What is the time complexity of Number of Subarrays That Match a Pattern I?

The optimal solution has a time complexity of O(n*m).

What data structures are used to solve Number of Subarrays That Match a Pattern I?

This problem uses Array, Rolling Hash, String Matching, Hash Function. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Subarrays That Match a Pattern I?

Insight 1: Transforming the original `nums` array into an array representing the relative differences between adjacent elements simplifies the matching process. Insight 2: Iterating through the `nums` array with a sliding window approach, comparing the generated 'difference' array to the given `pattern`, allows us to identify matching subarrays.

Which companies ask LeetCode 3034. Number of Subarrays That Match a Pattern I in interviews?

This problem is frequently asked at Autodesk, Capital One, Visa.

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Number of Subarrays That Match a Pattern I - LeetCode 3034 Solution

Q: How to solve Number of Subarrays That Match a Pattern I?

The solution involves first transforming the input `nums` array into an array called `nums_pattern` where each element represents the relative difference between adjacent elements in `nums`. Specifically, `nums_pattern[i]` is 1 if `nums[i+1] > nums[i]`, 0 if `nums[i+1] == nums[i]`, and -1 if `nums[i+1] < nums[i]`. After constructing `nums_pattern`, the solution iterates through `nums_pattern` using a sliding window of the same length as `pattern`. In each iteration, it compares the current window in `nums_pattern` with `pattern`. If they are identical, it increments the count. This process continues until all possible windows have been checked.

Number of Subarrays That Match a Pattern I - Complete Solution Guide

Number of Subarrays That Match a Pattern I is LeetCode problem 3034, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 0-indexed integer array nums of size n , and a 0-indexed integer array pattern of size m consisting of integers -1 , 0 , and 1 . A subarray nums[i..j] of size m + 1 is said to match the pattern if the following conditions hold for each element pattern[k] : nums[i + k + 1] > nums[i + k] if pattern[k] == 1 . nums[i + k + 1] == nums[i + k] if pattern[k] == 0 . nums[i + k + 1] < nums[i + k] if pattern[k] == -1 . Return the count of subarrays in nums that match the pattern . Example 1

Detailed Explanation

The problem asks us to find the number of subarrays within a given integer array `nums` that match a specified pattern. The `pattern` is an array of integers representing the relative relationships between consecutive elements in a potential matching subarray. A `pattern` value of 1 means the next element in the `nums` subarray should be greater than the current element. 0 means the next element should be equal, and -1 means the next element should be less than the current. We need to count how many subarrays of length `pattern.length + 1` in `nums` satisfy these relative conditions.

Solution Approach

The solution involves first transforming the input `nums` array into an array called `nums_pattern` where each element represents the relative difference between adjacent elements in `nums`. Specifically, `nums_pattern[i]` is 1 if `nums[i+1] > nums[i]`, 0 if `nums[i+1] == nums[i]`, and -1 if `nums[i+1] < nums[i]`. After constructing `nums_pattern`, the solution iterates through `nums_pattern` using a sliding window of the same length as `pattern`. In each iteration, it compares the current window in `nums_pattern` with `pattern`. If they are identical, it increments the count. This process continues until all possible windows have been checked.

Step-by-Step Algorithm

Step 1: Create a new array `nums_pattern` of size `n-1`. This array will store the relative differences between consecutive elements in `nums`.
Step 2: Populate the `nums_pattern` array. For each `i` from 0 to `n-2`, calculate the difference between `nums[i+1]` and `nums[i]`. Store 1 if `nums[i+1] > nums[i]`, 0 if `nums[i+1] == nums[i]`, and -1 if `nums[i+1] < nums[i]` in `nums_pattern[i]`.
Step 3: Initialize a counter `count` to 0. This counter will store the number of matching subarrays.
Step 4: Iterate through `nums_pattern` using a sliding window of length `m` (the length of the `pattern` array). The number of possible starting positions for the window is `n - m`.
Step 5: For each window, compare the elements within the window to the elements in the `pattern` array. If all elements match, increment the `count`.
Step 6: Return the final `count`.

Key Insights

Insight 1: Transforming the original `nums` array into an array representing the relative differences between adjacent elements simplifies the matching process.
Insight 2: Iterating through the `nums` array with a sliding window approach, comparing the generated 'difference' array to the given `pattern`, allows us to identify matching subarrays.
Insight 3: The constraints are relatively small (n <= 100), suggesting a brute-force or O(n*m) solution is acceptable.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(n)

Topics

This problem involves: Array, Rolling Hash, String Matching, Hash Function.

