---
title: '1143. Longest Common Subsequence'
description: Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0
icon: dot
topics:
  - { name: "String", slug: "string" }
  - { name: "Dynamic Programming", slug: "dynamic-programming" }
  - { name: "Longest Common Subsequence", slug: "longest-common-subsequence" }
sidebar:
  label: 'Longest Common Subsequence'
  badge: 'Medium'
---

### 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.

## Solution

```py
class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        dp = [[0 for j in range(len(text2) + 1)] for i in range(len(text1) + 1)]

        for i in range(len(text1) - 1, -1, -1):
            for j in range(len(text2) - 1, -1, -1):
                if text1[i] == text2[j]:
                    dp[i][j] = 1 + dp[i + 1][j + 1]
                else:
                    dp[i][j] = max(dp[i][j + 1], dp[i + 1][j])

        return dp[0][0]
```