Assignment Description: The goal of this assignment is to work with hash functions and understand the performance of the Division method using open addressing techniques discussed in the slides. Page 2 Write one simple program that uses a fixed set of 50 unique keys stored in an array as follows (Important: hard-code the array content in your program and make sure you have same exact key values below in the same order): int[] keys = {1234, 8234, 7867, 1009, 5438, 4312, 3420, 9487, 5418, 5299,  5078, 8239, 1208, 5098, 5195, 5329, 4543, 3344, 7698, 5412,  5567, 5672, 7934, 1254, 6091, 8732, 3095, 1975, 3843, 5589,  5439, 8907, 4097, 3096, 4310, 5298, 9156, 3895, 6673, 7871,  5787, 9289, 4553, 7822, 8755, 3398, 6774, 8289, 7665, 5523}; The program allows the user to select a hash function, from the menu, to be invoked on the set of keys. Keep it simple as follows: —–MAIN MENU—– 0 – Exit Program 1 – Run HF1 (Division with Linear Probing) 2 – Run HF2 (Division with Quadratic Probing) 3 – Run HF3 (Division with Double Hashing) 4 – Run HF4 (Student-Designed Function) The hash functions are defined below. To keep the implementation simple, design the hash table (call it Table) (of size 50) as a 2D array (50 rows and 2 columns) (int[][] Table = new int[50][2];) The first column stores the keys while the second column stores number of probes used to resolve collisions. After calling the hash function from the menu, the output of the program should display the hash table followed by the sum of all probe values in the table. Declare a separate method in your class, say sumProbes(), to perform this calculation and return the sum of all probes in the table (second column of the table). Note that the total number of probes a hashing function generates indicates the performance level of the function – The smaller the sum of probes the better the hash function. HF1: Declare a separate method HF1() that implements the Division method discussed in the slides with Linear Probing for collision resolution. HF2: Declare a separate method HF2() that implements the Division method discussed in the slides with Quadratic Probing for collision resolution. HF3: Declare a separate method HF3() that implements the Division method discussed in the slides with Double Hashing for collision resolution. For the second hashing function, use the following function and increment (see example in slides)  H2 (key) = 30 – key % 25;  Increment is (key % 50) + j * H2 (key) for j=1, 2, 3, 4, … Note: It is possible that HF3 will not be able to determine and empty index in the hash table for a give key, especially when very few empty entries remain in the hash table. I this care and to avoid entering into an infinite loop, limit number of attempt to locate a key in the hash table to no more than 50 tries. In such case, printout a message like this example: “Unable to has key 43654 to the table”. Page 3 Note this phenomenon happen due to not applying Load Factoring to our table. HF4: Declare a separate method HF4() that implements a hash function of your own design. The sky is your limit. You can come up with your own hash function or take and improve one of the above three functions by either using a different hashing method (other than Division method) or a different collision resolution method. Aim to come up with a function that beats the above three function (i.e., your function generates smallest number of probes for the given set of keys). Note: See the note in HF3 and apply it to you HF4 if necessary. The assignment is very specific and it must be implemented as specified. Any deviation from these requirements is not acceptable and receives no points. No exceptions. Only complete and correct code receives credit. Code must compile and run on its own as received. Using code from outside sources receives NO credit. Format the output following the sample run below. Sample output for format illustration purpose only (Our table is of size 50 elements) Hash table resulted from HF2: Index Key probes ————————  0 4576 0  1 9876 2  2 2341 0  3 8722 3  4 9988 4  5 1111 0  6 3443 1  7 4444 0  8 7788 1  9 2321 0 ———————— Sum of probe values = 11 probes.