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Longest Common Subsequence⚓︎

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Description⚓︎

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⚓︎

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class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int n = text1.length(), m = text2.length();
        text1 = ' ' + text1;
        text2 = ' ' + text2;
        vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0)); 
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                if (text1[i] == text2[j])
                    dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
            }
        }
        return dp[n][m];
    }
};

Other way of writing:

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class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int n = text1.length(), m = text2.length();
        vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0)); 
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                if (text1[i - 1] == text2[j - 1])
                    dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
            }
        }
        return dp[n][m];
    }
};
  • Time complexity: \(O(nm)\);
  • Space complexity: \(O(nm)\).