Formatted question description: https://leetcode.ca/all/1566.html

# 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 = 3Output:trueExplanation: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 = 2Output:trueExplanation:The pattern(1,2)of length 2 is repeated 2 consecutive times. Another valid pattern(2,1) isalso repeated 2 times.

**Example 3:**

Input:arr = [1,2,1,2,1,3], m = 2, k = 3Output:falseExplanation: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 = 2Output:falseExplanation: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 = 3Output:falseExplanation: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/
// 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;
}
};
```

Java

```
class Solution {
public boolean containsPattern(int[] arr, int m, int k) {
int length = arr.length;
int patternLength = m * k;
for (int i = patternLength; i <= length; i++) {
int start = i - patternLength;
if (checkPattern(arr, m, k, start))
return true;
}
return false;
}
public boolean checkPattern(int[] arr, int m, int k, int start) {
for (int i = 0; i < m; i++) {
int index = start + i;
int num = arr[index];
for (int j = 1; j < k; j++) {
int nextIndex = index + m * j;
int nextNum = arr[nextIndex];
if (nextNum != num)
return false;
}
}
return true;
}
}
```