What is the time complexity of K Radius Subarray Averages?

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

What data structures are used to solve K Radius Subarray Averages?

This problem uses Array, Sliding Window. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving K Radius Subarray Averages?

Insight 1: Using a sliding window technique is crucial for efficiently computing the sum of elements within the k-radius range for each index without recomputing the sum from scratch each time. Insight 2: Handling edge cases where `k` is 0 or the window size exceeds the array size is essential to avoid errors and produce the correct output.

Which companies ask LeetCode 2090. K Radius Subarray Averages in interviews?

This problem is frequently asked at Duolingo.

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K Radius Subarray Averages - LeetCode 2090 Solution

K Radius Subarray Averages - Complete Solution Guide

K Radius Subarray Averages is LeetCode problem 2090, 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 array nums of n integers, and an integer k . The k-radius average for a subarray of nums centered at some index i with the radius k is the average of all elements in nums between the indices i - k and i + k ( inclusive ). If there are less than k elements before or after the index i , then the k-radius average is -1 . Build and return an array avgs of length n where avgs[i] is the k-radius average for the subarray centered at index i . The average of x elements is the s

Detailed Explanation

The problem asks us to calculate the k-radius average for each index in a given array `nums`. For each index `i`, the k-radius average is the average of all elements within the range `i - k` to `i + k` (inclusive). If there are fewer than `k` elements before or after index `i`, the k-radius average for that index is -1. The division used to calculate the average should be integer division (truncating towards zero). The problem requires returning an array `avgs` where `avgs[i]` is the k-radius average for index `i` in `nums`.

Solution Approach

The solution employs a sliding window approach to efficiently calculate the k-radius average for each index. It initializes an array `avgs` with -1 for each index. It handles the edge case of `k` being 0 by returning the original array. If the window size (2 * k + 1) is larger than the array size, it also returns the `avgs` array initialized with -1. Otherwise, it calculates the initial sum of the first window, computes the average, and then slides the window across the array, updating the sum and the corresponding average in the `avgs` array.

Step-by-Step Algorithm

Step 1: Initialize an array `avgs` of the same length as `nums` and fill it with -1.
Step 2: Handle the base case where `k` is 0. If `k` is 0, return the original `nums` array.
Step 3: Calculate the window size as `2 * k + 1`. If `window_size` is greater than the length of `nums`, return the `avgs` array (filled with -1s).
Step 4: Calculate the sum of the first `window_size` elements of `nums` and store it in `current_sum`. Use `long` data type to avoid potential overflow.
Step 5: Calculate the average for the index `k` and store it in `avgs[k]` using integer division.
Step 6: Iterate through the rest of the array, sliding the window by one position at a time. For each iteration:
Step 7: Update `current_sum` by adding the new element at the right edge of the window and subtracting the element that just left the left edge of the window.
Step 8: Calculate the average for the current window's center index `i - k` and store it in `avgs[i - k]` using integer division.
Step 9: Return the `avgs` array.

Key Insights

Insight 1: Using a sliding window technique is crucial for efficiently computing the sum of elements within the k-radius range for each index without recomputing the sum from scratch each time.
Insight 2: Handling edge cases where `k` is 0 or the window size exceeds the array size is essential to avoid errors and produce the correct output.
Insight 3: Using `long` data type for the running sum is important to prevent integer overflow, especially when dealing with larger input numbers and window sizes.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Sliding Window.

Companies

Asked at: Duolingo.

K Radius Subarray Averages | LeetCode 2090

Medium

Array Sliding Window

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int[] getAverages(int[] nums, int k) {
        int n = nums.length;
        int[] avgs = new int[n];
        for (int i = 0; i < n; i++) {
            avgs[i] = -1;
        }

        if (k == 0) {
            return nums;
        }

        int window_size = 2 * k + 1;
        
        if (window_size > n) {
            return avgs;
        }
        
        long current_sum = 0;
        for(int i = 0; i < window_size; i++){
            current_sum += nums[i];
        }
        avgs[k] = (int)(current_sum / window_size);
        
        for (int i = window_size; i < n; i++) {
            current_sum += nums[i] - nums[i - window_size];
            avgs[i - k] = (int)(current_sum / window_size);
        }
        
        return avgs;
    }
}

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Problem Info

Description

Problem 2090: K Radius Subarray Averages - Detailed explanation and solution approaches.

