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Ivanshal [37]
3 years ago
13

Write a program, TwoDimentionalGrid. Ask the user to enter the size of the 2 dimensional array. Here we are not doing any input

validation. We are assuming, user is going to enter a number greater than 1. Create a 2D array with the same row and column number. Now print the array as a number grid starting at 1. Example: At place row number 3 and column number 2, number would be 6 (3*2). Any number would be the product of it's row number and column number. Hint: Use setw 5. Expected Output: Here I am attaching 2 expected output. Make sure your code is running for both input.
Computers and Technology
1 answer:
Nitella [24]3 years ago
4 0

Answer:

The program in Java is as follows:

import java.util.*;

public class Main{

public static void main(String[] args) {

 Scanner input = new Scanner(System.in);

 System.out.print("Array size: ");

 int n = input.nextInt();

 int[][] myarray = new int[n][n];

 for(int i =0;i<n;i++){

     for(int j =0;j<n;j++){

         myarray[i][j] = i * j;      }  }

 for(int i =0;i<n;i++){

     for(int j =0;j<n;j++){

         System.out.print(myarray[i][j]+" ");      }

     System.out.println();  }

}}

Explanation:

This prompts the user for the array size

 System.out.print("Array size: ");

This gets input for the array size

 int n = input.nextInt();

This declares the array

 int[][] myarray = new int[n][n];

This iterates through the rows

 for(int i =0;i<n;i++){

This iterates through the columns

     for(int j =0;j<n;j++){

This populates the array by multiplying the row and column

         myarray[i][j] = i * j;      }  }

This iterates through the rows

 for(int i =0;i<n;i++){

This iterates through the columns

     for(int j =0;j<n;j++){

This prints each array element

        System.out.print(myarray[i][j]+" ");      }

This prints a new line at the end of each row

     System.out.println();  }

}

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Answer:

True: In binary search algorithm, we follow the below steps sequentially:

Input: A sorted array  B[1,2,...n] of n items and one item x to be searched.

Output: The index of x in B if exists in B, 0 otherwise.

  1. low=1
  2. high=n
  3. while( low < high )
  4.  {      mid=low + (high-low)/2
  5.         if( B[mid]==x)
  6.          {
  7.             return(mid)  //returns mid as the index of x
  8.           }
  9.          else
  10.          {
  11.              if( B[mid] < x)      //takes only right half of the array
  12.               {
  13.                 low=mid+1
  14.               }
  15.              else               // takes only the left half of the array
  16.               {
  17.                high=mid-1
  18.               }
  19.           }
  20.  }
  21. return( 0 )

Explanation:

For each iteration the line number 11 or line number 15 will be executed.

Both lines, cut the array size to half of it and takes as the input for next iteration.

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