In the dynamic programming approach for LCS, the base

Solution

Detailed Answer and Dry Run: Let's dry run the lcs_length function with text1 = "" and text2 = "ABCD". 1.  Get lengths:     m = len(text1) = 0     n = len(text2) = 4 2.  Initialize dp table:     dp = [[0] * (n + 1) for _ in range(m + 1)]     This becomes dp = [[0] * (4 + 1) for _ in range(0 + 1)]     So, dp will be 1 x 5 (1 row, 5 columns), initialized to all zeros:     dp = [[0, 0, 0, 0, 0]]     This table represents:     dp[0][0] (LCS of "" and "") = 0     dp[0][1] (LCS of "" and "A") = 0     dp[0][2] (LCS of "" and "AB") = 0     dp[0][3] (LCS of "" and "ABC") = 0     dp[0][4] (LCS of "" and "ABCD") = 0 3.  Fill the dp table:     The outer loop is for i in range(1, m + 1).     Since m = 0, range(1, 0 + 1) is range(1, 1), which is an empty range.     Therefore, the outer loop (and consequently the inner loop) will not execute at all. 4.  Return result:     The function will directly proceed to return dp[m][n].     This is return dp[0][4]. From our initialized dp table, dp[0][4] is 0. The final result returned by lcs_length("", "ABCD") will be 0. This is correct because the Longest Common Subsequence of an empty string and any other string is always an empty string, which has a length of 0. This demonstrates how the initialization of the dp table (specifically the first row and column to zeros) effectively handles base cases where one or both strings are empty.