2.)Recurrences. a.)Provide a Sample code for the recurrences below. i.) T(n)=T(n/5)+T(3n/5)+n ii.)T(n)=T(2n/5)+T(3n/5)+n iii.) T(n)=4T(n/2)+n^3
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- def longest_unique_substring(s: str) -> str:"""Given a string <s>, return the longest unique substring that occurs within<s>.A unique substring is a substring within <s> which DOES NOT have anyrepeating characters. As an example, "xd" is unique but "xxd" is not.If there are two equal length unique substrings within <s>, return the onethat starts first (i.e., begins at a smaller index).tips:In order to get your function to run fast, consider using a dictionary tostore the indexes of previously seen characters, from there, you canfollow a set of rules based on each new character you see to determinethe length of the longest unique substring seen so far.>>> longest_unique_substring('aab')'ab' >>> longest_unique_substring('abcabcbb')'abc'""" RESTRICTIONS: - Do not add any imports and do it on python .Do not use recursion. Do not use break/continue.Do not use try-except statements.def longest_unique_substring(s: str) -> str:"""Given a string <s>, return the longest unique substring that occurs within<s>.A unique substring is a substring within <s> which DOES NOT have anyrepeating characters. As an example, "xd" is unique but "xxd" is not.If there are two equal length unique substrings within <s>, return the onethat starts first (i.e., begins at a smaller index).tips:In order to get your function to run fast, consider using a dictionary tostore the indexes of previously seen characters, from there, you canfollow a set of rules based on each new character you see to determinethe length of the longest unique substring seen so far.>>> longest_unique_substring('aab')'ab'""" RESTRICTIONS: - Do not add any imports and do it on python .Do not use recursion. Do not use break/continue.Do not use try-except statements.Describe each of the following sets by listing its elements:a. {x | x ∈ N and x2 − 5x + 6 = 0}b. {x | x ∈ R and x2 = 19}c. {x | x ∈ N and x2 − 2x − 8 = 0}}d. {x | x ∈ Z and x ∗ x = 4}
- Implement the following function, without using any data structure or #include <bits/stdc++.h> /* Given two vectors of chars, check if the two vectors are permutations of each other, i.e., they contains same values, in same or different order.e.g., V1=[‘a’,’b’,’a’] and V2=[‘b’,’a’,’a’] stores same multi-set of data points: i.e., both contains two ‘a’, and one ‘b’. e.g., V3=[‘a’,’c’,’t’,’a’] and V4=[‘a’,’c’,’t’] are not same multi-set. V3 contains two ‘a’s, while V4 has only one ‘a’. Note: when considering multiset, the number of occurrences matters. @param list1, list2: two vectors of chars @pre: list1, list2 have been initialized @post: return true if list1 and list2 stores same values (in same or different order); return false, if not. */ bool SameMultiSet (vector<char> list1, vector<char> list2) THIS IS NOT THE CORRECT SOLUTION / include headers #include <bits/stdc++.h> // deinfe the namespace using namespace std; // function fo rchecking same bool…def find_root4(x, epsilon): ''' IN PYTHON Assume: x, epsilon are floating point numbers and epsilon > 0 Use bisection search to find the following root of x such that If x >=0, return y such that x - epsilon <= y ** 2 <= x + epsilon Else, return y such that x - epsilon <= y ** 7 <= x + epsilon Note: You must use bisection search to implement the function. ''' passWith the following C++ code, implement the “indirect sum” method of computing a sum. You will need to provide an implementation inside both the setup() and sum() functions in the sum_indirect.cpp file in the code harness. Here, setup() consists of initializing an array of length N to contain random numbers in the range 0..N-1 (hint: use lrand48() % N). --- #include <algorithm> #include <chrono> #include <iomanip> #include <iostream> #include <random> #include <vector> #include <string.h> void setup(int64_t N, uint64_t A[]) { printf(" inside sum_indirect problem_setup, N=%lld ", N); } int64_t sum(int64_t N, uint64_t A[]) { printf(" inside sum_indirect perform_sum, N=%lld ", N); return 0; }
- Answer the given question with a proper explanation and step-by-step solution. #include<stdio.h>#include<stdlib.h>#include"NoteSynth.c" typedef struct BST_Node_Struct{ double key;double freq;int bar;double index;/*** TO DO:* Complete the definition of the BST_Node_Struct***/ } BST_Node; BST_Node *newBST_Node(double freq, int bar, double index){ /*** TO DO:* Complete this function to allocate and initialize a* new BST_Node. You should make sure the function sets* initial values for the data in the BST_Node that can* never occur in an actual musical note from a score!****/ return NULL;} BST_Node *BST_insert(BST_Node *root, BST_Node *new_node){/** This function inserts a new node into the BST. The* node must already have been initialized with valid* note data, and must have its unique key.** The insert function must check that no other node* exists in the BST with the same key. If a node with* the same key exists, it must print out a message* using the following format…Write aJava function given an array of integers nums [passed by reference] and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.Write a function that takes in a list of integers L and an integar n. The function returns the absolute value of the sum of the every other nth element of L. Make sure not to use an index outside the possible locations within L.
- Write a recursive function named binarySearch that accepts a reference to a sorted vector of integers and an integer target value and uses a recursive binary search algorithm to find and return an index at which that target value is found in the vector. If the target value is not found in the vector, return -1. The following code shows some example calls and their expected return values: // index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16vector<int>v {-4, 2, 7, 10, 15, 20, 22, 25, 30, 36, 42, 50, 56, 68, 85, 92, 103};intindex=binarySearch(v, 42); // 10intindex=binarySearch(v, 66); // -1 You should assume that the vector's elements are already sorted; you do not need to handle the case of an unsorted vector. Your function must be recursive and must use a binary search algorithm. Do not use loops or auxiliary data structures.Implement the following function, without using any data structure. /* Given two vectors of chars, check if the two vectors are permutations of each other, i.e., they contains same values, in same or different order.e.g., V1=[‘a’,’b’,’a’] and V2=[‘b’,’a’,’a’] stores same multi-set of data points: i.e., both contains two ‘a’, and one ‘b’. e.g., V3=[‘a’,’c’,’t’,’a’] and V4=[‘a’,’c’,’t’] are not same multi-set. V3 contains two ‘a’s, while V4 has only one ‘a’. Note: when considering multiset, the number of occurrences matters. @param list1, list2: two vectors of chars @pre: list1, list2 have been initialized @post: return true if list1 and list2 stores same values (in same or different order); return false, if not. */ bool SameMultiSet (vector<char> list1, vector<char> list2)Write the code for the following function."""def unmask_chars(s1: str, s2: str, mask: str) -> str:'''Return a new string where the character at index i iss1[i] if mask[i] is 0and s2[i] if mask[i] is 1.len(s1), len(s2), and len(mask) are equal.>>> unmask_chars('cat', 'bat', '001')'cat'>>> unmask_chars('a', 'b', '0')'a'>>> unmask_chars('apple', 'graph', '01011')'arpph''''# TODO: Write your code here