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

Solution 1

Python Code
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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

Java Code
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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;
    }
}

C++ Code
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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;
    }
};

Go Code
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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
}

TypeScript Code
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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;
}