GMU Spring 2023 – CS 211 – Project 1DescriptionThe purpose of this assignment is to get practice with strings, arrays, loops and dynamic memory allocation.Instructions...

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GMU Spring 2023 – CS 211 – Project 1 Description The purpose of this assignment is to get practice with strings, arrays, loops and dynamic memory allocation. Instructions   Documentation : Comments are not required in this assignment but it’s a good practice to add comments to any piece of code that is not obvious what it does.  Imports : You may not import any classes or packages.  You are not allowed to use any language construct that hasn’t been covered in class yet. 1 Grading • • If your code doesn’t pass all the compliance checks in Gradescope, it means that something fundamental is wrong and we won’t be able to grade it unless you +x it. You will get zero points if you don’t +x and reupload your code. Task You will implement the following methods in Java. Every method is self-contained; there are no dependencies among the methods. You’re free to create helper methods if you want though. Make sure you do not modify the method signatures that are provided, otherwise your code will not pass the compliance checks and we won’t be able to grade it. 2 public static int stringValue(String word) It returns the value of a string which is the sum of the values of its characters. The value of each character is the ASCII code of the character (e.g. A has a value of 65, B has a value of 66, a has a value of 97, etc.) with a few exceptions: • Double-letters (e.g. ee, mm, ll, tt, etc.) are valued once, not twice. • The space character has a value of zero • The value of each number-character is multiplied by the largest value of the non-number- characters in the string. To simplify things a bit, assume that the word will not include, and your program will not be tested with, any characters that have an ASCII code lower than 40 or higher than 123. Examples: word return solution Mason 510 77+97+115+111+110 Ma son 510 77+97+0+115+111+110 mass 321 109+97+115 50 Cent 12110 53*116+48*116+0+67+101+110+116 public static double expValue(int x, double precision) The calculation of the ex function appears a lot in applications and it’s considered quite an expensive computation. One way to calculate it is with the following formula: e x =∑k=0 ∞ x k k ! =1+x+ x 2 2 + x 3 6 + x 4 24 + ... As you can tell, this series can go for ever. After some point though, the precision of the result doesn’t improve much. Therefore, we can stop adding terms when the improvement is less than the precision parameter. In other words, if adding a certain term changes the result by less than the value of precision, then we do not add this term and we stop the series. The method returns the sum of the series. 3 https://en.wikipedia.org/wiki/ASCII public static int mirrorNum(int num) The method accepts an integer and returns the mirrored integer. In case of a negative number, it retains the sign. You may not use any classes from any package, not even the Math and String classes, to implement the mirrorNum method Examples: num return 4 4 98014 41089 -593 -395 12300 321 public static boolean raisedNum(long num) There are some numbers whose value is the sum of x y+ yx where x and y are integers greater than 1. Let's call these numbers raised numbers. Write a method that accepts a long and returns true if the number is raised, and false otherwise. Examples: input return solution 6 false 17 true 23 +32 593 true 92 +29 1125 false 4 public static int[][] smallestSubarray(int[][] array, int sum) It takes a rectangular array and returns a square-size subarray of it whose values have a sum that is equal or larger to sum. A subarray can’t be smaller than 2x2. If there are more than one square-size subarrays that satisfy the requirements, the method must return the square that has the smallest size. If two or more equally-sized squares satisfy the requirements, it must return the square whose values have the highest sum. Assume that the sum has such a value that always warrants a solution (i.e. no need to check or worry about this). Examples: array sum return {{0,1,2},{-4,5,6},{7,8,3}} 5 {{5,6},{8,3}} {{0,1,2},{-4,5,6},{7,8,3}} 23 {{0,1,2},{-4,5,6},{7,8,3}} public static void replaceElement(int[][] array, int elem, int[] newElem) It takes a two-dimensional array and modi+es it in-place (i.e. the method doesn’t return anything). Every occurrence of elem is replaced by the items contained in newElem. The modi+cation must happen in-place, i.e. the memory address of the whole array must remain the same; but the memory addresses of its rows can change of course. Examples: array before method invocation elem newElem array after method invocation {{1,2,3,4,5},{6,7,8}} 2 {0} {{1,0,3,4,5},{6,7,8}} {{1,2,3,4,5},{5,4,3,2}} 5 {-5,5} {{1,2,3,4,-5,5},{-5,5,4,3,2}} {{0,1,2,3,4,5},{5,4,3,4}} 4 {1,2,3} {{0,1,2,3,1,2,3,5},{5,1,2,3,3,1,2,3}} 5 public static int[][] removeDuplicates(int[][] array) It takes as input a two-dimensional array and returns a new two-dimensional array where every sequence of duplicate numbers is replaced by a single occurrence of the respective number. Keep in mind that duplicate numbers can extend in more than one rows though. Examples: input array returned array {{1,2,3,4,5},{6,7,8,9}} {{1,2,3,4,5},{6,7,8,9}} {{1,2,2,2,3,4,5,5},{6,7,8,8,8,9}} {{1,2,3,4,5},{6,7,8,9}} {{1,2,2,2,3,4,5,5},{5,5,5,5},{5,5,9}} {{1,2,3,4,5},{9}} public static int[] vortex(int[][] array) It takes as input a two-dimensional rectangular array, traverses its contents in a clockwise spiral order starting from index (0,0), and returns the result as a single-dimensional array. Examples: input array returned array {{1,2,3},{4,5,6},{7,8,9}} {1,2,3,6,9,8,7,4,5} {{1,2},{3,4},{5,6},{7,8},{9,10}} {1,2,4,6,8,10,9,7,5,3} {{1,2,3,4,5},{6,7,8,9,10},{11,12,13,14,15}, {16,17,18,19,20}} {1,2,3,4,5,10,15,20,19,18,17,16,11,6,7,8,9,14,13,12} 6 Changelog Description Instructions Submission Grading Testing Task
Answered 11 days AfterFeb 06, 2023

Answer To: GMU Spring 2023 – CS 211 – Project 1DescriptionThe purpose of this assignment is to get ...

Vikas answered on Feb 11 2023
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