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Java Array Exercises: Find the duplicate values of an array of string values

Java Array: Exercise-13 with Solution

Write a Java program to find the duplicate values of an array of string values.

Pictorial Presentation:

Java Array Exercises: Find the duplicate values of an array of string values

Sample Solution:

Java Code:

public class Exercise13 {
public static void main(String[] args) 
    {
        String[] my_array = {"bcd", "abd", "jude", "bcd", "oiu", "gzw", "oiu"};
 
        for (int i = 0; i < my_array.length-1; i++)
        {
            for (int j = i+1; j < my_array.length; j++)
            {
                if( (my_array[i].equals(my_array[j])) && (i != j) )
                {
                    System.out.println("Duplicate Element is : "+my_array[j]);
                }
            }
        }
    }    
}

Sample Output:

Duplicate Element is : bcd                                                                                    
Duplicate Element is : oiu

Flowchart:

Flowchart: Java exercises: Find the duplicate values of an array of string values

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Java Code Editor:

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Previous: Write a Java program to find the duplicate values of an array of integer values.
Next: Write a Java program to find the common elements between two arrays (string values).

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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