What is the time complexity of Maximum Sum of Subsequence With Non-adjacent Elements?

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

What data structures are used to solve Maximum Sum of Subsequence With Non-adjacent Elements?

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

What are the key insights for solving Maximum Sum of Subsequence With Non-adjacent Elements?

Insight 1: Dynamic Programming can be used to efficiently calculate the maximum sum of a non-adjacent subsequence. Insight 2: Segment Tree can be utilized to handle the updates on the array and maintain the dynamic programming state for each segment.

Which companies ask LeetCode 3165. Maximum Sum of Subsequence With Non-adjacent Elements in interviews?

This problem is frequently asked at Infosys.

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Maximum Sum of Subsequence With Non-adjacent Elements - LeetCode 3165 Solution

Maximum Sum of Subsequence With Non-adjacent Elements - Complete Solution Guide

Maximum Sum of Subsequence With Non-adjacent Elements is LeetCode problem 3165, 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 array nums consisting of integers. You are also given a 2D array queries , where queries[i] = [pos i , x i ] . For query i , we first set nums[pos i ] equal to x i , then we calculate the answer to query i which is the maximum sum of a subsequence of nums where no two adjacent elements are selected . Return the sum of the answers to all queries. Since the final answer may be very large, return it modulo 10 9 + 7 . A subsequence is an array that can be derived from another array

Detailed Explanation

The problem asks us to compute the sum of maximum subsequence sums after each update query on an array. For each query, we update a specific element in the array and then calculate the maximum sum of a subsequence with the constraint that no two adjacent elements can be selected in the subsequence. We need to return the sum of these maximum subsequence sums modulo 10^9 + 7.

Solution Approach

The provided solutions utilize a Segment Tree data structure to efficiently handle updates and calculate the maximum non-adjacent subsequence sum after each query. Each node in the Segment Tree stores a matrix that represents the DP states for the corresponding segment of the array. The `merge` operation combines the matrices of two child nodes to calculate the matrix for the parent node. The update operation modifies the matrix of the leaf node corresponding to the updated element and propagates the changes up the tree. For each query, after updating the array, we query the root of the segment tree which contains the result after the updates and sum to get the final answer.

Step-by-Step Algorithm

Step 1: Build the Segment Tree: Initialize the segment tree, where each leaf node represents an element of the input array. Each node stores a 2x2 matrix representing the maximum subsequence sum ending with including or excluding the last element of the segment. Initialize the leaf nodes with appropriate values.
Step 2: Implement the merge operation: Define a function to merge the matrices of two child nodes. The merging calculates the new matrix values for the parent node based on the maximum possible sums by including or excluding the last element from the left and right subsegments.
Step 3: Implement the update operation: When a query arrives, update the corresponding leaf node in the segment tree with the new value. Then, propagate this update up to the root by recomputing the merged values for each parent node along the path.
Step 4: Calculate the answer for each query: After the update, the root of the segment tree holds the matrix representing the entire updated array. Extract the maximum possible subsequence sum from the root node's matrix. Add this result to the total sum (modulo 10^9 + 7).
Step 5: Return the total sum: After processing all queries, return the computed total sum of maximum subsequence sums modulo 10^9 + 7.

Key Insights

Insight 1: Dynamic Programming can be used to efficiently calculate the maximum sum of a non-adjacent subsequence.
Insight 2: Segment Tree can be utilized to handle the updates on the array and maintain the dynamic programming state for each segment.
Insight 3: Representing the DP states as a matrix allows for efficient merging of states from different segments in the segment tree.

Complexity Analysis

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

Space Complexity: O(n)

Topics

This problem involves: Array, Divide and Conquer, Dynamic Programming, Segment Tree.

Companies

Asked at: Infosys.