Companies

Asked at: Autodesk, Capital One, Visa.

Number of Subarrays That Match a Pattern I | LeetCode 3034

Medium

Array Rolling Hash String Matching Hash Function

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int countMatchingSubarrays(int[] nums, int[] pattern) {
        int n = nums.length;
        int m = pattern.length;
        
        List<Integer> nums_pattern = new ArrayList<>();
        for (int i = 0; i < n - 1; i++) {
            nums_pattern.add((nums[i + 1] > nums[i]) ? 1 : ((nums[i + 1] < nums[i]) ? -1 : 0));
        }
        
        int count = 0;
        
        int num_windows = n - m;
        
        for (int i = 0; i < num_windows; i++) {
            boolean match = true;
            for (int j = 0; j < m; j++) {
                if (nums_pattern.get(i + j) != pattern[j]) {
                    match = false;
                    break;
                }
            }
            if (match) {
                count++;
            }
        }
        
        return count;
    }
}

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Problem Info

Description

Problem 3034: Number of Subarrays That Match a Pattern I - Detailed explanation and solution approaches.

Topics

Array Rolling Hash String Matching Hash Function

Companies

AutodeskCapital OneVisa

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:3034

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

6 steps

1
Step 1: Create a new array `nums_pattern` of size `n-1`. This array will store the relative differences between consecutive elements in `nums`.
2
Step 2: Populate the `nums_pattern` array. For each `i` from 0 to `n-2`, calculate the difference between `nums[i+1]` and `nums[i]`. Store 1 if `nums[i+1] > nums[i]`, 0 if `nums[i+1] == nums[i]`, and -1 if `nums[i+1] < nums[i]` in `nums_pattern[i]`.
3
Step 3: Initialize a counter `count` to 0. This counter will store the number of matching subarrays.
4
Step 4: Iterate through `nums_pattern` using a sliding window of length `m` (the length of the `pattern` array). The number of possible starting positions for the window is `n - m`.
5
Step 5: For each window, compare the elements within the window to the elements in the `pattern` array. If all elements match, increment the `count`.
6
Step 6: Return the final `count`.

💡

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*m)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loops: one loop to iterate through the possible subarrays and another loop to compare the current subarray with the `pattern`. The space complexity is determined by the `nums_pattern` array, which has a length proportional to the length of the `nums` array.

📖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

Rolling Hash

Instead of directly comparing subarrays, compute a hash value for both the pattern and each subarray of nums_pattern. Compare the hash values to identify potential matches. This can reduce comparison time in some cases.

Time: O(n) on average, O(n*m) in worst case (due to potential hash collisions).Space: O(n)

Trade-offs: Rolling hash offers potential performance gains if the hash function is well-chosen and collisions are rare. However, the implementation is more complex, and the worst-case time complexity can still be O(n*m) if there are many hash collisions. The main solution provides similar time complexity with a much simpler code, due to the constraints. It's thus a better approach.

✨

Key Insights

1Insight 1: Transforming the original `nums` array into an array representing the relative differences between adjacent elements simplifies the matching process.
2Insight 2: Iterating through the `nums` array with a sliding window approach, comparing the generated 'difference' array to the given `pattern`, allows us to identify matching subarrays.
3Insight 3: The constraints are relatively small (n <= 100), suggesting a brute-force or O(n*m) solution is acceptable.

📚

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: When `nums` and `pattern` are very short. The algorithm handles this gracefully because the loops will iterate a minimal number of times, and the comparisons will still be valid.
Edge case 2: When `pattern` is longer than `nums`. This case is prevented due to the problem constraint `1 <= m < n`, so `pattern.length` is always less than `nums.length`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not correctly mapping the relative differences to -1, 0, and 1. Ensure to accurately represent the relationships (greater, equal, less) between adjacent elements.

✗Common mistake 2: Off-by-one errors when calculating the starting and ending indices for the sliding window. Double-check the loop conditions to avoid accessing elements outside the bounds of the array.

✅

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: Stock market analysis. Identifying patterns of price increases, decreases, and stability over time. The `nums` array could represent stock prices, and the `pattern` could represent a desired trend.

🏦Real-world scenario 2: Signal processing. Detecting specific signal patterns within a larger audio or sensor data stream. The `nums` array could represent signal amplitudes, and the `pattern` could represent a known signal signature.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: November 2, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Number of Subarrays That Match a Pattern I"?

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

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