# Longest Common Subsequence

## Problem

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.

{% hint style="info" %}
For example:

```
Input: text1 = "abcde", text2 = "ace" 
Output: 3  
Explanation: The longest common subsequence 
is "ace" and its length is 3.
```

```
Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence 
is "abc" and its length is 3.
```

```
Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common 
subsequence, so the result is 0.
```

{% endhint %}

### Thought Process

![](https://1063826111-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-MGdx41c9p2PMgIHbUTK%2F-MNRLOTvvAgc1gXBht8B%2F-MNRPfjqQW5K5U8jEVNa%2FIMG_6A0AE76F06EE-1.jpeg?alt=media\&token=3ad7564d-44fd-4222-908b-00c6413962e4)

## Solution

```
class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        dp = [[0]*(len(text2)+1) for i in range(len(text1)+1)]
        print(dp)
        
        for i in range(1,len(text1)+1):
            for j in range(1,len(text2)+1):
                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]) #removing of letters
                    
        return dp[-1][-1]
        
#Time: O(mn)
#Space: O(mn)
```