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____ [38]
2 years ago
9

Write a class called Counter that represents a simple tally counter, which might be used to count people as they enter a room. T

he Counter class should contain a single integer as instance data, representing the count. Write a constructor to initialize the count to zero. Write a method called click that increments the count and another method called get Count that returns the current count. Include a method called reset that resets the counter to zero. Finally, create a driver class called CounterTest that creates two Counter objects and tests their methods.
Computers and Technology
1 answer:
Sonja [21]2 years ago
4 0

Answer:

Code is in java

Explanation:

Code is self explanatory, added come comments to explain more about the code.

public class CounterDriver {

   public static void main(String[] args){

       // Creating counter first object

       Counter counter1 = new Counter();

       //performing 2 clicks

       counter1.click();

       counter1.click();

       // displaying counter 1 click count

       System.out.println("Total Number of counter1 clicks :"+counter1.getCount());

       // resetting counter 1 click count

       counter1.reset();

       // again displaying click count which will be 0 after reset

       System.out.println("Total Number of counter1 clicks :"+counter1.getCount());

      // Same operation goes with counter 2  

       // the only difference is that it performs 3 clicks

       Counter counter2 = new Counter();

       counter2.click();

       counter2.click();

       counter2.click();

       System.out.println("Total Number of counter2 clicks :"+counter2.getCount());

       counter2.reset();

       System.out.println("Total Number of counter2 clicks :"+counter2.getCount());

   }

}

class Counter{

   int count;

// defining constructor which will initialize count variable to 0

   public Counter(){

       this.count=0;

   }

// adding click method to increment count whenever it's called

   public void click(){

       this.count++;

   }

// getCount method will return total count

   public int getCount(){

       return this.count;

   }

// reset method will set count to 0

   public void reset(){

       this.count = 0;

   }

}

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Explanation:

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Answer:

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Explanation:

a)

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Answer:

Check the explanation

Explanation:

  •    Each row of nxn array A consists of 1’s and 0’s such that , in any row of A, all the 1’s come before any 0’s in that row.
  •    Use binary search algorithm to find the index of the last 1 in a row.
  •    Perform this process for each row.
  •    Now, searching for last occurrence of 1 in a row will take O (log n) time.
  •    There are n such rows, therefore total time will be O (n log n).

Complexity analysis:

   The method would be to use binary search for each row to find the first zero starting with index of A[i][n/2+1].

   Let’s say j=n/2.

   The number of 1’s in a row would be j+1.

   This would take O (log n).

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   As there are n rows the complexity would be O (n log n).

Pseudo-code:

A = [[1,0,0,0],[0,0,0,0],[1,1,1,1],[1,1,0,0]]

n=4

c=0

for i in range(n): # Loop in rows

  j = n/2 # Search from middle index

  while j>0: # Loop in column

      if(A[i][j]==0): # search for first zero

          if(A[i][j-1]==1): # confirm first zero

              c = c+j # add 1's count to c

              break

          else: # reduce index by 1 or j/2

              if(j/2 == 0):

                  j = j-1

              else:

                  j = j - j/2

      else: # increase index by 1 or j/2

      if(j/2 == 0):

      j = j+1

      else:

          j = j + j/2

      if(j==n): # For all 1's

      c = c+n

      break  

print c

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