Detect Pattern of Length M Repeated K or More Times Python/Java solution

Both Python and Java implementations follow the same logic: iterate from index 0 to n - m - 1, compare arr[i] with arr[i + m], and maintain a running match counter. When the counter reaches m * (k - 1), return true. Otherwise continue scanning until the array ends.

How to solve Detect Pattern of Length M Repeated K or More Times in O(n)?

Scan the array while comparing arr[i] with arr[i + m]. If the values match, increment a counter; otherwise reset it. Once the counter reaches m * (k - 1), the pattern has repeated k times consecutively. This avoids checking every block separately and completes in a single linear pass.

What is the best approach for Detect Pattern of Length M Repeated K or More Times?

The most efficient approach uses modulo-style comparison. Iterate through the array and compare arr[i] with arr[i + m]. Count consecutive matches and check whether the count reaches m * (k - 1). This confirms that the pattern of length m repeats k times. The algorithm runs in O(n) time and uses O(1) space.

Is Detect Pattern of Length M Repeated K or More Times asked at Google/Amazon/Meta?

Pattern detection and repeated sequence problems frequently appear in interviews at companies like Amazon, Google, and Meta. While this exact question may not always appear, the underlying idea of recognizing repeated subarrays and optimizing comparisons is a common array interview pattern.

What data structure is used in Detect Pattern of Length M Repeated K or More Times?

The problem mainly uses arrays and index-based comparisons. No additional data structures are required because the optimized approach only compares elements separated by a fixed distance m while scanning the array once.

What is the time complexity of Detect Pattern of Length M Repeated K or More Times?

The optimal solution runs in O(n) time with O(1) extra space. Each element is compared with the element m positions ahead exactly once. A brute-force sliding window approach may take O(n * m * k) time because it repeatedly compares entire pattern segments.

Description Hints 1 Solution Video 6:25

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

Q: Is Detect Pattern of Length M Repeated K or More Times easy or hard?

Detect Pattern of Length M Repeated K or More Times is classified as an Easy problem on LeetCode with an acceptance rate around 43%. The main challenge is recognizing that repeated block comparison can be simplified using offset comparisons for an O(n) solution.

Easy43.9% acceptance1 company

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

Asked at 1 company

Hrt

Topics

Array Enumeration

Hints open one at a time

Hint 1

Use a three-layer loop to check all possible patterns by iterating through all possible starting positions, all indexes less than m, and if the character at the index is repeated k times.

6:25

Still stuck?Read 3 approaches, or watch a walkthrough.

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