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- write a c++ program that randomly generate a 20 X20 two-dimensional array, board, of type int. An element baord [i][j] is a peak )eaither a max or a min) if all its neighbors (there should be either 3, 5, or 8 neighbors for any cell) are less than board [i][j], or greater than board [i][j]. the program should output all elemts in board, with thier indices, which are peak. it should also output if a peak is a max or a min. use the below code to write the defintion of the functions. #include <iostream>#include <cstdlib>#include <ctime>#include <iomanip> using namespace std; const int ROWS = 20;const int COLS = 20; void fillBoard(int b[][COLS]);void showBoard(int b[][COLS]);void findPeaks(int b[][COLS]); int main(){int board[ROWS][COLS];fillBoard(board);showBoard(board);findPeaks(board,ROWS);return 0;} void fillBoard(int b[][COLS]){srand(time(0));for (int i =0; i < ROWS; i++)for (int j = 0; j < COLS; j++)b[i][j] = rand();}In C, implement the function that gets an array of ints and its length, and returns the maximum of the absolute values. int max_abs (const int* arr, int len); For example: On input [1, -3, 7, 4] the function should return 7. On input [2, -3, 2, -4] the function should return 4. On input [0, -3, 2] the function should return 3.Consider an array A of N values (N entered by the user). Write a C program that creates a second array B such that the i-th element of B, i.e., B[i] is equal to A[0] *A[1] * A[2] * ... A[i]; In other words:- B[0] = A[0]- B[1] = A[0] * A[1];- B[2] = A[0] * A[1] * A[2];- B[3] = A[0] * A[1] * A[2] * A[3]- ....etc.For this exercise, we would like to implement two solutions and compare their complexity /performance in terms of computation time.Solution 1: Implement blindly the following algorithms: for each element B[i], compute A[0] *A[1] * A[2] * .... *A[i] and save the result in B[i];To test the solution, create an array A of 1000 elements and initialize each of its elements with arandom value (instead of asking the user to . To do that, use the function rand, which, every timecalled, returns a random number Questions:- How long does your program take to produce the solution?- Analyze the solution above and explain why it is taking long time. What are the mainfactors that affect how long…
- Write a function/ program that produces the following: int findDifference(const string a1[], int n1, const string a2[], int n2); Return the position of the first corresponding elements of a1 and a2 that are not equal. n1 is the number of interesting elements in a1, and n2 is the number of interesting elements in a2. If the arrays are equal up to the point where one or both runs out, return the smaller of n1 and n2. Notwithstanding each function's behavior described below, all functions that return an int must return −1 if they are passed any bad arguments (e.g. a negative array size, or a position that would require looking at the contents of an element past the last element we're interested in). Unless otherwise noted, passing 0 to the function as the array size is not itself an error; it merely indicates the function should examine no elements of the array. Here's an example: string cast[5] = { "samwell", "jon", "margaery", "daenerys", "tyrion" }; string roles[4] = { "samwell",…Implement the Tic-Tac-Toe game in C++ using functions and 2D array of 3X3 size. Rules of the Game a) The game is to be played between computer and user.b) One of the players chooses ‘O’ and the other ‘X’ to mark their respective cells.c) The game starts with one of the players and the game ends when one of the players has onewhole row/ column/ diagonal filled with his/her respective character (‘O’ or ‘X’).d) If no one wins, then the game is said to be draw.In C programming Mathematically, given a function f, we recursively define fk(n) as follows: if k = 1, f1(n) = f(n). Otherwise, for k > 1, fk(n) = f(fk-1(n)). Assume that there is an existing function f, which takes in a single integer and returns an integer. Write a recursive function fcomp, which takes in both n and k (k > 0), and returns fk(n). int f(int n);int fcomp(int n, int k){
- A set of integers 0..MAX may be implemented using an array of boolean values. This particular implementation is called a bit-vector implementation of a Set. Since C doesn’t have the Boolean data type, then substitute TRUE with 1 and FALSE with 0, or define 1 and 0 as TRUE and FALSE, respectively.For example, if the integer 3 is an element of the set, then the array element indexed by 3 is TRUE. On the other hand, if 3 is not an element, then the array element indexed by 3 is FALSE. For example, if the integer 3 is an element of the set, then the array element indexed by 3 is TRUE. On the other hand, if 3 is not an element, then the array element indexed by 3 is FALSE.For example: if s = {3,4,6,8}, the array looks like this: Implement a programmer-defined data type called BitSet to represent a set as follows:typedef int BitSet[MAX];Implement the following functions:void initialize(BitSet s); - set all array elements to FALSEvoid add(int elem,BitSet s);- set the item indexed by elem to…Write a program in java for the below given problem statement: Given 2 sorted arrays (in increasing order), find a path through the intersection that produces themaximum sum and return the maximum sum. That is, we can switch from one array to another array only atcommon elements. If no intersection element is present, we need to take the sum of all elements from thearray with greater sum.Sample Input:61 5 10 15 20 2552 4 5 9 15Sample Output :81Explanation :We start from array 2 and take sum till 5 (sum = 11). Then we'll switch to array at element 10 and take till15. So sum = 36. Now, no elements are left in array after 15, so we'll continue in array 1. Hence the sum is81.Write a C++ program that declares and initializes an integer array of size 100 with randomintegers between 1 and 10 inclusive using initializeArray (from question 1) and counts the occurrence of eachnumber in the array and graphs the information in the form of a bar chart or histogram—each number isprinted, then a bar consisting of that many asterisks is printed beside the number.Here is a sample run of the program:Number Value Histogram1 12 ************2 5 *****3 3 ***4 16 ****************5 7 *******6 6 ******7 9 ********8 1 *9 15 *************** 10 10 **********
- Implement a range function for a dynamic array which returns a new dynamic array that is a subset of the original. input parameters: array - (the array and any related parameters) start - index of the first elementend - index of the last elementInterval - An integer number specifying the incrementation of index This function returns a new dynamic array containing the elements from the start thru the end indices of the original array.All array indexing must be done using pointer arithmetic. For example, given the array: 49 96 99 47 76 29 22 16 30 22 If the start and end positions were 5 and 9 with step 2, return a new dynamic array: 29 16 22 Please use following main to test your function. int main(){int *p = new int[10]{49,96,99,47,76,29,22,16,30,22}; int *q = range(p,10,5,9,2);for(int i=0;i<3;i++) cout<<q[i]<<" "; // print 29 16 22 cout<<endl;delete [] q;q = range(p,10,1,8,3); for(int i=0;i<3;i++)cout<<q[i]<<" "; // print 96 76 16 cout<<endl;…Write a C++ program that asks the user to enter the number of students in the class followed their grades out of 100 using the keyboard. After this, the program prints the grades in reverse order. Use a dynamic array of doubles to store the entered grades. Modify the previous C++ program in question above such that it does not ask the user about the number of students, however, the program will terminate grades input when a negative grade is entered at the keyboard. Use a vector of doubles instead of a dynamic array of doubles.Given an integer n and an array a of length n, your task is to apply the following mutation to a: Array a mutates into a new array b of length n. For each i from 0 to n - 1, b[i] = a[i - 1] + a[i] + a[i + 1]. If some element in the sum a[i - 1] + a[i] + a[i + 1] does not exist, it should be set to 0. For example, b[0] should be equal to 0 + a[0] + a[1].