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.

Approach 1: DP

LCS is a classic problem. Let $dp[i][j]$ be the LCS for string $text1$ ends at index $i$ and string $text2$ ends at index $j$ . If $text1[i] == text2[j]$ , then $dp[i][j]$ would be $1 + dp[i - 1][j - 1]$ . Otherwise, we target the largest LCS if we skip one character from either $text1$ or $text2$ , i.e. $dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])$ .

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int n = text1.size(), m = text2.size();
        vector<vector<int>> dp(n + 1, vector<int>(m + 1));
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; 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[n][m];
    }
};

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length();
        int n = text2.length();
        // dp[][] array for storing records of every charcters
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                if (i == 0 || j ==0) {
                    // setting first row and first column to be zero(initial readings)
                    dp[i][j] = 0;
                } else if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
                    /*
                    if match found, then store value of previous diagonal element(dp[i - 1][j - 1])
                    and increase the value by 1 i.e. a new character match is found
                    */
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    /*
                    otherwise, choose maximum of either previous element, either in
                    row(dp[i][j -1]) or column(dp[i][j - 1])
                    */
                    dp[i][j] = Math.max(dp[i][j - 1],dp[i - 1][j]);
                }
            }
        }
        // dp[m][n] would hold the value of the LCS obtained
        return dp[m][n];
    }
}

/**
 * @param {string} text1
 * @param {string} text2
 * @return {number}
 */
var longestCommonSubsequence = function (text1, text2) {
  let n = text1.length;
  let m = text2.length;
  let dp = new Array(n + 1).fill(0).map((x) => new Array(m + 1).fill(0));
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      if (text1[i - 1] != text2[j - 1]) {
        dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
      } else {
        dp[i][j] = 1 + dp[i - 1][j - 1];
      }
    }
  }
  return dp[n][m];
};

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        dp = [[0 for _ in range(len(text2) + 1)] for _ in range(len(text1) + 1)]

        for i in range(len(text1) - 1, -1, -1):
            for j in range(len(text2) - 1, -1, -1):
                if text1[i] == text2[j]:
                    dp[i][j] = 1 + dp[i + 1][j + 1]
                else:
                    dp[i][j] = max(dp[i + 1][j], dp[i][j + 1])
        return dp[0][0]

1143 - Longest Common Subsequence (Medium)

Problem Link

Problem Statement

Approach 1: DP

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