What is the time complexity of Maximize Subarray Sum After Removing All Occurrences of One Element?

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

What data structures are used to solve Maximize Subarray Sum After Removing All Occurrences of One Element?

This problem uses Array, Dynamic Programming, Segment Tree. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximize Subarray Sum After Removing All Occurrences of One Element?

Insight 1: We need to consider the maximum subarray sum of the original array as well as all possible arrays resulting from removing occurrences of a single element. Insight 2: The problem is efficiently solved by precomputing maximum subarray sums using a Segment Tree, enabling fast querying for different subarrays after removing a value.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Maximize Subarray Sum After Removing All Occurrences of One Element - LeetCode 3410 Solution

Q: Which companies ask LeetCode 3410. Maximize Subarray Sum After Removing All Occurrences of One Element in interviews?

This problem is frequently asked at Rubrik.

Maximize Subarray Sum After Removing All Occurrences of One Element - Complete Solution Guide

Maximize Subarray Sum After Removing All Occurrences of One Element is LeetCode problem 3410, 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 an integer array nums . You can do the following operation on the array at most once: Choose any integer x such that nums remains non-empty on removing all occurrences of x . Remove all occurrences of x from the array. Return the maximum subarray sum across all possible resulting arrays. Example 1: Input: nums = [-3,2,-2,-1,3,-2,3] Output: 7 Explanation: We can have the following arrays after at most one operation: The original array is nums = [ -3, 2, -2, -1, 3, -2, 3 ] . The maxi

Detailed Explanation

The problem asks us to find the maximum subarray sum of a given integer array `nums`. We are allowed to perform at most one operation: choosing any integer `x` present in `nums` and removing *all* occurrences of `x` from the array. The goal is to maximize the subarray sum after potentially removing all occurrences of one chosen element `x`. The array must remain non-empty after the removal operation. If no operation is performed, the entire array is considered, and its maximum subarray sum is calculated. The constraints state that the array length is between 1 and 10^5, and the element values range from -10^6 to 10^6.

Solution Approach

The provided solution uses a segment tree to efficiently calculate the maximum subarray sum. The segment tree is built to store subarray sum information for all possible subarrays within the input array. The algorithm iterates through each unique value in the input array. For each unique value, it considers the array that results from removing all instances of that value. This is achieved by querying the segment tree for the segments that remain after the removal. The algorithm then computes the maximum subarray sum of the remaining segments and updates the overall maximum subarray sum if needed. Special cases like full segments, starts/ends with prefixes/suffixes of other segments are also considered.

Step-by-Step Algorithm

Step 1: Build a segment tree on the original array `nums`. Each node in the segment tree stores the total sum (`s`), maximum prefix sum (`p`), maximum suffix sum (`u`), and maximum subarray sum (`m`) for the corresponding subarray.
Step 2: Calculate the maximum subarray sum of the original array using the segment tree's root node. Initialize `ans` to store the global maximum, setting it to the result obtained in this step.
Step 3: Create a map `val_to_indices` to store the indices of each unique value in the input array. This helps in efficiently identifying the boundaries of segments remaining after removing a value.
Step 4: Iterate through the `val_to_indices` map. For each value `val`, retrieve its indices. If removing `val` would leave the array empty (all elements are `val`), skip this value.
Step 5: Construct the boundaries of the segments that remain after removing `val`. The boundaries are the indices of `val` plus -1 and n. These boundries separate segments that remain. Query these segments from the segment tree.
Step 6: Evaluate the max subarray sums in the segments that resulted from Step 5. Consider the case where we can take full segments, prefixes, and suffixes. Keep track of an overall maximum.
Step 7: Return the overall maximum `ans` found across all possible scenarios (original array and arrays with one value removed).

Key Insights

Insight 1: We need to consider the maximum subarray sum of the original array as well as all possible arrays resulting from removing occurrences of a single element.
Insight 2: The problem is efficiently solved by precomputing maximum subarray sums using a Segment Tree, enabling fast querying for different subarrays after removing a value.
Insight 3: Instead of physically removing elements from the array, we can use a divide-and-conquer strategy combined with segment tree queries to effectively find the max subarray sum in the remaining segments after a value is 'virtually' removed. Careful consideration of overlapping segment sums is crucial.
Insight 4: Kadane's Algorithm can be used on the original array to find max subarray sum without removal for initial comparison and optimization.

Complexity Analysis

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

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Segment Tree.

Companies

Asked at: Rubrik.

