Answered: Input: An odd integer B, and a set A=…

Input: An odd integer B, and a set A= {a_1, .. a_2n} of 2n distinct positive integers. Question: Decide whether A can be partitioned into n disjoint pairs (a_i, a_j), where 1 si

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: [ "((()))", "(()())", "(())()", "()(())", "()()()" ] """ def generate_parenthesis_v1(n): defadd_pair(res, s, left, right): ifleft==0andright==0: res.append(s) return ifright>0: add_pair(res, s+")", left, right-1) ifleft>0: add_pair(res, s+"(", left-1, right+1) res= [] add_pair(res, "", n, 0) returnres def generate_parenthesis_v2(n): defadd_pair(res, s, left, right): ifleft==0andright==0: res.append(s) ifleft>0: add_pair(res, s+"(", left-1, right) ifright>0andleft<right: add_pair(res, s+")", left, right-1).
in C PROGRAMMING LANGUAGE AND COMMENT EVERY LINE SO I CAN UNDERSTAND EVERY STEP PLEASE, A C program can represent a real polynomial p(X) of degree n as an array of the real coefficients al, al, ..., an (an ‡ 0). p(X) = a0 + a1X + a2 X2 + . . .+ anXn Write a program that inputs a polynomial of maximum degree 8 and then evaluates the polynomial at various values of x. Include a function get_poly that fills the array of coefficients and sets the degree of the polynomial, and a function eval_poly that evaluates a polynomial at a given value of x. Use these function prototypes: void get_poly ( double coeffIl, int* degreep); double eval_poly( const double coeffIl, int degree, double x);
Read in two integers n and m (n, m < 50). Read n integers in an array A. Read m integers in anarray B. Then do the following (write separate programs for each, only the reading part iscommon).a) Find if there are any two elements x, y in A and an element z in B, such that x + y = z.b) Copy in another array C all elements that are in both A and B (intersection).c) Copy in another array C all elements that are in either A and B (union).d) Copy in another array C all elements that are in A but not in B (difference).c programm please tell all the answer
Let A = {1, 2, 3}, B = {{1}, 2, {1, 3}}, C = {1, {1}, 3}. Compute the following sets: 2A∩C − B, (A ∩ B) × (A − C), (B ∪ C) − A.
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…
Implement the sieve of Eratosthenes: a function for computing prime numbers, known to the ancient Greeks. Choose an integer n. This function will compute all prime numbers up to, and including, n. First insert all numbers from 1 to n into a set. Then erase all multiples of 2 (except 2); that is, 4, 6, 8, 10, 12, ... . Erase all multiples of 3, that is, 6, 9, 12, 15, ... . Go up to ?⎯⎯√n. The remaining numbers are all primes. Make sure that you use functions in your solution, including a main function. Define def sieveOfEratosthenes(n) that receives the maximum value and returns a list of all prime numbers less than or equal to this value in main() print the list of prime numbers.
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)
Here is --Given an array of strings strs, group the anagrams together. You can return the answer in any order. An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once. Example 1: Input: strs = ["eat","tea","tan","ate","nat","bat"] Output: [["bat"],["nat","tan"],["ate","eat","tea"]] Example 2: Input: strs = [""] Output: [[""]]. Asap.
Implement the following function without using any additional data structure, and without sorting the input vectors. /* Given two vectors of integers, check if the two vectors contain same set of values: e.g., V1=[3,4,10,4,10,11] and V2=[3,3,4, 11, 10] stores same set int: {3, 4, 10, 11}. Note that duplicates are removed when considering set */ @param list1, list2: two vectors of integers@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 SameSet (const vector<int> & list1, const vector<int> & list2) {//for each value in list1, check if it appears in list2, if not, return false //for each value in list2, check if the value appears in list1, if not return false //return true}
6. Read in two integers n and m (n, m < 50). Read n integers in an array A. Read m integers in anarray B. Then do the following (write separate programs for each, only the reading part iscommon).a) Find if there are any two elements x, y in A and an element z in B, such that x + y = z.b) Copy in another array C all elements that are in both A and B (intersection).c) Copy in another array C all elements that are in either A and B (union).d) Copy in another array C all elements that are in A but not in B (difference).Do In c Program
code in python And Vs Xor Shobhit is trying to impress Kriti. Kriti is stuck in a problem and Shobhit wants to solve it to impress her but he can't. So he asks for your help. You are given a positive integer NN, and an array AA of positive integers. The task is to calculate the number of such pairs (i,j)(i,j) where: i<ji<j AiAi &AjAj≥≥AiAiAjAj, where & denotes the bitwise AND operation, and @ denotes the bitwise XOR operation Input Format First-line contains an integer TT denoting the number of test cases. The next line contains a single positive integer NN. The next line contains NN non-negative integers representing the array AA. Output Format Print the number of pairs. Constraints 1≤T≤31≤T≤3 1≤N≤1051≤N≤105 1≤Ai≤1061≤Ai≤106 Time Limit 1 second Example Sample Input 3 5 1 4 3 7 10 3 1 1 1 1 1 Sample Output 1 3 0 Sample Testcase Explanation For the first test case, there is only one pair: (4,7). For the second test case, all pairs are selected. For the Third…
Write the following three functions: a) The function gets an array A of length n of ints, and a boolean predicate pred. It returns the smallest index i such that pred(A[i])==true.If no such element is not found, the function returns -1.int find(int* A, int n, bool (*pred)(int)); b) The function gets an array A of length n of ints, and a function f. It applies f to each element of A.void map(int* A, int n, int (*f)(int)); c) The function gets an array A of length n of ints, and a function f. The function f gets 2 ints and works as follows: Start with accumulator = A[0] For i=1...length-1 compute accumulator=f(accumulator, A[i]) Return accumulator For example, if f computes the sum of the two inputs, then reduce() will compute the sum of the entire array.int reduce(int* A, int n, int (*f)(int,int));