The given time complexity is: m T(m-1) + ca > 1 cb = 0 expanding using substitution: m T(m-1) + ca m[(m-1) T(m-2) + ca]+ ca m(m-1) T(m-2) + mca + ca m(m-1) [(m-2) T(m-3) + ca] + mca + ca m(m-1)(m-2) T(m-3) + m(m-1)ca + mca + ca what is the time complexity? form an expression for adding all the ca
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The given time complexity is:
m T(m-1) + ca > 1
cb = 0
expanding using substitution:
m T(m-1) + ca
m[(m-1) T(m-2) + ca]+ ca
m(m-1) T(m-2) + mca + ca
m(m-1) [(m-2) T(m-3) + ca] + mca + ca
m(m-1)(m-2) T(m-3) + m(m-1)ca + mca + ca
what is the time complexity?
form an expression for adding all the ca
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- Given two sequences X and Y, we want to find Z i.e., the longest common subsequence (LCS) of both X and Y. Because we know that subsequences of interest are of length >= a and <= b, we threshold the LCS definition to |Z| in [a, b] (where |Z| is the sequence length). In python state the recursive formula for this problem. Implement the algorithm with bottom-up approach dynamic programming and backtracking. Create an iterative function that prints the longest common subsequence(LCS).Assume that each of the expressions in the following table has processing time T(n) to solve a problem of size n. Identify the dominant term(s) having the growing increase in n and specify the Big-Oh complexity. Expression Dominant term(s) Big-Oh A.1 8 + 0.081n2 + 0.040n A.2 100 + n log2 n A.3 100n + 0.081 log3 n + n3 A.4 0.081 log4 n + 81nDetermine the time complexity for f1 and fr: def f1(n): 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 if n <= 1: return 1 else: return f1(n − 1) + f1(n − 2) def fr(n): t = [0 for i in range(n + 1)] for i in range (n + 1): if i <= 1: t[i] = 1 else: t[i] = t[n 1] + t[n - 2] return t[n] -
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