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

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Detect Pattern of Length M Repeated K or More Times?

This problem uses Array, Enumeration. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Detect Pattern of Length M Repeated K or More Times?

Insight 1: The solution involves iterating through all possible starting positions of a pattern of length `m` in the array. Insight 2: Efficiently checking for the repetition of the pattern involves nested loops or slicing (in Python). The key is to avoid overlapping subarrays when checking for repetition.

Which companies ask LeetCode 1566. Detect Pattern of Length M Repeated K or More Times in interviews?

This problem is frequently asked at Hudson River Trading.

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Detect Pattern of Length M Repeated K or More Times - LeetCode 1566 Solution

Detect Pattern of Length M Repeated K or More Times - Complete Solution Guide

Detect Pattern of Length M Repeated K or More Times is LeetCode problem 1566, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 Explanatio

Detailed Explanation

The problem asks to determine if an array of integers contains a pattern of a specified length that repeats a certain number of times consecutively. The pattern must be a contiguous subarray. The input consists of an integer array `arr`, an integer `m` representing the length of the pattern, and an integer `k` representing the minimum number of times the pattern must repeat. The output is a boolean value: `true` if such a pattern exists, and `false` otherwise. Importantly, the repetitions must be consecutive and non-overlapping.

Solution Approach

The provided solutions use a brute-force approach. They iterate through all possible starting indices of a pattern of length `m`. For each starting index, they check how many times the pattern repeats consecutively. If the count of repetitions is greater than or equal to `k`, the function returns `true`. Otherwise, it continues checking other potential patterns. The Python solution leverages slicing for more concise pattern comparison.

Step-by-Step Algorithm

Step 1: Iterate through all possible starting indices `i` from 0 up to `n - m * k`, where `n` is the length of the array. This ensures there's enough space for `k` repetitions of a pattern of length `m`.
Step 2: Extract the pattern `pattern` of length `m` starting at index `i`.
Step 3: Iterate through the array, starting at `i + m`, checking for consecutive occurrences of `pattern` at intervals of `m`. Count the number of repetitions.
Step 4: If the `count` of repetitions is greater than or equal to `k`, return `true`.
Step 5: If the loop completes without finding a pattern with `k` or more repetitions, return `false`.

Key Insights

Insight 1: The solution involves iterating through all possible starting positions of a pattern of length `m` in the array.
Insight 2: Efficiently checking for the repetition of the pattern involves nested loops or slicing (in Python). The key is to avoid overlapping subarrays when checking for repetition.
Insight 3: The algorithm needs to handle edge cases where the array length is smaller than `m * k`, or where `m` or `k` are invalid (although constraints prevent this in this problem).

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(m)

Topics

This problem involves: Array, Enumeration.

Companies

Asked at: Hudson River Trading.

Detect Pattern of Length M Repeated K or More Times | LeetCode 1566

Easy

Array Enumeration

Time: O(n^2)Space: O(m)

Solution Code

Available in:

class Solution {
    public boolean containsPattern(int[] arr, int m, int k) {
        int n = arr.length;
        for (int i = 0; i <= n - m * k; i++) {
            boolean found = true;
            for (int j = 1; j < k; j++) {
                for (int l = 0; l < m; l++) {
                    if (arr[i + l] != arr[i + j * m + l]) {
                        found = false;
                        break;
                    }
                }
                if (!found) break;
            }
            if (found) return true;
        }
        return false;
    }
}

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Problem Info

Description

Problem 1566: Detect Pattern of Length M Repeated K or More Times - Detailed explanation and solution approaches.

Topics

Array Enumeration

Companies

Hudson River Trading

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1566

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Iterate through all possible starting indices `i` from 0 up to `n - m * k`, where `n` is the length of the array. This ensures there's enough space for `k` repetitions of a pattern of length `m`.
2
Step 2: Extract the pattern `pattern` of length `m` starting at index `i`.
3
Step 3: Iterate through the array, starting at `i + m`, checking for consecutive occurrences of `pattern` at intervals of `m`. Count the number of repetitions.
4
Step 4: If the `count` of repetitions is greater than or equal to `k`, return `true`.
5
Step 5: If the loop completes without finding a pattern with `k` or more repetitions, return `false`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(m)

📝

Why?

The outer loop iterates `O(n)` times. The inner loop checking for pattern repetitions can iterate up to `O(k)` times in the worst case. Inside this inner loop, comparing the pattern involves `O(m)` operations for each comparison. Therefore, the overall time complexity is O(n*m*k). The space complexity is O(m) because we store the pattern of length `m`.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Optimized Brute Force

Instead of comparing entire subarrays, we could compare individual elements using a single loop which would slightly improve the constant factor of the time complexity.

Time: O(n*m*k)Space: O(1)

Trade-offs: This optimized brute force approach might offer a small performance improvement by avoiding some array copying operations implicit in subarray comparison, but the overall complexity remains the same.

✨

Key Insights

1Insight 1: The solution involves iterating through all possible starting positions of a pattern of length `m` in the array.
2Insight 2: Efficiently checking for the repetition of the pattern involves nested loops or slicing (in Python). The key is to avoid overlapping subarrays when checking for repetition.
3Insight 3: The algorithm needs to handle edge cases where the array length is smaller than `m * k`, or where `m` or `k` are invalid (although constraints prevent this in this problem).

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: An empty array will correctly return `false` because the condition `arrSize < m * k` in the C solution (and similar checks in other solutions) handles this. The loop condition `i <= n - m * k` in the other solutions also effectively handles this.
Edge case 2: A pattern that almost repeats `k` times but is broken at the end. The solution correctly handles this by checking only consecutive repetitions.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling overlapping subarrays when counting repetitions. The pattern must repeat consecutively without overlap.

✗Common mistake 2: Failing to consider all possible starting positions of the pattern in the array, leading to missing valid patterns.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Identifying repeating sequences in DNA or protein sequences for biological research.

🏦Real-world scenario 2: Detecting repeating patterns in network traffic analysis for anomaly detection and security.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 5, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

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

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether enumeration techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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