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Ksivusya [100]
2 years ago
7

Define a class named ComparableSquare that extends Square (defined above) and implements Comparable. Implement the compareTo met

hod to compare the Squares on the basis of area. Write a test class to find the larger of two instances of ComparableSquareobjects.
Computers and Technology
1 answer:
Alexxx [7]2 years ago
3 0

Answer:

/*********************************************************************************

* (The ComparableCircle class) Define a class named ComparableCircle that        *

* extends Circle and implements Comparable. Draw the UML diagram and implement   *

* the compareTo method to compare the circles on the basis of area. Write a test *

* class to find the larger of two instances of ComparableCircle objects.         *

*********************************************************************************/

public class Exercise_13_06 {

/** Main method */

public static void main(String[] args) {

 // Create two instances of ComparableCircle objects

 ComparableCircle comparableCircle1 = new ComparableCircle(12.5);

 ComparableCircle comparableCircle2 = new ComparableCircle(18.3);

 // Display comparableCircles

 System.out.println("\nComparableCircle1:");

 System.out.println(comparableCircle1);

 System.out.println("\nComparableCircle2:");

 System.out.println(comparableCircle2);

 // Find and display the larger of the two ComparableCircle objects

 System.out.println((comparableCircle1.compareTo(comparableCircle2) == 1  

  ? "\nComparableCircle1 " : "\nComparableCircle2 ") +  

  "is the larger of the two Circles");

}

}

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Answer: First lets solve the Prerequisite part

Lets say we have an input array of N numbers {3,2,5,0,5}. We have to  find number of ways to divide this array into 3 contiguous parts having equal sum. So the output for the above input array will be 2 as there are 2 ways to divide the array. One is (3,2),(5),(0,5) and the other is (3,2),(5,0),(5).

Following are the steps to achieve the above outcome.

  • Let p and q point to the index of array such that sum of array elements from 0 to p-1 is equal to sum of array elements from p to q which is equal to the sum of array elements from q+1 to N-1.  
  • If we see the array we can tell that the sum of 3 contiguous parts is 5. So the condition would be that sum of all array elements should be equal to 5 or sum of each contiguous part is equal to sum of all array elements divided by 5.
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Now lets solve the main task

We have an array A of size N and a number K. where A[]= {1,6,3,4,7} N=5 and K=3. We have to find if its possible to partition A into 3 contiguous subarrays such that sum of elements in each subarray is the same. It is possible in this example. Here we have 3 partitions (1,6),(3,4),(7) and sum of each of subarrays is same (1+6) (3+4) (7) which is 7.

Following are the steps to achieve the above outcome.

  • In order create K contiguous subarrays where each subarray has equal sum, first check the condition that sum of all elements in the given array should be divisible by K. Lets name another array as arrsum that will be the size of array A. Traverse A from first to  last index and keep adding current element of A with previous value in arrsum. Example A contains (1,6,3,4,7} and arrsum has {1,7,10,14,21}
  • If the above condition holds, now check the condition that each subarray or partition has equal sum. Suppose we represent sum1 to sum of all element in given array and sum2 of sum of each partition then: sum2 = sum2 / K.
  • Compare arrsum to subarray, begining from index 0 and when it becomes equal to sum2 this means that end of one subarray is reached. Lets say index q is pointing to that subarray.
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A easier function Partition for this task:

int Partition(int A[], int N, int k) // A arra y of size N and number k

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   for(int j = 0; j < N; j++)  //Loop that calculates sum of A

  sum = sum + A[j];        

  if(sum % k != 0) //checks condition that sum of all elements of A should be //divisible by k

   return 0;        

   sum = sum / k;  

   int sum2 = 0;  //represents sum of subarray

  for(int j = 0; j < N; j++) // Loop on subarrays

  {      sum2=sum2 + A[j];  

   if(sum2 == sum)    { //these lines locates subarrays and sum of elements //of subarrays should be equal

       sum2 = 0;  

       count++;  }  }  

/*calculate count of subarrays whose

sum is equal to (sum of complete array/ k.)

if count == k print Yes else print No*/

if(count == k)    

return 1;  

   else

   return 0;  }

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