What is the time complexity of Maximum Segment Sum After Removals?

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

What data structures are used to solve Maximum Segment Sum After Removals?

This problem uses Array, Union Find, Prefix Sum, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Segment Sum After Removals?

Insight 1: Processing the queries in reverse order allows us to think about adding elements back into the array instead of removing them, which is easier to manage. Insight 2: The Union Find data structure is well-suited for efficiently tracking contiguous segments of numbers and merging them when adjacent elements are 'added back'.

Which companies ask LeetCode 2382. Maximum Segment Sum After Removals in interviews?

This problem is frequently asked at Infosys.

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Maximum Segment Sum After Removals - LeetCode 2382 Solution

Maximum Segment Sum After Removals - Complete Solution Guide

Maximum Segment Sum After Removals is LeetCode problem 2382, a Hard 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 two 0-indexed integer arrays nums and removeQueries , both of length n . For the i th query, the element in nums at the index removeQueries[i] is removed, splitting nums into different segments. A segment is a contiguous sequence of positive integers in nums . A segment sum is the sum of every element in a segment. Return an integer array answer , of length n , where answer[i] is the maximum segment sum after applying the i th removal. Note: The same index will not be removed more

Detailed Explanation

The problem asks us to simulate removing elements from an array `nums` based on the indices provided in `removeQueries`. After each removal, we need to find the maximum sum of any contiguous segment of *positive* integers remaining in the array. The removal process is done in reverse order, and we need to return an array `answer` where `answer[i]` is the maximum segment sum after the i-th removal.

Solution Approach

The provided solution uses a Union Find data structure to efficiently track the segments. The `removeQueries` are processed in reverse order. For each query, the corresponding element in `nums` is considered 'added back' to the array. We then check if the element has adjacent elements that are also present. If so, we use the `union` function to merge these segments. After each 'addition', the maximum segment sum is updated. Since we're adding in reverse order of removal, the maximum segment sum calculated at each step represents the maximum segment sum *after* that removal, allowing us to build the `answer` array.

Step-by-Step Algorithm

Step 1: Initialize the Union Find data structure: `parent` array (initially each element is its own parent), `size` array (initially each set has size 1), and `sums` array (initially each element is its own sum). Also initialize a `present` array to keep track of which indices are currently present, an `answer` array to store the results, and `max_sum` to keep track of maximum segment sum so far.
Step 2: Iterate through `removeQueries` in reverse order (from `n-1` to `0`).
Step 3: For each query `i`, store the current `max_sum` in `answer[i]` before applying the removal at index `idx = removeQueries[i]`.
Step 4: 'Add' the element back into the array by setting `present[idx] = True`.
Step 5: Check if the element has adjacent present elements. If `idx > 0` and `present[idx - 1]` is true, perform `union(idx, idx - 1)`. If `idx < n - 1` and `present[idx + 1]` is true, perform `union(idx, idx + 1)`.
Step 6: Update `max_sum` with the maximum of the current `max_sum` and the segment sum of the set that `idx` belongs to (i.e., `sums[find(idx)]`).
Step 7: After iterating through all queries, return the `answer` array.

Key Insights

Insight 1: Processing the queries in reverse order allows us to think about adding elements back into the array instead of removing them, which is easier to manage.
Insight 2: The Union Find data structure is well-suited for efficiently tracking contiguous segments of numbers and merging them when adjacent elements are 'added back'.
Insight 3: The use of the `present` array to track which elements are currently present in the array is crucial for the Union Find operations and ensures correct segment identification.

Complexity Analysis

Time Complexity: O(n*alpha(n))

Space Complexity: O(n)

Topics

This problem involves: Array, Union Find, Prefix Sum, Ordered Set.

Companies

Asked at: Infosys.

Maximum Segment Sum After Removals | LeetCode 2382

Hard

Array Union Find Prefix Sum Ordered Set

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

Solution Code

Available in:

class Solution {
    public long[] maximumSegmentSum(int[] nums, int[] removeQueries) {
        int n = nums.length;
        int[] parent = new int[n];
        int[] size = new int[n];
        long[] sums = new long[n];
        
        for (int i = 0; i < n; i++) {
            parent[i] = i;
            size[i] = 1;
            sums[i] = nums[i];
        }
        
        
        
        boolean[] present = new boolean[n];
        long[] answer = new long[n];
        long max_sum = 0;
        
        for (int i = n - 1; i >= 0; i--) {
            answer[i] = max_sum;
            
            int idx = removeQueries[i];
            present[idx] = true;
            
            if (idx > 0 && present[idx - 1]) {
                union(idx, idx - 1, parent, size, sums);
            }
            
            if (idx < n - 1 && present[idx + 1]) {
                union(idx, idx + 1, parent, size, sums);
            }
            
            max_sum = Math.max(max_sum, sums[find(idx, parent)]);
            
