What is the brute force approach and complexity for Max Sum Increasing Subsequence?

Recurse over the items: skip one, or take it if it is bigger than the last one taken, adding its value with time complexity O(2^N) and space complexity O(N).

What is the optimal approach and complexity for Max Sum Increasing Subsequence?

msis[i] = arr[i] + the best msis[j] over earlier smaller j; the answer is the largest msis with optimal time complexity O(N^2) and space complexity O(N).

Max Sum Increasing Subsequence

Q: How do you identify the pattern for Max Sum Increasing Subsequence?

Look for key signals in the prompt: "subsequence" indicates Items can be skipped, so the chain is any set of positions in order.; "strictly increasing" indicates Each chosen item must be bigger than the one before it in the chain. Equal values are not allowed.; "largest sum" indicates Each item adds its value. A longer chain is not always better, since a short chain can have a bigger sum..

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Problem Statement

You are given an array arr of positive integers.

Pick a subsequence whose values are strictly increasing, and add up its values. Return the largest sum that any such subsequence can give. The answer is a sum, not a count of items.

The function is int maxSumIS(vector<int>& arr).

Input

An array of positive integers arr.

Output

One number: the largest sum of a strictly increasing subsequence.

Example 1

Input: arr = [1, 101, 2, 3, 100]

Output: 106

The chain 1, 2, 3, 100 sums to 106. The chain 1, 101 sums to 102, which is less.

Example 2

Input: arr = [4, 1, 2, 3]

Output: 6

The chain 1, 2, 3 sums to 6. The single 4 gives only 4.

Example 3

Input: arr = [4, 1, 2, 4]

Output: 7

The chain 1, 2, 4 sums to 7. Two 4s cannot be in a strictly increasing chain.

Constraints

• 1 ≤ arr.size() ≤ 10^3
• 1 ≤ arr[i] ≤ 10^5

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“subsequence”

→ Items can be skipped, so the chain is any set of positions in order.

“strictly increasing”

→ Each chosen item must be bigger than the one before it in the chain. Equal values are not allowed.

“largest sum”

→ Each item adds its value. A longer chain is not always better, since a short chain can have a bigger sum.

Approach Complexity Ladder

Brute Force — take or skip each itemO(2^N)O(N)

Recurse over the items: skip one, or take it if it is bigger than the last one taken, adding its value

Optimal — msis[i]: the best sum of a chain that ends at arr[i]O(N^2)O(N)

msis[i] = arr[i] + the best msis[j] over earlier smaller j; the answer is the largest msis

Brute Force Intuition

Walk through the items from left to right. For each one, either skip it, or take it when it is bigger than the last value taken, and add its value to the sum. The answer is the best sum over all the paths.

Recognition rule for this sub-pattern:

O(n²): try every earlier j. O(n log n): binary search the tails array.

Remember this · DP — Longest Increasing Subsequence

Trigger

Longest chain where each item must be bigger/fit inside the previous one (numbers, pairs, envelopes, words).

Think

Does each element extend the best chain ending at some smaller earlier element?

Key question

Sort first? Then what does "increasing" mean for these items?

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