Maximum Sum of Subsequence With Non-adjacent Elements | LeetCode 3165

Hard

Array Divide and Conquer Dynamic Programming Segment Tree

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maximumSumSubsequence(int[] nums, int[][] queries) {
        int n = nums.length;
        int MOD = (int) 1e9 + 7;
        double neg_inf = -Double.MAX_VALUE;

        double[][] tree = new double[4 * n][4];

        // Initialize tree array
        for (int i = 0; i < 4 * n; i++) {
            Arrays.fill(tree[i], 0.0);
        }

        class Matrix {
            double m11, m12, m21, m22;

            public Matrix(double m11, double m12, double m21, double m22) {
                this.m11 = m11;
                this.m12 = m12;
                this.m21 = m21;
                this.m22 = m22;
            }
        }

        Matrix merge(Matrix mat_L, Matrix mat_R) {
            double l11 = mat_L.m11, l12 = mat_L.m12, l21 = mat_L.m21, l22 = mat_L.m22;
            double r11 = mat_R.m11, r12 = mat_R.m12, r21 = mat_R.m21, r22 = mat_R.m22;

            double c11 = Math.max(l11 + r11, l12 + r21);
            double c12 = Math.max(l11 + r12, l12 + r22);
            double c21 = Math.max(l21 + r11, l22 + r21);
            double c22 = Math.max(l21 + r12, l22 + r22);

            return new Matrix(c11, c12, c21, c22);
        }

        void build(int v, int tl, int tr) {
            if (tl == tr) {
                double val = nums[tl];
                tree[v][0] = 0;
                tree[v][1] = val;
                tree[v][2] = 0;
                tree[v][3] = neg_inf;
                return;
            }

            int tm = (tl + tr) / 2;
            build(2 * v + 1, tl, tm);
            build(2 * v + 2, tm + 1, tr);

            Matrix leftMatrix = new Matrix(tree[2 * v + 1][0], tree[2 * v + 1][1], tree[2 * v + 1][2], tree[2 * v + 1][3]);
            Matrix rightMatrix = new Matrix(tree[2 * v + 2][0], tree[2 * v + 2][1], tree[2 * v + 2][2], tree[2 * v + 2][3]);
            Matrix mergedMatrix = merge(leftMatrix, rightMatrix);

            tree[v][0] = mergedMatrix.m11;
            tree[v][1] = mergedMatrix.m12;
            tree[v][2] = mergedMatrix.m21;
            tree[v][3] = mergedMatrix.m22;
        }

        void update(int v, int tl, int tr, int pos, int new_val) {
            if (tl == tr) {
                tree[v][0] = 0;
                tree[v][1] = new_val;
                tree[v][2] = 0;
                tree[v][3] = neg_inf;
                return;
            }

            int tm = (tl + tr) / 2;
            if (pos <= tm) {
                update(2 * v + 1, tl, tm, pos, new_val);
            } else {
                update(2 * v + 2, tm + 1, tr, pos, new_val);
            }

            Matrix leftMatrix = new Matrix(tree[2 * v + 1][0], tree[2 * v + 1][1], tree[2 * v + 1][2], tree[2 * v + 1][3]);
            Matrix rightMatrix = new Matrix(tree[2 * v + 2][0], tree[2 * v + 2][1], tree[2 * v + 2][2], tree[2 * v + 2][3]);
            Matrix mergedMatrix = merge(leftMatrix, rightMatrix);

            tree[v][0] = mergedMatrix.m11;
            tree[v][1] = mergedMatrix.m12;
            tree[v][2] = mergedMatrix.m21;
            tree[v][3] = mergedMatrix.m22;
        }

        if (n > 0) {
            build(0, 0, n - 1);
        }

        long total_sum = 0;

        for (int[] query : queries) {
            int pos = query[0];
            int val = query[1];
            if (n > 0) {
                update(0, 0, n - 1, pos, val);

                double m11 = tree[0][0];
                double m12 = tree[0][1];

                double ans_q = Math.max(0, Math.max(m11, m12));
                total_sum = (total_sum + (long)ans_q) % MOD;
            }
        }

        return (int) total_sum;
    }
}

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Problem Info

Description

Problem 3165: Maximum Sum of Subsequence With Non-adjacent Elements - Detailed explanation and solution approaches.

