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

#### QUESTION:

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
``````

#### EXPLANATION:

easy的题目, 只要进行for循环, 判断是否有k个相同的数值即可.

#### SOLUTION:

``````class Solution {
fun containsPattern(arr: IntArray, m: Int, k: Int): Boolean {
if (arr.size < m || arr.size.div(m) < k ) return false

for (i in 0..arr.size - m * k) {
var count: Int = 1
var j = i + m
while (j < arr.size - m + 1) {
if (containsPatternHelper(arr, i, j, m)) {
count++
j += m -1
} else count = 0

if (count >= k) return true
j++
}
}
return false
}

fun containsPatternHelper(arr: IntArray, i: Int, j: Int, m: Int): Boolean {
for (p in 0 until m) {
if (arr[i+p] != arr[j+p]) return false
}
return true
}
}
``````