Maximize Subarray Sum After Removing All Occurrences of One Element | LeetCode 3410

Hard

Array Dynamic Programming Segment Tree

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

Solution Code

Available in:

class Solution {
    public long maxSubarraySum(int[] nums) {
        int n = nums.length;
        long neg_inf = Long.MIN_VALUE;

        class Node {
            long s, p, u, m;

            Node() {
                this.s = 0;
                this.p = neg_inf;
                this.u = neg_inf;
                this.m = neg_inf;
            }

            Node(long s, long p, long u, long m) {
                this.s = s;
                this.p = p;
                this.u = u;
                this.m = m;
            }
        }

        Node[] tree = new Node[4 * n];
        for (int i = 0; i < 4 * n; i++) {
            tree[i] = new Node();
        }


        Node merge(Node left, Node right) {
            if (left.m == neg_inf) return right;
            if (right.m == neg_inf) return left;

            long s = left.s + right.s;
            long p = Math.max(left.p, left.s + right.p);
            long u = Math.max(right.u, right.s + left.u);
            long m = Math.max(Math.max(left.m, right.m), left.u + right.p);
            return new Node(s, p, u, m);
        }

        void build(int v, int tl, int tr) {
            if (tl == tr) {
                tree[v] = new Node(nums[tl], nums[tl], nums[tl], nums[tl]);
            } else {
                int tm = (tl + tr) / 2;
                build(2 * v, tl, tm);
                build(2 * v + 1, tm + 1, tr);
                tree[v] = merge(tree[2 * v], tree[2 * v + 1]);
            }
        }

        Node query(int v, int tl, int tr, int l, int r) {
            if (l > r) {
                return new Node();
            }
            if (l == tl && r == tr) {
                return tree[v];
            }
            int tm = (tl + tr) / 2;
            Node left_res = query(2 * v, tl, tm, l, Math.min(r, tm));
            Node right_res = query(2 * v + 1, tm + 1, tr, Math.max(l, tm + 1), r);
            return merge(left_res, right_res);
        }

        long ans;
        if (n > 0) {
            build(1, 0, n - 1);
            ans = query(1, 0, n - 1, 0, n - 1).m;
        } else { // Should not happen based on constraints
            return 0;
        }

        java.util.Map<Integer, java.util.List<Integer>> val_to_indices = new java.util.HashMap<>();
        for (int i = 0; i < n; i++) {
            if (!val_to_indices.containsKey(nums[i])) {
                val_to_indices.put(nums[i], new java.util.ArrayList<>());
            }
            val_to_indices.get(nums[i]).add(i);
        }

        for (java.util.Map.Entry<Integer, java.util.List<Integer>> entry : val_to_indices.entrySet()) {
            int val = entry.getKey();
            java.util.List<Integer> indices = entry.getValue();
            if (indices.size() == n) {
                continue;
            }

            java.util.List<Integer> boundaries = new java.util.ArrayList<>();
            boundaries.add(-1);
            boundaries.addAll(indices);
            boundaries.add(n);

            java.util.List<Node> segs_props = new java.util.ArrayList<>();
            for (int i = 0; i < boundaries.size() - 1; i++) {
                int start = boundaries.get(i) + 1;
                int end = boundaries.get(i + 1) - 1;
                if (start <= end) {
                    segs_props.add(query(1, 0, n - 1, start, end));
                }
            }

            if (segs_props.isEmpty()) {
                continue;
            }

            int m = segs_props.size();
            long[] S = new long[m];
            long[] L = new long[m];
            long[] R = new long[m];
            long[] M = new long[m];

            for (int i = 0; i < m; i++) {
                S[i] = segs_props.get(i).s;
                L[i] = segs_props.get(i).p;
                R[i] = segs_props.get(i).u;
                M[i] = segs_props.get(i).m;
            }

            long max_m = neg_inf;
            for(int i = 0; i < m; i++){
                max_m = Math.max(max_m, M[i]);
            }
            ans = Math.max(ans, max_m);

            if (m > 0) {
                // Case: full segments
                long current_s = S[0];
                long max_s = S[0];
                for (int i = 1; i < m; i++) {
                    current_s = Math.max(S[i], S[i] + current_s);
                    max_s = Math.max(max_s, current_s);
                }
                ans = Math.max(ans, max_s);

                // Case: starts full, ends with prefix
                current_s = S[0];
                for (int i = 1; i < m; i++) {
                    ans = Math.max(ans, L[i] + current_s);
                    current_s = Math.max(S[i], S[i] + current_s);
                }

                // Case: starts with suffix, ends full
                current_s = S[m - 1];
                for (int i = m - 2; i >= 0; i--) {
                    ans = Math.max(ans, R[i] + current_s);
                    current_s = Math.max(S[i], S[i] + current_s);
                }

                // Case: starts with suffix, ends with prefix
                if (m > 1) {
                    long[] ps = new long[m + 1];
                    for (int i = 0; i < m; i++) {
                        ps[i + 1] = ps[i] + S[i];
                    }

                    long max_R_minus_ps = R[0] - ps[1];
                    for (int j = 1; j < m; j++) {
                        ans = Math.max(ans, L[j] + ps[j] + max_R_minus_ps);
                        max_R_minus_ps = Math.max(max_R_minus_ps, R[j] - ps[j + 1]);
                    }
                }
            }
        }

        return ans;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 3410: Maximize Subarray Sum After Removing All Occurrences of One Element - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Segment Tree

