## Description

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.

## Solutions

### 1. DP

```
# Time: O(mn)
# Space: O(mn)
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
if not text1 or not text2:
return 0
n1, n2 = len(text1), len(text2)
dp = [[0 for _ in range(n2 + 1)] for _ in range(n1 + 1)]
for i in range(1, n1 + 1):
for j in range(1, n2 + 1):
if text1[i - 1] == text2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
else:
dp[i][j] = max(dp[i][j-1], dp[i-1][j])
return dp[n1][n2]
# Runtime: 440 ms, faster than 60.67%
# Memory Usage: 21.3 MB, less than 100.00%
```