What is the time complexity of Maximum Sum of Distinct Subarrays With Length K?

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

What data structures are used to solve Maximum Sum of Distinct Subarrays With Length K?

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

What are the key insights for solving Maximum Sum of Distinct Subarrays With Length K?

Insight 1: The sliding window technique is suitable for efficiently processing contiguous subarrays of a fixed length (k). Insight 2: A hash table (or similar data structure like a dictionary or map) is essential to track the frequency of elements within the current window, allowing us to quickly determine if all elements are distinct.

Which companies ask LeetCode 2461. Maximum Sum of Distinct Subarrays With Length K in interviews?

This problem is frequently asked at IBM, Nvidia.

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Maximum Sum of Distinct Subarrays With Length K - LeetCode 2461 Solution

Maximum Sum of Distinct Subarrays With Length K - Complete Solution Guide

Maximum Sum of Distinct Subarrays With Length K is LeetCode problem 2461, 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 an integer array nums and an integer k . Find the maximum subarray sum of all the subarrays of nums that meet the following conditions: The length of the subarray is k , and All the elements of the subarray are distinct . Return the maximum subarray sum of all the subarrays that meet the conditions . If no subarray meets the conditions, return 0 . A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [1,5,4,2,9,9,9], k = 3 Output: 15 Ex

Detailed Explanation

The problem asks us to find the maximum sum of a subarray of length `k` within a given array `nums`, with the constraint that all elements within the subarray must be distinct (unique). If no such subarray exists, we return 0. A subarray is a contiguous segment of the original array.

Solution Approach

The solution uses the sliding window technique combined with a hash table to track the frequency of elements within the current window. The window slides one element at a time. For each window, we check if all elements are distinct by verifying that the number of unique elements (size of the hash table) equals the window size `k`. If the elements are distinct, we update the maximum sum found so far. The hash table is updated when the window slides by adding the new element and removing the outgoing element.

Step-by-Step Algorithm

Step 1: Initialize a hash table (e.g., `counts` in Python, `HashMap` in Java, `unordered_map` in C++) to store the frequencies of elements within the current window.
Step 2: Calculate the sum of the first `k` elements and populate the `counts` hash table with these elements and their frequencies.
Step 3: Initialize `max_sum` to 0. If all elements in the initial window are distinct (i.e., `counts.size() == k`), set `max_sum` to the sum of the initial window.
Step 4: Iterate through the remaining elements of the array, starting from index `k`. In each iteration (sliding the window):
Step 5: Add the new element (rightmost element of the new window) to the `counts` hash table and update the running sum (`current_sum`).
Step 6: Remove the leftmost element of the previous window from the `counts` hash table and update `current_sum`. If the frequency of the removed element becomes 0, remove the element from the hash table entirely.
Step 7: Check if all elements in the current window are distinct (i.e., `counts.size() == k`). If so, update `max_sum` with the maximum of the current `max_sum` and `current_sum`.
Step 8: Return `max_sum`.

Key Insights

Insight 1: The sliding window technique is suitable for efficiently processing contiguous subarrays of a fixed length (k).
Insight 2: A hash table (or similar data structure like a dictionary or map) is essential to track the frequency of elements within the current window, allowing us to quickly determine if all elements are distinct.
Insight 3: Maintaining a running sum of the elements in the current window reduces the computational cost of re-calculating the sum for each new window position.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(k)

Topics

This problem involves: Array, Hash Table, Sliding Window.

Companies

Asked at: IBM, Nvidia.

Maximum Sum of Distinct Subarrays With Length K | LeetCode 2461

Medium

Array Hash Table Sliding Window

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public long maximumSubarraySum(int[] nums, int k) {
        if (nums.length < k) {
            return 0;
        }
        
        Map<Integer, Integer> counts = new HashMap<>();
        long current_sum = 0;
        
        for (int i = 0; i < k; i++) {
            counts.put(nums[i], counts.getOrDefault(nums[i], 0) + 1);
            current_sum += nums[i];
        }
        
        long max_sum = 0;
        if (counts.size() == k) {
            max_sum = current_sum;
        }
        
        for (int i = k; i < nums.length; i++) {
            int right_num = nums[i];
            int left_num = nums[i - k];
            
            current_sum += right_num;
            counts.put(right_num, counts.getOrDefault(right_num, 0) + 1);
            
            current_sum -= left_num;
            counts.put(left_num, counts.getOrDefault(left_num, 0) - 1);
            if (counts.get(left_num) == 0) {
                counts.remove(left_num);
            }

            if (counts.size() == k) {
                max_sum = Math.max(max_sum, current_sum);
            }
        }
        
        return max_sum;
    }
}

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Problem Info

Description

Problem 2461: Maximum Sum of Distinct Subarrays With Length K - Detailed explanation and solution approaches.

