Consider the recursive algorithm below for computing the sum of the first n cubes: S(n) = 13 + 23 + … + n3 ALGORITHM S(n) //n is a positive integer if (n==1) return 1 else return S(n – 1) + (n*n*n);     Solve the recurrence relation in your answer for the previous question. The runtime efficiency of the algorithm is?    A.  O(n)  B.  O(n2)  C.  O(n3)  D.  None of the above

write a computer program that produces the desired output from the given input. Input: Elements in a finite set SOutput: Elements in ℘(S)Algorithm: Use recursion.

Which among the following is the recursive definition of Factorial , i.e., n! ? f(0) = 1, f(n) = (n)f(n – 1), where n E Z andn 2 1. %D f(0) = 1, f (n) = (n + 1)f(n – 1), where n E Z and n > 1. f(0) = 0, f(n) = (n)f(n – 1), where n E Z and n > 1. %3D f (0) = 0, f(n) = (n + 1)f(n – 1), where n E Z and n 2 1.

Find the running time for each of the following algorithms. Show work by finding a table of values for each while loop, writing the summations, then solving. Be sure to show work on both the upper bound and lower bound, justify the split, and check that the bounds differ by only a constant factor. Use asymptotic notation to write the answer. b) Func3(n) 1 2 3 4 5 6 7 8 9 10 11 S← 0; i+n; while (i 8) do s+ s + i - j; j← lj/⁹]; end i + 4 x i; end return (s);

can you please fix the AssertionError in this: (please don't use try and except or break/continue or recursion) def long_new(s: str) -> str: longest_string = 0longest_substring = s[0]i = 0inputs = {s[i]: i}j = i + 1while j < len(s):while j < len(s) and (inputs.get(s[j], -1) == -1):inputs[s[j]] = jj = j + 1if j - 1 > longest_string:longest_string = j - ilongest_substring = s[i:j]if j >= len(s):return longest_substringi = inputs[s[j]] + 1inputs = {}j = i + 1if j - i > longest_string:longest_string = j - ilongest_substring = s[i:j]return longest_substring print(long_new(aab)) The print statement should print 'ab'

Please show and explain so I can nderstand.

Please show work to the answer.

Given an input string, a non-empty substring subs, and a non-negative number n, compute recursively and return true if and only if at least n copies of subs occur in the string, possibly with overlapping. Do not use loops (do not write for or while anywhere in your code). Do not use any regular expressions and methods such as matches, split, replaceAll. Test case 1: checkSubstring("abcxxxxabc", "abc", 2) true Test case 2: checkSubstring("abcxxxxabc", "abc", 3) false Test case 3: checkSubstring ("ababaxxxaba", "aba", 3) true

Consider the following recursive algorithm, where // denotes integer division: 3//2 = 1, 5//2 = 2, etc. F(n):if n <= 1:    returnF(n//2)for i from 0 to n    for j from 0 to n//2        print(i+j)    Let function T(n) denote the running time of this algorithm. Derive T(n) and prove its worst case timecomplexity

Use Binary Search, Recursive (Algorithm 2.1) to search for the integer 120 in the following list (array) of integers. Show the actions (execution) step by step. No code needed. Solution: 12 34 37 45 57 82 99 120 134 1. Compute the middle index from initial low (1) and high (9) indices: mid = Continue... A