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 (Java)

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int n = text1.length();
        int m = text2.length();
        int[][] dp = new int[n + 1][m + 1];
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (text1.charAt(i - 1) == text2.charAt(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]);
                }
            }
        }
        return dp[n][m];
    }
}

Solution (Javascript)

/**
 * @param {string} text1
 * @param {string} text2
 * @return {number}
 */
var longestCommonSubsequence = function(text1, text2) {
    // Create dp table
    const dp = Array(text1.length+1).fill(0).map(() => Array(text2.length+1).fill(0))
    for(let i = 1; i < dp.length; i++) {
        for(let j = 1; j < dp[i].length; j++) {
            // If the letters match, look diagonally to get the max subsequence before this letter and add one
            if(text1[i-1]===text2[j-1]){
                dp[i][j] = dp[i-1][j-1] + 1
            } else {
                // If there is no match, set the cell to the previous current longest subsequence
                dp[i][j] = Math.max(dp[i][j-1], dp[i-1][j])
            }
        }
    }
    return dp[text1.length][text2.length]
};

Explain:

nope.

Complexity:

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