What is the time complexity of Maximum Erasure Value?

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

What data structures are used to solve Maximum Erasure Value?

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 Erasure Value?

Insight 1: The sliding window technique is suitable for efficiently finding subarrays that satisfy a given condition. Insight 2: A hash set (or similar data structure) is useful to keep track of the elements currently within the window to check for uniqueness.

Which companies ask LeetCode 1695. Maximum Erasure Value in interviews?

This problem is frequently asked at Cashfree.

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Maximum Erasure Value - LeetCode 1695 Solution

Maximum Erasure Value - Complete Solution Guide

Maximum Erasure Value is LeetCode problem 1695, 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 array of positive integers nums and want to erase a subarray containing unique elements . The score you get by erasing the subarray is equal to the sum of its elements. Return the maximum score you can get by erasing exactly one subarray. An array b is called to be a subarray of a if it forms a contiguous subsequence of a , that is, if it is equal to a[l],a[l+1],...,a[r] for some (l,r) . Example 1: Input: nums = [4,2,4,5,6] Output: 17 Explanation: The optimal subarray here is [2

Detailed Explanation

The problem asks us to find a subarray within a given array of positive integers that contains only unique elements. Our goal is to maximize the sum of the elements within this unique subarray. We need to return this maximum possible sum. A subarray is a contiguous sequence of elements within the array. The constraints specify that the input array's length is between 1 and 100,000, and each element's value is between 1 and 10,000.

Solution Approach

The provided solution uses a sliding window approach along with a hash set to solve this problem. The window expands to the right, including new elements, while shrinking from the left whenever a duplicate element is encountered within the window. The hash set is used to efficiently check for the presence of elements within the current window. The current sum of the elements in the window is maintained, and the maximum sum encountered during the process is tracked and eventually returned.

Step-by-Step Algorithm

Step 1: Initialize the left pointer of the window to 0, the maximum score to 0, the current score to 0, and an empty hash set to store the elements currently in the window.
Step 2: Iterate through the array using the right pointer of the window.
Step 3: For each element at the right pointer, check if it already exists in the hash set.
Step 4: If the element exists in the hash set (duplicate found), shrink the window from the left by removing elements from the left until the duplicate element is no longer in the window. Update the current score accordingly.
Step 5: Add the current element at the right pointer to the hash set and update the current score.
Step 6: Update the maximum score with the maximum between the current maximum score and the current score.
Step 7: Continue steps 2-6 until the right pointer reaches the end of the array.
Step 8: Return the maximum score.

Key Insights

Insight 1: The sliding window technique is suitable for efficiently finding subarrays that satisfy a given condition.
Insight 2: A hash set (or similar data structure) is useful to keep track of the elements currently within the window to check for uniqueness.
Insight 3: When a duplicate is found, shrink the window from the left until the duplicate is removed, maintaining the uniqueness condition.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Cashfree.

Maximum Erasure Value | LeetCode 1695

Medium

Array Hash Table Sliding Window

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

Solution Code

Available in:

import java.util.HashSet;
import java.util.Set;

class Solution {
    public int maximumUniqueSubarray(int[] nums) {
        int left = 0;
        int max_score = 0;
        int current_score = 0;
        Set<Integer> seen = new HashSet<>();

        for (int right = 0; right < nums.length; right++) {
            int num_right = nums[right];

            while (seen.contains(num_right)) {
                int num_left = nums[left];
                seen.remove(num_left);
                current_score -= num_left;
                left += 1;
            }

            seen.add(num_right);
            current_score += num_right;

            max_score = Math.max(max_score, current_score);
        }

        return max_score;
    }
}

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Problem Info

Description

Problem 1695: Maximum Erasure Value - Detailed explanation and solution approaches.

Topics

Array Hash Table Sliding Window

Companies

Cashfree

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1695

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 the left pointer of the window to 0, the maximum score to 0, the current score to 0, and an empty hash set to store the elements currently in the window.
2
Step 2: Iterate through the array using the right pointer of the window.
3
Step 3: For each element at the right pointer, check if it already exists in the hash set.
4
Step 4: If the element exists in the hash set (duplicate found), shrink the window from the left by removing elements from the left until the duplicate element is no longer in the window. Update the current score accordingly.
5
Step 5: Add the current element at the right pointer to the hash set and update the current score.
6
Step 6: Update the maximum score with the maximum between the current maximum score and the current score.
7
Step 7: Continue steps 2-6 until the right pointer reaches the end of the array.
8
Step 8: Return the maximum score.

💡

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 linear because each element of the input array is visited at most a constant number of times. The space complexity is also linear because, in the worst-case scenario, the hash set (or similar data structure) can potentially store all elements of the input 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 and Hash Map

Calculate the prefix sum of the array. Use a hash map to store the last seen index of each number. Use the sliding window approach as before, but instead of maintaining the current sum on the fly, calculate the sum of the current window using prefix sums.

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

Trade-offs: The prefix sum approach has similar time and space complexity. The benefit would be that the current sum is computed directly from the prefix sum in O(1) time within the window. However, it requires pre-computation of prefix sum array and still needs to maintain the uniqueness using a hash map.

✨

Key Insights

1Insight 1: The sliding window technique is suitable for efficiently finding subarrays that satisfy a given condition.
2Insight 2: A hash set (or similar data structure) is useful to keep track of the elements currently within the window to check for uniqueness.
3Insight 3: When a duplicate is found, shrink the window from the left until the duplicate is removed, maintaining the uniqueness condition.

📚

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: Empty array. The code handles this implicitly since the loop won't execute and 0 will be returned, which is the correct result for an empty array.
Edge case 2: Array with all duplicate elements. The window will slide one element at a time, and the maximum score will be the value of the largest single element.
Edge case 3: Array with all unique elements. The entire array will be the unique subarray, and the maximum score will be the sum of all elements.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not properly shrinking the window when a duplicate element is found, leading to incorrect subarrays and incorrect sums.

✗Common mistake 2: Incorrectly calculating the current sum, especially during the shrinking of the window.

✗Common mistake 3: Forgetting to consider all possible subarrays by not sliding the window correctly.

✅

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: Analyzing data streams to find periods of unique activity for anomaly detection.

🏦Real-world scenario 2: Optimizing resource allocation by identifying contiguous blocks of unique resources to maximize utilization.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: August 15, 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 13

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

@AlgoMasterJul 6

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

@InterviewProJun 29

❓

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 Erasure Value"?

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.

🤖 Need help with this problem?