# LeetCode 1143. Longest Common Subsequence

## Description

https://leetcode.com/problems/longest-common-subsequence/

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.
```

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.