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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1101_1200.s1143_longest_common_subsequence;

// #Medium #Top_100_Liked_Questions #String #Dynamic_Programming
// #Algorithm_II_Day_17_Dynamic_Programming #Dynamic_Programming_I_Day_19
// #Udemy_Dynamic_Programming #Big_O_Time_O(n*m)_Space_O(n*m)
// #2023_06_01_Time_33_ms_(46.23%)_Space_48.2_MB_(90.63%)

/**
 * 1143 - Longest Common Subsequence\.
 *
 * Medium
 *
 * 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.
**/
public 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];
    }
}