We build a 2D DP table where dp[i][j] is the LCS length of text1[:i] and text2[:j]. If characters match, we extend the LCS by 1; otherwise, we take the max of excluding either character.

1. Create a (m+1) x (n+1) table initialized to 0.
2. For each pair (i, j):
   • 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]).
3. Return dp[m][n].

• Time Complexity: O(M * N).
• Space Complexity: O(M * N).

def longest_common_subsequence(text1, text2):
    m, n = len(text1), len(text2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    
    for i in range(1, m + 1):
        for j in range(1, n + 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])
    
    return dp[m][n]