1566. Detect Pattern of Length M Repeated K or More Times

Description

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.

 

Constraints:

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

Solution

detect-pattern-of-length-m-repeated-k-or-more-times.py

class Solution:
    def containsPattern(self, arr: List[int], m: int, k: int) -> bool:
        n = len(arr)
        res = 0

        for i in range(n-m):

            if arr[i] != arr[i+m]:
                res = 0

            res += (arr[i] == arr[i+m])

            if res == (k-1)*m:
                return True

        return False

detect-pattern-of-length-m-repeated-k-or-more-times.cpp

class Solution {
public:
    bool containsPattern(vector<int>& arr, int m, int k) {
        int cnt=0;
        for(int i=0;i+m < arr.size(); i++){

            if(arr[i]!=arr[i+m]){
              cnt=0;  
            }
            cnt += (arr[i] == arr[i+m]);
            if(cnt == (k-1)*m)
                return true;

        }
        return false;
    }
};