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    Question

    Consider the standard dynamic programming approach to

    find the length of the Longest Common Subsequence (LCS) of two strings, text1 and text2. The dp table is initialized with dimensions (len(text1) + 1) x (len(text2) + 1). Given text1 = "ABC" and text2 = "AXBY", what will be the value of dp[0][3] and dp[2][0] after the initialization and the first row/column filling steps of the following Python code? def lcs_length(text1, text2):     m = len(text1)     n = len(text2)     dp = [[0] * (n + 1) for _ in range(m + 1)]     # Initialization (first row and column are already 0 by default in Python list comprehension)     # for i in range(m + 1):     #     dp[i][0] = 0     # for j in range(n + 1):     #     dp[0][j] = 0     # Filling the DP table     for i in range(1, m + 1):         for j in range(1, n + 1):             if text1[i - 1] == text2[j - 1]:                 dp[i][j] = 1 + dp[i - 1][j - 1]             else:                 dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])     return dp[m][n] # For this question, we only care about the state after initialization # and before the main loops start filling values beyond the first row/column. # The Python list comprehension [[0] * (n + 1) for _ in range(m + 1)] # already initializes all cells to 0.
    A dp[0][3] = 0, dp[2][0] = 0 Correct Answer Incorrect Answer
    B dp[0][3] = 3, dp[2][0] = 2 Correct Answer Incorrect Answer
    C dp[0][3] = 0, dp[2][0] = 2 Correct Answer Incorrect Answer
    D dp[0][3] = 3, dp[2][0] = 0 Correct Answer Incorrect Answer

    Solution

    Detailed Answer and Dry Run: The Longest Common Subsequence (LCS) dynamic programming approach uses a 2D table, dp, where dp[i][j] stores the length of the LCS of text1[0...i-1] and text2[0...j-1]. For text1 = "ABC" (length m=3) and text2 = "AXBY" (length n=4), the dp table will have dimensions (m+1) x (n+1), which is 4 x 5. The Python code initializes the dp table using a list comprehension: dp = [[0] * (n + 1) for _ in range(m + 1)]. This creates a table where all elements are initially set to 0. Let's visualize the dp table after this initialization:     ""  A   X   B   Y ""  [0, 0, 0, 0, 0] A   [0, 0, 0, 0, 0] B   [0, 0, 0, 0, 0] C   [0, 0, 0, 0, 0] The first row (dp[0][j]) represents the LCS length when text1 is an empty string (""). The LCS of an empty string and any prefix of text2 is always 0. Similarly, the first column (dp[i][0]) represents the LCS length when text2 is an empty string. The LCS of any prefix of text1 and an empty string is also 0. Therefore, after initialization, dp[0][3] (LCS of "" and "AXB") will be 0, and dp[2][0] (LCS of "AB" and "") will also be 0. The main loops for i in range(1, m + 1) and for j in range(1, n + 1) then proceed to fill the rest of the table, but the question specifically asks for the values after initialization and the first row/column filling, which are inherently 0 due to the Python initialization.

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