1143. Longest Common Subsequence

/*
 * @lc app=leetcode id=1143 lang=cpp
 *
 * [1143] Longest Common Subsequence
 *
 * https://leetcode.com/problems/longest-common-subsequence/description/
 *
 * algorithms
 * Medium (57.74%)
 * Likes:    13643
 * Dislikes: 199
 * Total Accepted:    1.2M
 * Total Submissions: 2.1M
 * Testcase Example:  '"abcde"\n"ace"'
 *
 * 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.
 *
 *
 */

// @lc code=start
class Solution {
public:
    int longestCommonSubsequence(string text1, string text2)
    {
        int m = text1.size(), n = text2.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        for (int i = 1; i < m + 1; ++i)
        {
            for (int j = 1; j < n + 1; ++j)
            {
                if (text1[i - 1] == text2[j - 1])
                {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }
                else
                {
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[m][n];
    }
};
// @lc code=end

• T: $O(M \cdot N)$
• S: $O(M \cdot N)$