Question

Only part 1, makinf the 2nd and 3rd picture of code do the same thing using array

Purpose This homework is to learn the concept and usage of one-dimensional (ID) arrays. Part 1 (30 pts) Modify your HW6' (i.e., digit frequency histogram) in a new program (DigitFrequencyArray.java) to use a count array instead of 10 count variables. Your new program should be much shorter than your original code since there is no need for branching statements (if or switch). For reference, my entire program has only 20 lines of code. Part 2 (40 pts) Write a program (NumberFrequency.java) to generate random numbers between a user specified range, then count and display the frequencies of the most / least appeared numbers. Sample Output How many random numbers? 1000000 Enter the minimum number: 100 Enter the maximum number: 200 Generated 1000000 random numbers between 100 (inclusive) and 200 (exclusive) Number 186 has the most occurrences (10197). Number 117 has the least occurrences (9749). What to Submit Submit your Java source program (DigitFrequencyArray.java and NumberFrequency.java) on Blackboard before the due date.

Answer 1

//Java program to calculate the digit frequency for all digits in all generated random numbers

public class DigitFrequency {

   public static void main(String[] args) {

       int counts[] = new int[10]; // create an array of length 10 for digits [0-9]
       // initialize the counts to 0 for all digits
       for(int i=0;i<counts.length;i++)
           counts[i] = 0;
       int number;
       final int LOOP = 200;
       // loop 200 times
       for(int k=1;k<=LOOP;k++)
       {
           //generate a random integer [0,999]
           number = (int)(Math.random()*1000);
           // loop to increment the digits present in number
           while(number > 0)
           {
               // get the last digit from the number which is also the index for counts as index is from 0-9
               counts[number%10]++;
               number = number/10; // remove the last digit from number
           }
       }
      
       // loop to display the histogram
       for(int i=0;i<counts.length;i++)
       {
           System.out.printf("%d (%2d) : ",i,counts[i]);
           for(int j=0;j<counts[i];j++)
               System.out.print("*");
           System.out.println();
       }
   }

}
//end of program

Output:

9 (72): 8 (68): 7 (54):求求: 6 (59) : ** 5(52): 4 (58): 3 (63): 2 (61) 1(48): 月(42) * * * | 我并,并 非开并开 # , # # # # # # # # # # # # # * * * # # # # # # # 开开开光 开光 开光 开 光 开光 开光 开 我开开 # # # # # # 开光 开光 * * * * * % %%%%%%% %%%%%%% * % * % * % * % % * % % * # # # 光 开光 开光 开 光 开光 开 开开开 # # # # 开开开开 # # # # % # , # * % %%% * % * * %%% %%% 宪宪宪宪宪宪宪軍軍事来来来来来来来来来来来来来来来来来来宋玉宋玉米来来来来来来来来来 * * * # # # 光 开光 开光 开光 光 开光 开光 开 #, # # 开 # , # # # # # 开开 开光 开光 开 , 并 # # # # * * * * %%%% * * * * * * # # # # # 开光 开光 开 开光 开 | 北非 # # # # 开光 开光 # # # # %%%此外北 * %%%%%%% %%%%%%%% * * * * * * # # # 开光 开光 开光 开光 光 开光 开光 开 非非非 # # # # # # 开光 开光 * * * % * % %%%%%%%%% * % * % # # # # # # # # # * | 北非 * * * %%%%%%% * ********************** * * * 宝宝宝宝宝宝宝 * %%%%%%% 本来宋宋宋宋牢牢牢牢牢牢牢牢牢牢 * 宋宋宋宋宋宋宋宋宋宋宋宋宋宋宋宋宋宋 * * * * * * * * * * * * * * * *