## Description

https://leetcode.com/problems/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:3Explanation:The longest common subsequence is "ace" and its length is 3.

**Example 2:**

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

**Example 3:**

Input:text1 = "abc", text2 = "def"Output:0Explanation: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.

## Explanation

Use dynamic programming approach to solve this problem. f[i][j] represents the longest common subsequence between text1[0 : i] and text2[0 : j].

If text1[i] == text2[j], then the longest common subsequence just increase from previous position from both words by 1: f[i][j] = f[i – 1][j – 1] + 1.

Otherwise, if text1[i] != text2[j], the longest common subsequence would be the longest between text[i – 1] and text2[j] or text1[i] and text2[j – 1].

## Python Solution

```
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
m = len(text1)
n = len(text2)
f = []
for i in range(m + 1):
f.append([])
for j in range(n + 1):
f[i].append(0)
for i in range(m):
for j in range(n):
if text1[i] == text2[j]:
f[i + 1][j + 1] = f[i][j] + 1
else:
f[i + 1][j + 1] = max(f[i][j + 1], f[i + 1][j])
return f[m][n]
```

- Time complexity: O(MN). We’re solving M * N subproblems. Solving each subproblem is an O(1)operation.
- Space complexity: O(MN). We’re allocating a 2D array of size M * N to save the answers to subproblems.