Longest Common Subsequence.cpp

/*
Longest Common Subsequence
Solution
Given two strings text1 and text2, return the length of their longest common subsequence.

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. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A common subsequence of two strings is a subsequence that is common to both strings.

 

If there is no common subsequence, return 0.

 

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 <= 1000
1 <= text2.length <= 1000
The input strings consist of lowercase English characters only.
   Hide Hint #1  
Try dynamic programming. DP[i][j] represents the longest common subsequence of text1[0 ... i] & text2[0 ... j].
   Hide Hint #2  
DP[i][j] = DP[i - 1][j - 1] + 1 , if text1[i] == text2[j] DP[i][j] = max(DP[i - 1][j], DP[i][j - 1]) , otherwise
*/


class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int a[text1.length()+1][text2.length()+1];
        for(long long int i=0;i<text1.length()+1;i++)
        {
            a[i][0] = 0;
            // dp[i][0] = -1;
        }
        for(long long int i=0;i<text2.length()+1;i++)
        {
            a[0][i] = 0;
            // dp[0][i] = -1;
        }
        for(long long int i=1;i<text1.length()+1;i++)
        {
            for(long long int j=1;j<text2.length()+1;j++)
            {
                if(text1[i-1] == text2[j-1])
                {
                    a[i][j] = a[i-1][j-1] + 1;
                }else
                {
                    if(a[i-1][j] >= a[i][j-1])
                    {
                        a[i][j] = a[i-1][j];
                    }else
                    {
                        a[i][j] = a[i][j-1];                 
                    }
                }
            }
        }
        return a[text1.length()][text2.length()];
    }
};