Companies

Rubrik

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3410

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: Build a segment tree on the original array `nums`. Each node in the segment tree stores the total sum (`s`), maximum prefix sum (`p`), maximum suffix sum (`u`), and maximum subarray sum (`m`) for the corresponding subarray.
2
Step 2: Calculate the maximum subarray sum of the original array using the segment tree's root node. Initialize `ans` to store the global maximum, setting it to the result obtained in this step.
3
Step 3: Create a map `val_to_indices` to store the indices of each unique value in the input array. This helps in efficiently identifying the boundaries of segments remaining after removing a value.
4
Step 4: Iterate through the `val_to_indices` map. For each value `val`, retrieve its indices. If removing `val` would leave the array empty (all elements are `val`), skip this value.
5
Step 5: Construct the boundaries of the segments that remain after removing `val`. The boundaries are the indices of `val` plus -1 and n. These boundries separate segments that remain. Query these segments from the segment tree.
6
Step 6: Evaluate the max subarray sums in the segments that resulted from Step 5. Consider the case where we can take full segments, prefixes, and suffixes. Keep track of an overall maximum.
7
Step 7: Return the overall maximum `ans` found across all possible scenarios (original array and arrays with one value removed).

💡

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*log(n))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Building the segment tree takes O(n) time. Iterating through each distinct number is O(n) in the worst case. Inside the loop, finding indices of numbers will take O(n). Then building the vector of segment properties will take O(n*log(n)) due to the segment tree query. Calculating the max segments takes O(n). Overall time complexity is O(n*log(n)). The space complexity is O(n) due to the segment tree needing 4*n space, the hash map for indices uses O(n) space, and the boundary indices use O(n) space.

📖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

Kadane's Algorithm with Removal Simulation

For each unique value, physically remove all occurrences and then apply Kadane's Algorithm to find the maximum subarray sum of the resulting array. While simple to understand, this approach is less efficient.

Time: O(n^2) in the worst case if many values have many occurrencesSpace: O(n) for storing the modified array

Trade-offs: Easier to implement but less efficient. Suitable for smaller input sizes. The segment tree approch avoids the array copying and is more efficient for large arrays.

Dynamic Programming

A DP approach may be to calculate all possible subarray sums for various combinations of deleted elements. However, this can become exponentially complex since there may be multiple values, yielding O(2^n) complexity.

Time: O(2^n)Space: O(n)

Trade-offs: Not viable due to exponential complexity.

✨

Key Insights

1Insight 1: We need to consider the maximum subarray sum of the original array as well as all possible arrays resulting from removing occurrences of a single element.
2Insight 2: The problem is efficiently solved by precomputing maximum subarray sums using a Segment Tree, enabling fast querying for different subarrays after removing a value.
3Insight 3: Instead of physically removing elements from the array, we can use a divide-and-conquer strategy combined with segment tree queries to effectively find the max subarray sum in the remaining segments after a value is 'virtually' removed. Careful consideration of overlapping segment sums is crucial.
4Insight 4: Kadane's Algorithm can be used on the original array to find max subarray sum without removal for initial comparison and optimization.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty input array. The code explicitly handles this case by returning 0.
Edge case 2: Removing a value that leaves the array empty. The code checks for this and skips the corresponding value if it would result in an empty array.
Edge case 3: Array where no removal yields a better subarray sum than the original. The solution calculates the maximum subarray sum of the original array and compares all removed array scenarios against it.
Edge case 4: Array contains all the same number. This is handled in the case where indices size is n, so the code continues without the segment tree query. The segment tree must handle single element arrays correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to consider the original array's maximum subarray sum. Always start by calculating this as it may be the optimal solution.

✗Common mistake 2: Not handling the edge case where removing a value results in an empty array.

✗Common mistake 3: Incorrectly implementing the segment tree, leading to incorrect subarray sum calculations.

✗Common mistake 4: Inefficiently searching and removing all occurences of values. Use a hash map to store indices for faster removal.

✅

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: Financial data analysis: Identifying the most profitable time period (subarray) in a stock's history after potentially removing periods significantly affected by a specific outlier event.

🏦Real-world scenario 2: Signal processing: Finding the strongest signal within a noisy data stream after filtering out interference from a known source.

🎮Real-world scenario 3: Image processing: Detecting the most prominent feature (subarray of pixels) in an image after removing artifacts caused by sensor errors.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 18, 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

❓

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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 "Maximize Subarray Sum After Removing All Occurrences of One Element"?

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

🤖 Need help with this problem?