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

Description

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

Solutions

• class Solution {
public boolean containsPattern(int[] arr, int m, int k) {
int n = arr.length;
for (int i = 0; i <= n - m * k; ++i) {
int j = 0;
for (; j < m * k; ++j) {
if (arr[i + j] != arr[i + (j % m)]) {
break;
}
}
if (j == m * k) {
return true;
}
}
return false;
}
}

• class Solution {
public:
bool containsPattern(vector<int>& arr, int m, int k) {
int n = arr.size();
for (int i = 0; i <= n - m * k; ++i) {
int j = 0;
for (; j < m * k; ++j) {
if (arr[i + j] != arr[i + (j % m)]) {
break;
}
}
if (j == m * k) {
return true;
}
}
return false;
}
};

• class Solution:
def containsPattern(self, arr: List[int], m: int, k: int) -> bool:
n = len(arr)
for i in range(n - m * k + 1):
j = 0
while j < m * k:
if arr[i + j] != arr[i + (j % m)]:
break
j += 1
if j == m * k:
return True
return False


• func containsPattern(arr []int, m int, k int) bool {
n := len(arr)
for i := 0; i <= n-m*k; i++ {
j := 0
for ; j < m*k; j++ {
if arr[i+j] != arr[i+(j%m)] {
break
}
}
if j == m*k {
return true
}
}
return false
}

• function containsPattern(arr: number[], m: number, k: number): boolean {
const n = arr.length;
for (let i = 0; i <= n - m * k; ++i) {
let j = 0;
for (; j < m * k; ++j) {
if (arr[i + j] != arr[i + (j % m)]) {
break;
}
}
if (j == m * k) {
return true;
}
}
return false;
}