Topics

Array Divide and Conquer Dynamic Programming Segment Tree

Companies

Infosys

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3165

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

5 steps

1
Step 1: Build the Segment Tree: Initialize the segment tree, where each leaf node represents an element of the input array. Each node stores a 2x2 matrix representing the maximum subsequence sum ending with including or excluding the last element of the segment. Initialize the leaf nodes with appropriate values.
2
Step 2: Implement the merge operation: Define a function to merge the matrices of two child nodes. The merging calculates the new matrix values for the parent node based on the maximum possible sums by including or excluding the last element from the left and right subsegments.
3
Step 3: Implement the update operation: When a query arrives, update the corresponding leaf node in the segment tree with the new value. Then, propagate this update up to the root by recomputing the merged values for each parent node along the path.
4
Step 4: Calculate the answer for each query: After the update, the root of the segment tree holds the matrix representing the entire updated array. Extract the maximum possible subsequence sum from the root node's matrix. Add this result to the total sum (modulo 10^9 + 7).
5
Step 5: Return the total sum: After processing all queries, return the computed total sum of maximum subsequence sums modulo 10^9 + 7.

💡

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 + q * log(n))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Building the Segment Tree requires traversing the array once, thus takes O(n) time. The height of segment tree is log(n), and updating an element in the array requires modification of each node on the path from leaf to root, which is O(log(n)). The space complexity is O(n) because we store the segment tree which is proportional to size of n.

📖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

Dynamic Programming with Array Updates

A simpler approach would be to recompute the dynamic programming solution for the maximum non-adjacent subsequence sum from scratch after each query. This involves iterating through the array and updating the DP states.

Time: O(n * q), where n is the size of the array and q is the number of queries. For each of the q queries we traverse the array once.Space: O(n), to store the DP array of size n.

Trade-offs: This approach is simpler to implement but less efficient than the Segment Tree approach, especially when there are many queries. Segment Tree provides a better complexity with a cost of code complexity.

✨

Key Insights

1Insight 1: Dynamic Programming can be used to efficiently calculate the maximum sum of a non-adjacent subsequence.
2Insight 2: Segment Tree can be utilized to handle the updates on the array and maintain the dynamic programming state for each segment.
3Insight 3: Representing the DP states as a matrix allows for efficient merging of states from different segments in the segment tree.

📚

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 (n=0). The solutions handle this by checking if n > 0 before building and updating the segment tree. If the array is empty, the total sum remains 0.
Edge case 2: Array with all negative numbers. The algorithm correctly calculates the maximum subsequence sum as 0 (by selecting no elements).
Edge case 3: Large input numbers, leading to integer overflow. The solutions use `long long` to store intermediate sums to prevent overflow and perform modulo operations to keep the final result within the required range.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow when calculating sums, especially when dealing with large numbers. Solutions need to use `long long` and the modulo operator to prevent overflow.

✗Common mistake 2: Incorrectly implementing the `merge` operation for the matrix in the Segment Tree. The merging logic needs to correctly combine the DP states from the child nodes to obtain the correct state for the parent node.

✗Common mistake 3: Forgetting the edge case when the array is empty (n=0) or has all negative elements.

✅

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: Optimizing network configurations where selecting adjacent configurations can lead to conflicts. This algorithm can determine the optimal selection with maximum performance without adjacent conflicts.

🏦Real-world scenario 2: Financial modeling where selecting investment options with non-overlapping risk windows is required. Maximizing investment returns while avoiding consecutive risky investment periods.

💡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.

👨‍💻 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: September 14, 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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 Subsequence With Non-adjacent Elements"?

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

🤖 Need help with this problem?