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12345 [234]
3 years ago
6

There are N bulbs numbered from 1 to N, arranged in a row. The first bulb is plugged into the power socket and each successive b

ulb is connected to the previous one. (Second bulb to first, third to second) Initially, all the bulbs are turned off. A moment K (for K from 0 to N-1), the A[K]-th bulb is turned on. A bulbs shines if it is one and all the previous bulbs are turned on too. Write a function in java that given an Array A of N different integers from 1 to N, returns the number of moments for which every turned on bulb shines.
Computers and Technology
1 answer:
Ksivusya [100]3 years ago
3 0

Answer:

The code is given below with appropriate comments

Explanation:

// TestSolution class implementation

import java.util.Arrays;

public class TestSolution

{

  // solution function implementation

  public static int solution(int[] arr)

  {

      // declare the local variables

      int i, j, count = 0;

      boolean shines;

     

      // use the nested loops to count the number of moments for which every turned on bulb shines

      for (i = 0; i < arr.length; i++)

      {

          shines = true;

          for (j = i + 1; j < arr.length && shines; j++)

          {

              if (arr[i] > arr[j])

                  shines = false;

          }

          if (shines)

              count++;

      }

      // return the number of moments for which every turned on bulb shines

      return count;

     

  } // end of solution function

 

  // start main function

  public static void main(String[] args)

  {

      // create three arrays named A, B, and C

      int[] A = {2, 1, 3, 5, 4};

      int[] B = {2, 3, 4, 1, 5};

      int[] C = {1, 3, 4, 2, 5};

     

      // generate a random number N within the range range[1..100000]

      int N = 1 + (int)(Math.random() * 100000);

     

      // create an array named D of size N

      int[] D = new int[N];

     

      // fill the array D with the distinct random numbers within the range [1..N]

      int i = 0;

      while(i < N)

      {

          int num = 1 + (int)(Math.random() * N);          

          boolean found = false;

          for(int j = 0; j < i && !found; j++)

          {

              if(D[j] == num)

                  found = true;

          }

         

          if(!found)

          {

              D[i] = num;

              i++;

          }

      }          

     

      // print the elements and number of moments of the arrays A, B, and C

      System.out.println("Array A: " + Arrays.toString(A) + " and Moments: " + solution(A));

      System.out.println("Array B: " + Arrays.toString(B) + " and Moments: " + solution(B));

      System.out.println("Array C: " + Arrays.toString(C) + " and Moments: " + solution(C));

     

      // print the size and number of moments of the array D

      System.out.println("Size(N) of Array D: " + N + " and Moments: " + solution(D));

     

  } // end of main function

} // end of TestSolution class

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     #increament in the variable by 1

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     #increament in the variable by 1

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   #then increment in the variable by 1

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   #set discount of the items

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   # calculate bill after deducting discount

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   #print the final bill

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   #check the choice of the user is y or Y

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items = len(price)

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<u>Output</u>:

Enter the price (-1 to quit): 96

Is it a pet (Y / N)? y

Enter the price (-1 to quit): 69

Is it a pet (Y / N)? n

Enter the price (-1 to quit): 41

Is it a pet (Y / N)? n

Enter the price (-1 to quit): 52

Is it a pet (Y / N)? n

Enter the price (-1 to quit): 96

Is it a pet (Y / N)? n

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Is it a pet (Y / N)? n

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Explanation:

<u>The following are the description of the program</u>.

  • Firstly, define the function 'discount()' and pass the arguments 'prices', 'isPet' and 'nItems' in its parameter.
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  • Set the if-else conditional statement to check the pet is greater than equal to 1 and the items are greater than equal to 5 then, set the discount and print the bill after deduction.
  • Otherwise, print the bill without deduction.
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