1143. Longest Common Subsequence.cpp

/*
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.
*/

// Using dp matrix (LCS Calculation Method) (Unoptimised Code)
class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int dp[1001][1001] = {};            // initialising matrix for max length of string (constraints)
        for (int i = 0; i < text1.size(); ++i){
            for (int j = 0; j < text2.size(); ++j){
                if (text1[i] == text2[j])
                    dp[i + 1][j + 1] = 1 + dp[i][j];
                else
                    dp[i + 1][j + 1] = max(dp[i + 1][j], dp[i][j + 1]);
            }
        }
        return dp[text1.size()][text2.size()];
    }
};

// the above code uses unnecessary space

// Optimised Code (Using Only 2 Rows to calculate the max length of LCS)
class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int dp[2][1001] = {};
        for (int i = 0; i < text1.size(); ++i){
            for (int j = 0; j < text2.size(); ++j){
                if (text1[i] == text2[j])
                    dp[!(i % 2)][j + 1] = 1 + dp[i % 2][j]; // Using "%" so that the range does not go out of bound
                else
                    dp[!(i % 2)][j + 1] = max(dp[!(i % 2)][j], dp[i % 2][j + 1]);
            }
        }
        return dp[text1.size() % 2][text2.size()];
    }
};