What is the brute force approach and complexity for Longest Common Increasing Subsequence?

Try every subsequence of a; keep the ones that strictly increase and also appear in b with time complexity O(2^N · (N + M)) and space complexity O(N).

What is the optimal approach and complexity for Longest Common Increasing Subsequence?

Walk b once per value of a, keeping the best dp value seen so far with optimal time complexity O(N · M) and space complexity O(M).

How do you identify the pattern for Longest Common Increasing Subsequence?

Look for key signals in the prompt: "common subsequence of both arrays" indicates A chain must be matched in `a` and in `b`, in order. The state needs a position in `b`, not just a value.; "strictly increases" indicates Equal values cannot follow each other, so the test is `<`, not `<=`.; "longest" indicates Keep a length for each cell of `b`, and take the maximum at the end..

Longest Common Increasing Subsequence

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GeeksforGeeks

Problem Statement

You get two arrays, a and b. Find the length of the longest sequence that appears in both arrays as a subsequence, and whose values strictly increase.

A subsequence keeps the order of the values but may skip some of them. Return the length as an int from LCIS(a, b). If the arrays share no value, return 0.

Input

Two arrays a and b. Their sizes are between 1 and 1000, and their values are whole numbers.

Output

The length of the longest common increasing subsequence.

Example 1

Input: a = [3, 4, 9, 1], b = [5, 3, 8, 9, 10, 2, 1]

Output: 2

The longest increasing subsequence that is common is [3, 9], and its length is 2.

Example 2

Input: a = [1, 1, 4, 3], b = [1, 1, 3, 4]

Output: 2

[1, 1] is common, but it is not strictly increasing. The common increasing ones are [1, 4] and [1, 3], each of length 2.

Example 3

Input: a = [1], b = [2]

Output: 0

No value is shared, so there is no common subsequence to count.

Constraints

• 1 ≤ a.size(), b.size() ≤ 10^3
• 1 ≤ a[i], b[i] ≤ 10^4

Pattern Recognition Clues

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

“common subsequence of both arrays”

→ A chain must be matched in `a` and in `b`, in order. The state needs a position in `b`, not just a value.

“strictly increases”

→ Equal values cannot follow each other, so the test is `<`, not `<=`.

“longest”

→ Keep a length for each cell of `b`, and take the maximum at the end.

Approach Complexity Ladder

Brute Force — try every subsequence of aO(2^N · (N + M))O(N)

Try every subsequence of a; keep the ones that strictly increase and also appear in b

Better — scan every earlier cell of bO(N · M^2)O(M)

For each matching cell, scan all earlier cells of b for the best chain to extend

Optimal — one pass over b for each value of aO(N · M)O(M)

Walk b once per value of a, keeping the best dp value seen so far

Brute Force Intuition

Each subsequence of a is a choice of which values to take (2^N choices). For each one, check two things: it strictly increases, and it is a subsequence of b. Keep the longest one that passes both checks.

Recognition rule for this sub-pattern:

Same idea with a custom "can follow" rule; keep parent pointers to rebuild the chain.

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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