1143. Longest Common Subsequence

Problem

Given two strings text1 and text2, return **the length of their longest *common subsequence*. **If there is no common subsequence, return 0.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

• For example, "ace" is a subsequence of "abcde".

A common subsequence of two strings is a subsequence that is common to both strings.

  Example 1:

Input: text1 = "abcde", text2 = "ace" 
Output: 3  
Explanation: The longest common subsequence is "ace" and its length is 3.

Example 2:

Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc" and its length is 3.

Example 3:

Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common subsequence, so the result is 0.

  Constraints:

• 1 <= text1.length, text2.length <= 1000

• text1 and text2 consist of only lowercase English characters.

Solution

/**
 * @param {string} text1
 * @param {string} text2
 * @return {number}
 */
var longestCommonSubsequence = function(text1, text2) {
    var dp = Array(text1.length).fill(0).map(() => Array(text2.length));
    var dfs = function(i, j) {
        if (i === text1.length || j === text2.length) return 0;
        if (dp[i][j] !== undefined) return dp[i][j];
        if (text1[i] === text2[j]) {
            dp[i][j] = 1 + dfs(i + 1, j + 1);
        } else {
            dp[i][j] = Math.max(
                dfs(i + 1, j),
                dfs(i, j + 1),
                dfs(i + 1, j + 1),
            );
        }
        return dp[i][j];
    };
    dfs(0, 0, 0);
    return dp[0][0];
};

Explain:

nope.

Complexity:

• Time complexity : O(n * m).
• Space complexity : O(n * m).