Published: May 23, 2024

## Problem Description

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.

Constraints:

`1 <= text1.length, text2.length <= 1000`

`text1`

and`text2`

consist of only lowercase English characters.https://leetcode.com/problems/longest-common-subsequence/description/

## Examples

```
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.
```

## How to Solve

This is a typical dynamic programming problem.
The easiest way is to create 2D array to save states up to indices i, j where i, j are index of
each text. If `text1[i]`

is the same as `text2[j]`

, the state will be `matrix[i-1][j-1]+1`

.
If not the same, take the maximum of `matrix[i-1][j]`

and `matrix[i][j-1]`

.
The answer is in the bottom right of the matrix. For example, when “abcdef” and “acbcf” are given,
the algorithm works as in below:

```
| - a b c d e f
--+--------------------------
- | 0 0 0 0 0 0 0
a | 0 1 1 1 1 1 1
c | 0 1 1 2 2 2 2
b | 0 1 2 2 2 2 2
c | 0 1 1 3 3 3 3
f | 0 1 1 3 3 3 4
```

The solution here uses two 1D auxiliary arrays instead of 2D matrix. This is a memory performance tweak. Allocating 1D array is much faster compared to 2D matrix even though it needs two arrays.

## Solution

```
class LCS {
public:
int longestCommonSubsequence(string text1, string text2) {
int m = text1.length(), n = text2.length();
vector<int> prev(n + 1, 0);
for (int i = 1; i <= m; ++i) {
vector<int> cur(n + 1, 0);
for (int j = 1; j <= n; ++j) {
if (text1[i - 1] == text2[j - 1]) {
cur[j] = prev[j - 1] + 1;
} else {
cur[j] = max(cur[j - 1], prev[j]);
}
}
prev = cur;
}
return prev[n];
}
};
```

```
```

```
```

```
class LCS:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
if len(text1) < len(text2):
text1, text2 = text2, text1
m, n = len(text1), len(text2)
prev = [0 for _ in range(n+1)]
for i in range(m):
cur = [0 for _ in range(n+1)]
for j in range(n):
if text1[i] == text2[j]:
cur[j+1] = prev[j]+1
else:
cur[j+1] = max(prev[j+1], cur[j])
prev = cur
return cur[-1]
```

```
```

## Complexities

- Time:
`O(mn)`

- Space:
`O(n)`