1566

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.

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

Related Topics:
Array

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/detect-pattern-of-length-m-repeated-k-or-more-times/
// Author: github.com/lzl124631x
// Time: O(NMK)
// Space: O(1)
class Solution {
public:
    bool containsPattern(vector<int>& A, int m, int k) {
        int N = A.size();
        if (m * k > A.size()) return false;
        for (int i = 0; i <= N - m * k; ++i) {
            bool valid = true;
            for (int j = 1; j < k && valid; ++j) {
                for (int x = 0; x < m && valid; ++x) {
                    if (A[i + m * j + x] != A[i + x]) valid = false;
                }
            }
            if (valid) return true;
        }
        return false;
    }
};