Longest Common Subsequence Length

A subsequence is formed by deleting zero or more elements while preserving the order of the remaining elements.

このチャレンジで学べること

2 つの列に対する LCS 表を作る
一致とスキップの遷移を書く
メモリを 2 行分に減らす

問題

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.

制約

Each array has 0…1,000 integers between −1,000,000,000 and 1,000,000,000. Duplicates are allowed.

例

例 1

入力：
first = [1, 3, 2, 4]
second = [3, 1, 2, 4]

期待される出力：
expected = 3

入力：
std::vector<long long> first = {1, 3, 2, 4};
std::vector<long long> second = {3, 1, 2, 4};

期待される出力：
long long expected = 3;

入力：
long[] first = new long[] {1, 3, 2, 4};
long[] second = new long[] {3, 1, 2, 4};

期待される出力：
long expected = 3;

解説

Both contain [1, 2, 4] in that order, after deleting one value. The original arrays differ, so a length-four match is impossible.

例 2

入力：
first = [1, 2]
second = [3, 4]

期待される出力：
expected = 0

入力：
std::vector<long long> first = {1, 2};
std::vector<long long> second = {3, 4};

期待される出力：
long long expected = 0;

入力：
long[] first = new long[] {1, 2};
long[] second = new long[] {3, 4};

期待される出力：
long expected = 0;

解説

The arrays have no shared value, so their only common subsequence is empty.

解き方のヒント

まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。

ヒント 1

隣接ではなく順序を合わせる問題なので、直接の一致より飛ばし方が重要です。

ヒント 2

2つの配列の先頭部分を比べる動的計画法で解きます。

ヒント 3

要素が違うときは、どちらか一方を捨てた場合の良いほうを選びます。同じ値は複数回対応できます。

目安の計算量

時間 O(nm)、追加メモリ O(min(n, m))

参考資料

この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。

CSES: Longest Common Subsequence