Given an JavaScript object list of books that each have a pages attribute to define the number of pages in the book, find the product of all the pages in the object list of books using recursion (you must use recursion to solve this problem). Keep in mind: The input list object may be completely empty (ex. {}) The next attribute may not be defined

Rooks on a rampage def safe_squares_rooks(n, rooks): A generalized n-by-n chessboard has been invaded by a parliament of rooks, each rook represented as a two-tuple (row, column) of the row and the column of the square that the rook is in. Since we are again computer programmers instead of chess players and other normal folks, our rows and columns are numbered from 0 to n - 1. A chess rook covers all squares that are in the same row or in the same column. Given the board size n and the list of rooks on that board, count the number of empty squares that are safe, that is, are not covered by any rook. To achieve this in reasonable time and memory, you should count separately how many rows and columns on the board are safe from any rook. Because permuting the rows and columns does not change the answer to this question, you can imagine all these safe rows and columns to have been permuted to form an empty rectangle at the top left corner of the board. The area of that safe rectangle is…

Project Description: In this project you implement an ArrayStack ADT and use the stack for implementing the following methods: a) Reverse an array of Words: Accept an array of words as input parameter and return an array of words in reverse order. Use the method signature: public static String[] reverse Words (String[] wordList) Example Input: Bird Cat Dog Elephant Output: Elephant Dog Cat Bird

5. The following function indexCounter aims to find and return all the indices of the element x in a list L. if x is not in the list L, return an empty list. However, the implementation has a flaw, so that many times it will return error message. Can you fix the flaw? def indexCounter(L, x): indexList [] startIndex = 0 while startIndex < len(L): indexList.append (L.index (x, startIndex)) startIndex = L.index (x, startIndex) + 1 return indexList