In the dynamic programming solution for Matrix Chain

Solution

• Code Analysis: o n is the number of matrices. o len should iterate from the smallest possible chain length (2 matrices) up to the total number of matrices (n). o i should iterate through all possible starting points for a subproblem of length len. o j = i + len calculates the ending index of the subproblem. • Explanation of Correct Answer (B): len = 2; len < n; len++ and i = 0 o len loop:  len starts from 2 because a chain of length 1 (a single matrix) requires no multiplication. The smallest meaningful chain is two matrices.  len < n: The chain length len goes up to n-1 (if n is the number of matrices, then n-1 is the maximum chain length for subproblems, as j goes up to n-1). If n is the size of the arr (dimensions array), then n-1 is the number of matrices. The loop should go up to n-1 matrices, so len < n is correct. o i loop:  i starts from 0 because the first matrix can be at index 0.  i < n - len: This ensures that j = i + len does not exceed the bounds of the matrix array. If n is the number of matrices, then i can go up to n - len. For example, if len = n-1, i can only be 0.