How do you solve Detect Pattern of Length M Repeated K or More Times on LeetCode?

Detect Pattern of Length M Repeated K or More Times (LeetCode #1566) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Patterns can overlap, so we need to check consecutive segments carefully.

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

The optimal solution for Detect Pattern of Length M Repeated K or More Times runs in O(n) time and uses O(1) space. This approach only requires a single pass through the array, making it O(n) in time complexity.

What is the brute force solution for Detect Pattern of Length M Repeated K or More Times?

The brute force approach for Detect Pattern of Length M Repeated K or More Times is Brute Force, with O(n²) time complexity and O(1) space complexity. We will check every possible starting position in the array for a pattern of length m and count how many times it appears consecutively. If we find a pattern that appears k or more times, we return true. The optimal Optimal Solution approach improves this to O(n).

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

Detect Pattern of Length M Repeated K or More Times uses the following concepts: Array, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Detect Pattern of Length M Repeated K or More Times?

Clarify the definition of a pattern with the interviewer. Think about edge cases, like patterns at the end of the array. Explain your thought process as you code to show your understanding.

What are common mistakes when solving Detect Pattern of Length M Repeated K or More Times?

Not resetting the count when a mismatch occurs. Confusing the length of the pattern with the number of repetitions.

#1566

Detect Pattern of Length M Repeated K or More Times

Easy

ArrayEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of checking every possible pattern, we can use a sliding window approach to efficiently count consecutive patterns of length m.

⚙️

Algorithm

5 steps

1Step 1: Initialize a count variable to track how many times we've seen the current pattern.
2Step 2: Iterate through the array using a loop, checking for the current pattern starting from each index.
3Step 3: If the next segment matches the current pattern, increment the count.
4Step 4: If the count reaches k, return true immediately.
5Step 5: If no pattern is found after checking all, return false.

solution.py11 lines

1def containsPattern(arr, m, k):
2    n = len(arr)
3    count = 0
4    for i in range(n - m):
5        if arr[i:i + m] == arr[i + m:i + 2 * m]:
6            count += 1
7            if count >= k - 1:
8                return True
9        else:
10            count = 0
11    return False

ℹ

Complexity note: This approach only requires a single pass through the array, making it O(n) in time complexity.

1Patterns can overlap, so we need to check consecutive segments carefully.
2Using a sliding window can significantly reduce the number of checks needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.