LeetCode #3165 — HARD

Maximum Sum of Subsequence With Non-adjacent Elements

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given an array nums consisting of integers. You are also given a 2D array queries, where queries[i] = [posi, xi].

For query i, we first set nums[posi] equal to xi, 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 109 + 7.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [3,5,9], queries = [[1,-2],[0,-3]]

Output: 21

Explanation:
After the 1st query, nums = [3,-2,9] and the maximum sum of a subsequence with non-adjacent elements is 3 + 9 = 12.
After the 2nd query, nums = [-3,-2,9] and the maximum sum of a subsequence with non-adjacent elements is 9.

Example 2:

Input: nums = [0,-1], queries = [[0,-5]]

Output: 0

Explanation:
After the 1st query, nums = [-5,-1] and the maximum sum of a subsequence with non-adjacent elements is 0 (choosing an empty subsequence).

Constraints:

  • 1 <= nums.length <= 5 * 104
  • -105 <= nums[i] <= 105
  • 1 <= queries.length <= 5 * 104
  • queries[i] == [posi, xi]
  • 0 <= posi <= nums.length - 1
  • -105 <= xi <= 105
Patterns Used
📊Dynamic Programming→📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an array nums consisting of integers. You are also given a 2D array queries, where queries[i] = [posi, xi]. For query i, we first set nums[posi] equal to xi, 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 109 + 7. A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Segment Tree

Example 1

[3,5,9]
[[1,-2],[0,-3]]

Example 2

[0,-1]
[[0,-5]]
Step 02

Core Insight

What unlocks the optimal approach

  • Can you solve each query in <code>O(nums.length)</code> with dynamic programming?
  • In order to optimize, we will use segment tree where each node contains the maximum value of (front element has been chosen or not, back element has been chosen or not).
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3165: Maximum Sum of Subsequence With Non-adjacent Elements
class Node {
    int l, r;
    long s00, s01, s10, s11;

    Node(int l, int r) {
        this.l = l;
        this.r = r;
        this.s00 = this.s01 = this.s10 = this.s11 = 0;
    }
}

class SegmentTree {
    Node[] tr;

    SegmentTree(int n) {
        tr = new Node[n * 4];
        build(1, 1, n);
    }

    void build(int u, int l, int r) {
        tr[u] = new Node(l, r);
        if (l == r) {
            return;
        }
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
    }

    long query(int u, int l, int r) {
        if (tr[u].l >= l && tr[u].r <= r) {
            return tr[u].s11;
        }
        int mid = (tr[u].l + tr[u].r) >> 1;
        long ans = 0;
        if (r <= mid) {
            ans = query(u << 1, l, r);
        }
        if (l > mid) {
            ans = Math.max(ans, query(u << 1 | 1, l, r));
        }
        return ans;
    }

    void pushup(int u) {
        Node left = tr[u << 1];
        Node right = tr[u << 1 | 1];
        tr[u].s00 = Math.max(left.s00 + right.s10, left.s01 + right.s00);
        tr[u].s01 = Math.max(left.s00 + right.s11, left.s01 + right.s01);
        tr[u].s10 = Math.max(left.s10 + right.s10, left.s11 + right.s00);
        tr[u].s11 = Math.max(left.s10 + right.s11, left.s11 + right.s01);
    }

    void modify(int u, int x, int v) {
        if (tr[u].l == tr[u].r) {
            tr[u].s11 = Math.max(0, v);
            return;
        }
        int mid = (tr[u].l + tr[u].r) >> 1;
        if (x <= mid) {
            modify(u << 1, x, v);
        } else {
            modify(u << 1 | 1, x, v);
        }
        pushup(u);
    }
}

class Solution {
    public int maximumSumSubsequence(int[] nums, int[][] queries) {
        int n = nums.length;
        SegmentTree tree = new SegmentTree(n);
        for (int i = 0; i < n; ++i) {
            tree.modify(1, i + 1, nums[i]);
        }
        long ans = 0;
        final int mod = (int) 1e9 + 7;
        for (int[] q : queries) {
            tree.modify(1, q[0] + 1, q[1]);
            ans = (ans + tree.query(1, 1, n)) % mod;
        }
        return (int) ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O((n + q)
Space
O(n)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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