# LeetCode 1143. Longest Common Subsequence

## The Problem

Link to original problem on LeetCode.

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.

## 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.
```

## Constraints

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

`text1`

and`text2`

consist of only lowercase English characters.

## My Solution

### Bottom-Up Dynamic Programming

We can break the problem down into subproblems with dynamic programming. For any pair of characters in `text1`

or `text2`

, if they match, then we know we can add `1`

to the longest common subsequence we've found so far.

But how should we keep track of the longest subsequence we've seen? We can create a 2D array to make a table of values. One "column" for every letter in `text1`

, and one "row" for each letter of `text2`

. We'll add one extra row and column upfront too, just to have an initial `0`

to compare against in the subproblems. Example, where `text1 = 'coder'`

and `text2 = 'ace'`

:

```
c o d e r
0 0 0 0 0 0
a 0 0 0 0 0 0
c 0 1 1 1 1 1
e 0 1 1 1 2 2
```

For each letter in "coder", we see if "a" matches. It never does, so that whole row is `0`

. But when we go to "c", now we have `1`

match! The letter "c" doesn't match anything else in "coder" besides the first letter, so that `1`

just carries over to each position afterward. Finally, we check for "e". We continue carrying over our previous value of `1`

until we hit another match, and increment to `2`

.

In array form, the above table would look like:

`const dp = [`

[0, 0, 0, 0],

[0, 0, 1, 1],

[0, 0, 1, 1],

[0, 0, 1, 1],

[0, 0, 1, 2],

[0, 0, 1, 2],

]

We know that the very last value of this array, `dp[text.length][text2.length]`

will be the length of the longest common subsequence.

`function longestCommonSubsequence(text1: string, text2: string): number {`

// First we initialize an array to store our table of values.

// Since we're using dynamic programming, we'll call it dp.

// It had an outer array of the length of text1 + 1, and each

// value of that array is another array of length equal to

// the length of text2 + 1. We intitialize with all zeroes.

const dp: number[][] = Array.from(

{length: text1.length + 1},

() => Array.from(

{length: text2.length + 1},

() => 0

)

);

for (let i = 1; i <= text1.length; i++) {

for (let j = 1; j <= text2.length; j++) {

// Check if the characters at the i - 1 and j - 1

// indices of text1 and text2 respectively are the same.

if (text1.charAt(i - 1) === text2.charAt(j - 1)) {

// If the characters match, set dp[i][j] to 1 plus

// whatever was stored in dp for the previous indices.

dp[i][j] = 1 + dp[i - 1][j - 1];

} else {

// Otherwise, just set it to the maximum of the previous

// j - 1 value of i, or j value of i - 1.

dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);

}

}

}

// The final values in dp should have the correct length of

// the longest common subsequence.

return dp[text1.length][text2.length];

};