Detect Pattern of Length M Repeated K or More Times

Problem Statement:

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.



Solution:
class Solution:
    def containsPattern(self, arr: List[int], m: int, k: int) -> bool:
        
        ### O(nk); O(n)
        
        # initialize dict which will keep track of patterns
        patterns=defaultdict(int)
        
        # set left and right pointer
        left=0
        right=m
        
        while right<=len(arr):
            
            # if current pattern is previously visited and is consecutive to current pattern
            if tuple(arr[left:right]+[left-1]) in patterns:
                
                # update end index of current pattern and increment count
                patterns[tuple(arr[left:right]+[right-1])]=patterns[tuple(arr[left:right]+[left-1])]+1
                
                # check if cnt is greater or equal to k
                if patterns[tuple(arr[left:right]+[right-1])]>=k:
                    return True
            else:
                
                # set current pattern as visited with its ending index
                patterns[tuple(arr[left:right]+[right-1])]=1
                
            # increment left and right pointers
            left+=1
            right+=1
            
        return False