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Java Exercises: Test if an array of integers contains an element 10 next to 10 or an element 20 next to 20, but not both

Java Basic: Exercise-93 with Solution

Write a Java program to test if an array of integers contains an element 10 next to 10 or an element 20 next to 20, but not both.

Sample Solution:

Java Code:

import java.util.*; 
public class Exercise93 {
 public static void main(String[] args)
 {
    //int[] nums = {10, 10, 2, 4, 9};
	int[] nums = {10, 10, 2, 4, 20, 20};  
	int ctr_even = 0, ctr_odd = 0;
	System.out.println("Original Array: "+Arrays.toString(nums)); 
	    
    boolean found1010 = false;
    boolean found2020 = false;
      
    for(int i = 0; i < nums.length - 1; i++) {
        if(nums[i] == 10 && nums[i+1] == 10)
            found1010 = true;
                        
        if(nums[i] == 20 && nums[i+1] == 20)
            found2020 = true;
    }
   
	System.out.printf( String.valueOf(found1010 != found2020));	
	System.out.printf("\n");
  }
}

Sample Output:

Original Array: [10, 10, 2, 4, 20, 20]                                 
false 

Pictorial Presentation:

Java exercises: Test if an array of integers contains an element 10 next to 10 or an element 20 next to 20, but not both
Java exercises: Test if an array of integers contains an element 10 next to 10 or an element 20 next to 20, but not both

Flowchart:

Flowchart: Java exercises: Test if an array of integers contains an element 10 next to 10 or an element 20 next to 20, but not both

Java Code Editor:

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Previous: Write a Java program to count the number of even and odd elements in a given array of integers.
Next: Write a Java program to rearrange all the elements of an given array of integers so that all the odd numbers come before all the even numbers.

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Java: Tips of the Day

Try and catch:

Java is excellent at catching errors, but it can only recover gracefully if you tell it what to do. The cascading hierarchy of attempting to perform an action in Java starts with try, falls back to catch, and ends with finally. Should the try clause fail, then catch is invoked, and in the end, there's always finally to perform some sensible action regardless of the results. Here's an example:

try {
        cmd = parser.parse(opt, args); 
       
        if(cmd.hasOption("help")) {
                HelpFormatter helper = new HelpFormatter();
                helper.printHelp("Hello ", opt);
                System.exit(0);
                }
        else {
                if(cmd.hasOption("shell") || cmd.hasOption("s")) {
                String target = cmd.getOptionValue("tgt");
                } // else
        } // fi
} catch (ParseException err) {
        System.out.println(err);
        System.exit(1);
        } //catch
        finally {
                new Hello().helloWorld(opt);
        } //finally
} //try

It's a robust system that attempts to avoid irrecoverable errors or, at least, to provide you with the option to give useful feedback to the user. Use it often, and your users will thank you!

Ref: https://red.ht/3EZc9OC