# Solving Leetcode Interviews in Seconds with AI: Longest Common Subsequence


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1143" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.   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
	Here's a breakdown of the approach, followed by the Python code:

*   **Dynamic Programming:** Build a 2D table (DP table) where `dp[i][j]` stores the length of the longest common subsequence of `text1[0...i-1]` and `text2[0...j-1]`.
*   **Base Case:** The first row and column of the DP table are initialized to 0, representing the case where one of the strings is empty.
*   **Recursive Relation:** If `text1[i-1]` and `text2[j-1]` are equal, then `dp[i][j] = dp[i-1][j-1] + 1`.  Otherwise, `dp[i][j] = max(dp[i-1][j], dp[i][j-1])`.

*   **Runtime Complexity:** O(m\*n), where 'm' is the length of `text1` and 'n' is the length of `text2`.
*   **Storage Complexity:** O(m\*n)

	
	# Code
	```python
	def longestCommonSubsequence(text1: str, text2: str) -> int:
    """
    Finds the length of the longest common subsequence of two strings.

    Args:
        text1: The first string.
        text2: The second string.

    Returns:
        The length of the longest common subsequence.
    """

    n = len(text1)
    m = len(text2)

    # Initialize a DP table to store lengths of LCS for subproblems
    dp = [[0] * (m + 1) for _ in range(n + 1)]

    # Iterate through the strings to populate the DP table
    for i in range(1, n + 1):
        for j in range(1, m + 1):
            if text1[i - 1] == text2[j - 1]:
                dp[i][j] = dp[i - 1][j - 1] + 1
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

    # The bottom-right cell contains the length of the LCS of the entire strings
    return dp[n][m]
	```