Java Array Exercises: Print all the LEADERS in the array

Java Array: Exercise-39 with Solution

Write a Java program to print all the LEADERS in the array.

Note: An element is leader if it is greater than all the elements to its right side.

Pictorial Presentation:

Java Array Exercises: Print all the LEADERS in the array

Sample Solution:

Java Code:

import java.util.HashMap;
import java.util.Map;
import java.util.Iterator;
import java.util.Arrays; 

public class Main
 public static void main(String[] args)
      int arr[] = {10, 9, 14, 23, 15, 0, 9};
      int size = arr.length;
        for (int i = 0; i < size; i++) 
            int j;
            for (j = i + 1; j < size; j++) 
                if (arr[i] <= arr[j])
            if (j == size) 
                System.out.print(arr[i] + " ");

Sample Output:

23 15 9


Flowchart: Print all the LEADERS in the array

Visualize Java code execution (Python Tutor):

Java Code Editor:

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Previous: Write a Java program to get the majority element from a given array of integers containing duplicates.
Next: Write a Java program to find the two elements from a given array of positive and negative numbers such that their sum is closest to zero.

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Java: Tips of the Day

Different between parseInt() and valueOf() in java?

Well, the API for Integer.valueOf(String) does indeed say that the String is interpreted exactly as if it were given to Integer.parseInt(String). However, valueOf(String) returns a new Integer() object whereas parseInt(String) returns a primitive int.

If you want to enjoy the potential caching benefits of Integer.valueOf(int), you could also use this eyesore:

Integer k = Integer.valueOf(Integer.parseInt("123"))

Now, if what you want is the object and not the primitive, then using valueOf(String) may be more attractive than making a new object out of parseInt(String) because the former is consistently present across Integer, Long, Double, etc.

Ref: https://bit.ly/3vRuIPY