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Gnoma [55]
2 years ago
12

Create a class named BaseballGame that contains data fields for two team names and scores for each team in each of nine innings.

names should be an array of two strings and scores should be a two-dimensional array of type int; the first dimension indexes the team (0 or 1) and the second dimension indexes the inning. Create get and set methods for each field. The get and set methods for the scores should require a parameter that indicates which inning’s score is being assigned or retrieved. Do not allow an inning score to be set if all the previous innings have not already been set. If a user attempts to set an inning that is not yet available, issue an error message. Also include a method named display in DemoBaseballGame.java that determines the winner of the game after scores for the last inning have been entered. (For this exercise, assume that a game might end in a tie.) Create two subclasses from BaseballGame: HighSchoolBaseballGame and LittleLeagueBaseballGame. High school baseball games have seven innings, and Little League games have six innings. Ensure that scores for later innings cannot be accessed for objects of these subtypes.
Computers and Technology
1 answer:
m_a_m_a [10]2 years ago
6 0

Answer:

Check the explanation

Explanation:

BaseballGame:

public class BaseballGame {

  protected String[] names = new String[2];

  protected int[][] scores;

  protected int innings;

  public BaseballGame() {

      innings = 9;

      scores = new int[2][9];

      for(int i = 0; i < 9; i++)

          scores[1][i] = scores[0][i] = -1;

  }

 

  public String getName(int team) {

      return names[team];

  }

  public void setNames(int team, String name) {

      names[team] = name;

  }

 

  public int getScore(int team, int inning) throws Exception {

      if(team < 0 || team >= 2)

          throw new Exception("Team is ut of bounds.");

      if(inning < 0 || inning >= innings)

          throw new Exception("Inning is ut of bounds.");

     

      return scores[team][inning];

  }

  public void setScores(int team, int inning, int score) throws Exception {

      if(team < 0 || team >= 2)

          throw new Exception("Team is ut of bounds.");

      if(inning < 0 || inning >= innings)

          throw new Exception("Inning is ut of bounds.");

      if(score < 0)

          throw new Exception("Score is ut of bounds.");

      for(int i = 0; i < inning; i++)

          if(scores[team][i] == -1)

              throw new Exception("Previous scores are not set.");

     

      scores[team][inning] = score;

  }

 

}

HighSchoolBaseballGame:

public class HighSchoolBaseballGame extends BaseballGame {

  public HighSchoolBaseballGame() {

      innings = 7;

      scores = new int[2][7];

      for(int i = 0; i < 7; i++)

          scores[1][i] = scores[0][i] = -1;

  }

}

LittleLeagueBaseballGame:

public class LittleLeagueBaseballGame extends BaseballGame {

  public LittleLeagueBaseballGame() {

      innings = 6;

      scores = new int[2][6];

      for(int i = 0; i < 6; i++)

          scores[1][i] = scores[0][i] = -1;

  }

}

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Suppose that each row of an n×n array A consists of 1’s and 0’s such that, in any row i of A, all the 1’s come before any 0’s in
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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).

   An algorithm that divides by 2 until the number gets sufficiently small then it terminates in O (log n) steps.

   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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