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      Question

      What are the time and space complexities of the standard

      dynamic programming approach for finding the length of the Longest Common Subsequence (LCS) of two strings, text1 of length m and text2 of length n? Consider the following Python code: def lcs_length(text1, text2):     m = len(text1)     n = len(text2)     dp = [[0] * (n + 1) for _ in range(m + 1)] # Space complexity     for i in range(1, m + 1): # Outer loop runs m times         for j in range(1, n + 1): # Inner loop runs n times             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]
      A Time: O(m + n), Space: O(m + n) Correct Answer Incorrect Answer
      B Time: O(m * n), Space: O(m + n) Correct Answer Incorrect Answer
      C Time: O(m * n), Space: O(m * n) Correct Answer Incorrect Answer
      D Time: O(2^(m+n)), Space: O(m * n) Correct Answer Incorrect Answer

      Solution

      Detailed Answer and Dry Run: Let's analyze the time and space complexity of the provided Python code for LCS. Time Complexity: The core of the algorithm involves two nested loops: The outer loop iterates m times (from i = 1 to m). The inner loop iterates n times (from j = 1 to n). Inside the inner loop, a constant number of operations are performed: Character comparison (text1[i - 1] == text2[j - 1]). Arithmetic operations (addition, max). Array assignments. Therefore, the total number of operations is proportional to m * n. The time complexity is O(m * n). Space Complexity: The algorithm uses a 2D list (or array) named dp to store the results of subproblems. The dp table has m + 1 rows and n + 1 columns. The total number of cells in this table is (m + 1) * (n + 1), which simplifies to O(m * n). Each cell stores an integer. Thus, the space required to store this table is proportional to m * n. The space complexity is O(m * n). Dry Run Example (Conceptual): If text1 has length 5 and text2 has length 7: The dp table will be 6 x 8. The outer loop runs 5 times. The inner loop runs 7 times for each iteration of the outer loop. Total operations for filling the table: approximately 5 * 7 = 35 cell computations. Total space for the table: 6 * 8 = 48 integer cells. This confirms the O(m * n) time and space complexities.

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