Problem
A subsequence is formed by deleting zero or more elements while preserving the order of the remaining elements. Find the longest length shared by two arrays. Matching elements need not be adjacent. Return only the length, since several equally long subsequences may exist.
Implement longest_common_subsequence_length(first, second).
Input
first: The first sequence.
second: The second sequence.
Output
Return the maximum common subsequence length. If either array is empty, return 0.
Constraints
Each array has 0…1,000 integers between −1,000,000,000 and 1,000,000,000. Duplicates are allowed.
Examples
Input:
first = [1, 3, 2, 4]
second = [3, 1, 2, 4]
Expected output:
expected = 3Input:
std::vector<long long> first = {1, 3, 2, 4};
std::vector<long long> second = {3, 1, 2, 4};
Expected output:
long long expected = 3;Input:
long[] first = new long[] {1, 3, 2, 4};
long[] second = new long[] {3, 1, 2, 4};
Expected output:
long expected = 3;Explanation
Both contain [1, 2, 4] in that order, after deleting one value. The original arrays differ, so a length-four match is impossible.
Input:
first = [1, 2]
second = [3, 4]
Expected output:
expected = 0Input:
std::vector<long long> first = {1, 2};
std::vector<long long> second = {3, 4};
Expected output:
long long expected = 0;Input:
long[] first = new long[] {1, 2};
long[] second = new long[] {3, 4};
Expected output:
long expected = 0;Explanation
The arrays have no shared value, so their only common subsequence is empty.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
You are matching order, not adjacency, so skip choices matter more than direct alignment.
Hint 2
Use dynamic programming where each state compares prefixes of the two arrays.
Hint 3
When elements differ, take the better of dropping one from either sequence; repeated values can create many valid matches.
Expected complexity
O(nm) time, O(min(n, m)) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.