1143. Longest Common Subsequence

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.


Note: Having dp with index 1 makes the m-1, n-1 cases
really easy to handle for the 0-th index.

def longestCommonSubsequence(self, text1: str, text2: str):
   
    M, N = len(text1), len(text2)
    dp = [[0]*(N+1) for _ in range(M+1)]

    for m in range(1, M+1):
        for n in range(1, N+1):                
            if text1[m-1] == text2[n-1]:
                dp[m][n] = dp[m-1][n-1] + 1
            else:
                dp[m][n] = max(dp[m-1][n], dp[m][n-1])
               

    return dp[M][N]

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