## 1. Problem

1143. Longest Common Subsequence

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.

## 2. Solution

I solved this problem like this.

### 2.1. DP

If we use dp, we can solve this problem.

```
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
t1, t2 = len(text1), len(text2)
dp = [[0 for x in range(t2)] for _ in range(t1)]
for r in range(t1):
for c in range(t2):
left = dp[r][c-1] if c > 0 else 0
up = dp[r-1][c] if r > 0 else 0
left_up = dp[r-1][c-1] if c > 0 and r > 0 else 0
if text1[r] == text2[c]:
dp[r][c] = left_up + 1
else:
dp[r][c] = max(left, up)
return dp[-1][-1]
```