Topics

Array Hash Table Sliding Window

Companies

IBMNvidia

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2461

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

8 steps

1
Step 1: Initialize a hash table (e.g., `counts` in Python, `HashMap` in Java, `unordered_map` in C++) to store the frequencies of elements within the current window.
2
Step 2: Calculate the sum of the first `k` elements and populate the `counts` hash table with these elements and their frequencies.
3
Step 3: Initialize `max_sum` to 0. If all elements in the initial window are distinct (i.e., `counts.size() == k`), set `max_sum` to the sum of the initial window.
4
Step 4: Iterate through the remaining elements of the array, starting from index `k`. In each iteration (sliding the window):
5
Step 5: Add the new element (rightmost element of the new window) to the `counts` hash table and update the running sum (`current_sum`).
6
Step 6: Remove the leftmost element of the previous window from the `counts` hash table and update `current_sum`. If the frequency of the removed element becomes 0, remove the element from the hash table entirely.
7
Step 7: Check if all elements in the current window are distinct (i.e., `counts.size() == k`). If so, update `max_sum` with the maximum of the current `max_sum` and `current_sum`.
8
Step 8: Return `max_sum`.

💡

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(k)

📝

Why?

The sliding window technique ensures that each element is visited only a constant number of times. The hash table stores at most `k` elements, resulting in O(k) space. The time complexity stems from the single pass through the array, resulting in O(n) performance.

📖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

Brute-Force

Iterate through all possible subarrays of length `k` and, for each subarray, check if all elements are distinct. If so, calculate the sum and update the maximum sum.

Time: O(n*k), where n is the length of `nums`. O(n) to generate all subarrays and O(k) in the worst case to check if the subarray's elements are distinct.Space: O(k) in worst case if we make a copy of the sub-array.

Trade-offs: Simpler to implement but less efficient. Not suitable for large input arrays.

✨

Key Insights

1Insight 1: The sliding window technique is suitable for efficiently processing contiguous subarrays of a fixed length (k).
2Insight 2: A hash table (or similar data structure like a dictionary or map) is essential to track the frequency of elements within the current window, allowing us to quickly determine if all elements are distinct.
3Insight 3: Maintaining a running sum of the elements in the current window reduces the computational cost of re-calculating the sum for each new window position.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Sliding Window

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

Edge Cases

Edge case 1: `nums.length < k`: The problem states to return 0 if no subarray meets the conditions. The code correctly handles this case by checking `if len(nums) < k` at the beginning and returning 0.
Edge case 2: Array with duplicate elements such that no subarray of length `k` has distinct elements: The code handles this by checking `if len(counts) == k`. If this condition is never met, `max_sum` remains 0, which is the correct output.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle the case where `nums.length < k`, leading to incorrect results or errors.

✗Common mistake 2: Forgetting to update both the running sum (`current_sum`) and the hash table (`counts`) when the window slides.

✗Common mistake 3: Incorrectly implementing the hash table update logic, which could lead to miscounting element frequencies and an incorrect assessment of distinctness.

✅

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: Detecting fraudulent transactions within a stream of transactions, where you want to identify a sequence of `k` transactions that involve distinct accounts (indicating potential collusion).

🏦Real-world scenario 2: Analyzing network traffic to identify bursts of activity from `k` distinct IP addresses, which could indicate a distributed denial-of-service (DDoS) attack.

🎮Real-world scenario 3: Bioinformatics: finding sequence patterns of length 'k' with distinct elements in DNA or protein sequences. This can be helpful in identifying unique motifs.

💡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: September 30, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Sum of Distinct Subarrays With Length K"?

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

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