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:

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:

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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