        }
        
        return answer;
    }
    
    private int find(int i, int[] parent) {
        if (parent[i] == i) {
            return i;
        }
        parent[i] = find(parent[i], parent);
        return parent[i];
    }
    
    private void union(int i, int j, int[] parent, int[] size, long[] sums) {
        int root_i = find(i, parent);
        int root_j = find(j, parent);
        if (root_i != root_j) {
            if (size[root_i] < size[root_j]) {
                int temp = root_i;
                root_i = root_j;
                root_j = temp;
            }
            
            parent[root_j] = root_i;
            sums[root_i] += sums[root_j];
            size[root_i] += size[root_j];
        }
    }
}

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Problem Info

Description

Problem 2382: Maximum Segment Sum After Removals - Detailed explanation and solution approaches.

Topics

Array Union Find Prefix Sum Ordered Set

Companies

Infosys

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2382

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

7 steps

1
Step 1: Initialize the Union Find data structure: `parent` array (initially each element is its own parent), `size` array (initially each set has size 1), and `sums` array (initially each element is its own sum). Also initialize a `present` array to keep track of which indices are currently present, an `answer` array to store the results, and `max_sum` to keep track of maximum segment sum so far.
2
Step 2: Iterate through `removeQueries` in reverse order (from `n-1` to `0`).
3
Step 3: For each query `i`, store the current `max_sum` in `answer[i]` before applying the removal at index `idx = removeQueries[i]`.
4
Step 4: 'Add' the element back into the array by setting `present[idx] = True`.
5
Step 5: Check if the element has adjacent present elements. If `idx > 0` and `present[idx - 1]` is true, perform `union(idx, idx - 1)`. If `idx < n - 1` and `present[idx + 1]` is true, perform `union(idx, idx + 1)`.
6
Step 6: Update `max_sum` with the maximum of the current `max_sum` and the segment sum of the set that `idx` belongs to (i.e., `sums[find(idx)]`).
7
Step 7: After iterating through all queries, return the `answer` 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*alpha(n))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The Union Find data structure's `find` and `union` operations have amortized time complexity close to O(1) (specifically O(alpha(n))). The dominant factor is the loop iterating through removeQueries which takes O(n) and inside there are constant union find operations which lead to O(n*alpha(n)). We use 4 arrays of size 'n', hence the O(n) space complexity.

📖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

After each removal, iterate through the array to find all contiguous segments and calculate their sums, then find the maximum sum. This involves repeatedly scanning the array for segments.

Time: O(n^2) or O(n^3) depending on how segments are calculated. In worst case every removal will result in scanning of the array.Space: O(n) in the worst case where all the numbers are non-zero and stored as a separate segment, or O(1) if done in place.

Trade-offs: Simple to implement but significantly less efficient than the Union Find approach, especially for large input sizes. Only practical for very small values of 'n'.

✨

Key Insights

1Insight 1: Processing the queries in reverse order allows us to think about adding elements back into the array instead of removing them, which is easier to manage.
2Insight 2: The Union Find data structure is well-suited for efficiently tracking contiguous segments of numbers and merging them when adjacent elements are 'added back'.
3Insight 3: The use of the `present` array to track which elements are currently present in the array is crucial for the Union Find operations and ensures correct segment identification.

📚

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 all elements are removed, the maximum segment sum is 0. The algorithm correctly handles this because `max_sum` is initialized to 0 and updated appropriately.
Edge case 2: When an element is 'added back' but has no adjacent present elements, it forms a segment of its own. The Union Find structure correctly handles this as the element remains its own parent until it's merged with another segment.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not processing the queries in reverse order, which makes it significantly harder to track the segments. Removing elements directly and recalculating sums is inefficient.

✗Common mistake 2: Incorrectly implementing the Union Find data structure, especially the `find` and `union` functions, which can lead to incorrect results or performance issues. Forgetting path compression in `find` is a performance issue.

✗Common mistake 3: Failing to consider the case where an element 'added back' has no adjacent present elements. This can lead to missed segments and an incorrect maximum segment sum.

✅

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: Network connectivity analysis: Determining connected components in a network as links are restored.

🏦Real-world scenario 2: Image segmentation: Grouping pixels with similar properties into segments as more features are gradually included, simulating the merging process of image regions based on the reverse process.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: July 13, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 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 approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Maximum Segment Sum After Removals"?

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

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