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

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

 

My answer: basic idea to check pattern in length m, and since it must be consecutive, we just need to check if index and index + m is same with each other. If not, index ++; if yes, check index + 1 and index + 1 + m

class Solution {
    public boolean containsPattern(int[] arr, int m, int k) {
        if (arr == null || arr.length < m || arr.length < m * k) {
            return false;
        }
        int count = 1;
        for (int start = 0; start < arr.length - m ;) {
            boolean patternFound = hasPattern(arr, start, m);
            if (patternFound) {
                count += 1;
                start += m;
                if (count >= k) {
                    return true;
                }
            } else {
                count = 1;
                start ++;
            }
        }
        return false;
    }
    
    private boolean hasPattern(int[] arr, int start, int m) {
        for (int i = 0 ; i < m; i ++) {
            if ((start + i + m) >= arr.length ||
                arr[start + i] != arr[start + i + m]) {
                return false;
            }
        }
        return true;
    }
}