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Java: Find the numbers greater than the average of the numbers of a specified array

Java Basic: Exercise-162 with Solution

Write a Java program that finds numbers greater than the average of an array.

Pictorial Presentation:

Java Basic Exercises: Find the numbers greater than the average of the numbers of a specified array.

Sample Solution:

Java Code:

import java.util.*;
public class Solution {
  public static void main(String[] args) 
    {
      Integer nums[] = new Integer[]{1, 4, 17, 7, 25, 3, 100};
      int sum = 0;
	  System.out.println("Original Array: ");
	  System.out.println(Arrays.toString(nums));
	  for(int i = 0; i < nums.length; i++) {      
      sum = sum + nums[i];
    }
      double average = sum / nums.length;
      System.out.println("The average of the said array is: " + average);
      System.out.println("The numbers in the said array that are greater than the average are: ");
      for(int i = 0; i < nums.length; i++) {
      if(nums[i] > average) {
        System.out.println(nums[i]);
      }
    }
	}
}

Sample Output:

Original Array: 
[1, 4, 17, 7, 25, 3, 100]
The average of the said array is: 22.0
The numbers in the said array that are greater than the average are:
25
100

Flowchart:

Flowchart: Java exercises: Find the numbers greater than the average of the numbers of a specified array.

Java Code Editor:

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Previous: Write a Java program to find the kth smallest and largest element in a given array. Elements in the array can be in any order.
Next: Write a Java program that will accept an interger and convert it into a binary representation. Now count the number of bits which is equal to zero of the said binary represntation.

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