and the number of repetitions.

Return true if there exists a pattern of length m that is repeated k or more times, otherwise return false.

Example 1:
Input: arr = [1,2,4,4,4,4], m = 1, k = 3
Output: true
Explanation: The pattern (4) of length 1 is repeated 4 consecutive times. Notice that pattern can be repeated k or more times but not less.

Example 2:
Input: arr = [1,2,1,2,1,1,1,3], m = 2, k = 2
Output: true
Explanation: The pattern (1,2) of length 2 is repeated 2 consecutive times. Another valid pattern (2,1) is also repeated 2 times.

Example 3:
Input: arr = [1,2,1,2,1,3], m = 2, k = 3
Output: false
Explanation: The pattern (1,2) is of length 2 but is repeated only 2 times. There is no pattern of length 2 that is repeated 3 or more times.

Example 4:
Input: arr = [1,2,3,1,2], m = 2, k = 2
Output: false
Explanation: Notice that the pattern (1,2) exists twice but not consecutively, so it doesn’t count.

Example 5:
Input: arr = [2,2,2,2], m = 2, k = 3
Output: false
Explanation: The only pattern of length 2 is (2,2) however it’s repeated only twice. Notice that we do not count overlapping repetitions.

Constraints:
2 <= arr.length <= 100
1 <= arr[i] <= 100
1 <= m <= 100
2 <= k <= 100

Bruteforce Algorithm to Determine the Repeative String Pattern

Given the m and k are only less than 100 – we can totally do bruteforce algorithm. First loop the start index of the first pattern – then, we know the first pattern looks like – and we can construct k such patterns – and see if it equals to the sub array of the string.

The following bruteforce implemented in Python has O(N^2) time complexity.

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class Solution:
    def containsPattern(self, arr: List[int], m: int, k: int) -> bool:
        L = len(arr)
        for i in range(L - m * k + 1):
            p = arr[i:i+m]*k
            if p == arr[i:i+m*k]:
                return True
        return False

class Solution:
    def containsPattern(self, arr: List[int], m: int, k: int) -> bool:
        L = len(arr)
        for i in range(L - m * k + 1):
            p = arr[i:i+m]*k
            if p == arr[i:i+m*k]:
                return True
        return False

Better Bruteforce Algorithm to Detect Pattern of Length M Repeated K or More Times

We can do this in a better way. We can check if the value at index i and index i + m equals – and increment the counter. If the counter equals (k-1)*m then we have k times of m-size pattern.

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class Solution:
    def containsPattern(self, arr: List[int], m: int, k: int) -> bool:
        L = len(arr)
        cnt = 0
        for i in range(L - m):
            if arr[i] == arr[i + m]:
                cnt += 1
            else:
                cnt = 0
            if cnt == m * (k - 1):
                return True
        return False            

class Solution:
    def containsPattern(self, arr: List[int], m: int, k: int) -> bool:
        L = len(arr)
        cnt = 0
        for i in range(L - m):
            if arr[i] == arr[i + m]:
                cnt += 1
            else:
                cnt = 0
            if cnt == m * (k - 1):
                return True
        return False            

The complexity is O(N).

–EOF (The Ultimate Computing & Technology Blog) —