Topics

Array Sliding Window

Companies

Duolingo

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2090

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

9 steps

1
Step 1: Initialize an array `avgs` of the same length as `nums` and fill it with -1.
2
Step 2: Handle the base case where `k` is 0. If `k` is 0, return the original `nums` array.
3
Step 3: Calculate the window size as `2 * k + 1`. If `window_size` is greater than the length of `nums`, return the `avgs` array (filled with -1s).
4
Step 4: Calculate the sum of the first `window_size` elements of `nums` and store it in `current_sum`. Use `long` data type to avoid potential overflow.
5
Step 5: Calculate the average for the index `k` and store it in `avgs[k]` using integer division.
6
Step 6: Iterate through the rest of the array, sliding the window by one position at a time. For each iteration:
7
Step 7: Update `current_sum` by adding the new element at the right edge of the window and subtracting the element that just left the left edge of the window.
8
Step 8: Calculate the average for the current window's center index `i - k` and store it in `avgs[i - k]` using integer division.
9
Step 9: Return the `avgs` array.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because the algorithm iterates through the `nums` array a constant number of times. The initial sum calculation takes O(k) time, but since k is bounded by the input constraints, and the rest of the loop iterates n-k times, the overall time complexity simplifies to O(n). The space complexity is O(n) because the `avgs` array has the same length as 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

Prefix Sum

An alternative approach is to precompute the prefix sum of the array. Then, the sum of elements within a given range can be calculated in O(1) time by subtracting prefix sums. This could be combined with initial validation checks.

Time: O(n) for precomputing the prefix sum and O(n) for computing averages, so O(n) in total.Space: O(n) to store the prefix sum array.

Trade-offs: The prefix sum approach has the same time and space complexity as the sliding window approach. However, the sliding window approach may be slightly more efficient in practice due to its simplicity and reduced memory access overhead. Prefix sum could be useful in cases where you need to calculate multiple range sums efficiently after the initial calculation of the prefix sum array.

✨

Key Insights

1Insight 1: Using a sliding window technique is crucial for efficiently computing the sum of elements within the k-radius range for each index without recomputing the sum from scratch each time.
2Insight 2: Handling edge cases where `k` is 0 or the window size exceeds the array size is essential to avoid errors and produce the correct output.
3Insight 3: Using `long` data type for the running sum is important to prevent integer overflow, especially when dealing with larger input numbers and window sizes.

📚

Key Terms Explained

For Students

Array

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

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: `k = 0`. The solution correctly handles this case by returning the original `nums` array directly. This avoids unnecessary calculations.
Edge case 2: `2 * k + 1 > n`. The solution checks for this case and returns an array filled with -1s, as no valid k-radius averages can be calculated.
Edge case 3: `nums` is an empty array. The code doesn't explicitly handle this, but it works correctly because `n = 0` will lead to the conditions `k == 0` or `window_size > n` being met, which will either return `nums` or an array of -1s.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow when calculating the sum of elements in the window. Using the `long` data type for the `current_sum` variable is crucial to avoid this.

✗Common mistake 2: Incorrectly calculating the index for storing the average in the `avgs` array. Ensure that the average for the window centered at index `i` is stored in `avgs[i]`.

✗Common mistake 3: Not handling the edge cases where `k` is 0 or the window size is greater than the array size.

✅

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: Calculating moving averages for stock prices over a certain period. The k-radius average is essentially a moving average.

🏦Real-world scenario 2: Smoothing data in signal processing. The k-radius average can be used to reduce noise and highlight trends in a signal.

🎮Real-world scenario 3: Image processing: Applying a blurring filter to an image. Each pixel's color value is replaced by the average of the colors of its neighbors within a certain radius.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: September 19, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 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 do I determine the window size in sliding window problems?

For fixed-size windows (like max sum of k elements), window size is given. For variable-size windows (like smallest subarray with sum >= target), expand right pointer to meet condition, then shrink left pointer to find optimal. Track the best result during shrinking.

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 "K Radius Subarray Averages"?

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 sliding window techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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