# 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];};