# Longest Common Subsequence LeetCode Solution

Difficulty Level Medium

## Problem Statement

Longest Common Subsequence LeetCode Solution – Given two strings `text1` and `text2`, return the length of their longest common subsequenceIf there is no common subsequence, return `0`.

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"`.

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.```

## Explanation

• The idea is to use dynamic programming.
• Create a 2D DP array of n+1*m+1 dimensions
•  I am using the bottom up dp approach so i have iterated from text1.length()-1 to 0 and text2.length()-1 to 0
• if the characters match,we do 1+ the diagnal value dp[i+1][j+1]
• else we find the max between left and bottom value max(dp[i][j+1],dp[i+1][j])
• return the dp as it will have the value for longest common Subsequence

## Code

### Longest Common Subsequence Leetcode Java Solution:

```class Solution {
public int longestCommonSubsequence(String text1, String text2) {
int [][]dp = new int[text1.length()+1][text2.length()+1];
for(int i= text1.length()-1;i>=0;i--){
for(int j = text2.length()-1;j>=0;j--){
char ch1 = text1.charAt(i);
char ch2 = text2.charAt(j);
if(ch1==ch2) // diagnal
dp[i][j]= 1+dp[i+1][j+1];
else// left,down
dp[i][j] = Math.max(dp[i][j+1],dp[i+1][j]);

}
}
return dp;
}
}```

### C++ Solution:

```class Solution {
public:
int longestCommonSubsequence(string text1, string text2) {

//         // base case
//         if( text1.length() == 0 || text2.length() == 0) return 0;

//         // recursive call
//         if( text1 == text2) return 1 + longestCommonSubsequence( text1.substr(1) , text2.substr(1) );

//         else return max(  longestCommonSubsequence( text1.substr(1) , text2  ) ,  longestCommonSubsequence( text1  , text2.substr(1) ));

int n = text1.length();
int m = text2.length();

int dp[n+1][m+1];

for( int i = 0 ; i <= m ; i++) dp[i] = 0;
for( int i = 0 ; i <= n ; i++) dp[i] = 0;

for(int i = 1 ; i <= n ; i++){

for( int j = 1 ; j <= m ; j++){

if( text1[i-1] == text2[j-1]) dp[i][j] = 1 + dp[i-1][j-1];

else dp[i][j] = max( dp[i][j-1] , dp[i-1][j]) ;
}
}

return dp[n][m];

}
};```

## Complexity Analysis:

• Time Complexity: o(nm)
• Space Complexity :o(nm)
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