README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1500-1599/1566.Detect%20Pattern%20of%20Length%20M%20Repeated%20K%20or%20More%20Times/README_EN.md	1486	Weekly Contest 204 Q1	Array Enumeration

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

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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: Single Traversal

First, if the length of the array is less than $m \times k$, then there is definitely no pattern of length $m$ that repeats at least $k$ times, so we directly return $\textit{false}$.

Next, we define a variable $\textit{cnt}$ to record the current count of consecutive repetitions. If there are $(k - 1) \times m$ consecutive elements $a_i$ in the array such that $a_i = a_{i - m}$, then we have found a pattern of length $m$ that repeats at least $k$ times, and we return $\textit{true}$. Otherwise, we reset $\textit{cnt}$ to $0$ and continue traversing the array.

Finally, if we finish traversing the array without finding a pattern that meets the conditions, we return $\textit{false}$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Python3

class Solution:
    def containsPattern(self, arr: List[int], m: int, k: int) -> bool:
        if len(arr) < m * k:
            return False
        cnt, target = 0, (k - 1) * m
        for i in range(m, len(arr)):
            if arr[i] == arr[i - m]:
                cnt += 1
                if cnt == target:
                    return True
            else:
                cnt = 0
        return False

Java

class Solution {
    public boolean containsPattern(int[] arr, int m, int k) {
        if (arr.length < m * k) {
            return false;
        }
        int cnt = 0, target = (k - 1) * m;
        for (int i = m; i < arr.length; ++i) {
            if (arr[i] == arr[i - m]) {
                if (++cnt == target) {
                    return true;
                }
            } else {
                cnt = 0;
            }
        }
        return false;
    }
}

C++

class Solution {
public:
    bool containsPattern(vector<int>& arr, int m, int k) {
        if (arr.size() < m * k) {
            return false;
        }
        int cnt = 0, target = (k - 1) * m;
        for (int i = m; i < arr.size(); ++i) {
            if (arr[i] == arr[i - m]) {
                if (++cnt == target) {
                    return true;
                }
            } else {
                cnt = 0;
            }
        }
        return false;
    }
};

Go

func containsPattern(arr []int, m int, k int) bool {
	cnt, target := 0, (k-1)*m
	for i := m; i < len(arr); i++ {
		if arr[i] == arr[i-m] {
			cnt++
			if cnt == target {
				return true
			}
		} else {
			cnt = 0
		}
	}
	return false
}

TypeScript

function containsPattern(arr: number[], m: number, k: number): boolean {
    if (arr.length < m * k) {
        return false;
    }
    const target = (k - 1) * m;
    let cnt = 0;
    for (let i = m; i < arr.length; ++i) {
        if (arr[i] === arr[i - m]) {
            if (++cnt === target) {
                return true;
            }
        } else {
            cnt = 0;
        }
    }
    return false;
}