LeetCode-in-Java

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

Easy

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

import java.util.Arrays;

public class Solution {
    public boolean containsPattern(int[] arr, int m, int k) {
        for (int i = 0; i < arr.length - m; i++) {
            int[] pattern = Arrays.copyOfRange(arr, i, i + m);
            int times = 1;
            for (int j = i + m; j < arr.length; j += m) {
                int[] candidate = Arrays.copyOfRange(arr, j, j + m);
                if (Arrays.equals(pattern, candidate)) {
                    times++;
                    if (times == k) {
                        return true;
                    }
                } else {
                    break;
                }
            }
        }
        return false;
    }
}

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