LeetCode 1143. Longest Common Subsequence (javascript)

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.

Idea:

Dynamic Programing

Use dp[i][j] to represent the length of longest common sub-sequence of text1[0:i-1] and text2[0:j-1]
dp[i][j] = dp[i – 1][j – 1] + 1 if text1[i – 1] == text2[j – 1] else max(dp[i][j – 1], dp[i – 1][j])

Solution:

/**
 * @param {string} text1
 * @param {string} text2
 * @return {number}
 */
var longestCommonSubsequence = function(text1, text2) {
    // Use dp[i][j] to represent the length of longest common sub-sequence of text1[0:i-1] and text2[0:j-1]
    // dp[i][j] = dp[i – 1][j – 1] + 1 if text1[i – 1] == text2[j – 1] else max(dp[i][j – 1], dp[i – 1][j])
    let dp = Array.from(new Array(text1.length + 1), () => new Array(text2.length + 1));
    
    // final result dp[i + 1][j + 1] contains length of LCS  
    // for text1[0..i] and text2[0..j], so use <=
    for (let i = 0; i <= text1.length; i++) {
        for (let j = 0; j <= text2.length; j++) {
            // base case:
            if (i === 0 || j === 0) 
                dp[i][j] = 0;
            else if (text1[i - 1] === text2[j - 1])
                dp[i][j] = dp[i - 1][j - 1] + 1;
            else
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
        }
    }
    // dp[text1.length][text2.length] contains length of LCS  
    // for text1[0..text1.length-1] and text2[0..text2.length-1]
    return dp[text1.length][text2.length];
}