Java – Exam – Longest Odd-Even Sequence – Challenge!

As you probably know, for a long time (about 5 weeks) I started to learn Java from scratch. Today, I have decided to take a look at one of the exam problems at the Software University and I have spent some time on it, getting more of the QA tests were passed 🙂

Anyway, if you can come up with a better result, using my code, consider it a challenge.

Thus, let me show you what was the problem about – you have some input like this –

And your task is to give an output of the count of the numbers, which are sequential. In our case “sequential” has another meaning – it means “odd-even sequence” or “even-odd sequence”. E.g. in the given example, the highest count of the sequential numbers is “4”:

(-18) (55) (44) (3)

To make it more interesting, ZERO is counted as odd and even in the same time.

Ok, this is what I have come up with:

import java.util.*;

public class LongestOddEvenSequence {
	public static void main(String[] args) {

		Scanner scanner = new Scanner(System.in);
		String inputLine = scanner.nextLine();
		String[] numbers = inputLine.split("[ ()]+");
		int[] iArray = new int[numbers.length - 1];
		for (int i = 0; i < iArray.length; i++) {
			iArray[i] = Integer.parseInt(numbers[i + 1]);
		}

		int Counter = 0;
		int CurrentCounter = 0;
		int CounterA = 0;
		int CounterB = 0;
		boolean EvenNeeded = true;
		// Start with Odd
		for (int i = 0; i < iArray.length; i++) {
			if (EvenNeeded) {
				if (iArray[i] == 0) {
					CurrentCounter++;
					EvenNeeded = false;
				} else if (iArray[i] % 2 == 0) {
					CurrentCounter++;
					EvenNeeded = false;
				} else if (iArray[i] % 2 == 1) {
					if (CurrentCounter > Counter) {
						Counter = CurrentCounter;
					}
					EvenNeeded = true;
					CurrentCounter = 0;
				}
			} else {
				if (iArray[i] == 0) {
					CurrentCounter++;
					EvenNeeded = true;
				} else if (iArray[i] % 2 == 1) {
					CurrentCounter++;
					EvenNeeded = true;
				} else if (iArray[i] % 2 == 0) {
					if (CurrentCounter > Counter) {
						Counter = CurrentCounter;
					}
					EvenNeeded = false;
					CurrentCounter = 0;
				}
			}
		}
		if (CurrentCounter > Counter) {
			Counter = CurrentCounter;
		}
		CounterA = Counter;
		Counter = 0;
		CurrentCounter = 0;

		for (int i = 0; i < iArray.length; i++) {
			if (EvenNeeded == false) {
				if (iArray[i] == 0) {
					CurrentCounter++;
					EvenNeeded = true;
				} else if (iArray[i] % 2 == 1) {
					CurrentCounter++;
					EvenNeeded = true;
				} else if (iArray[i] % 2 == 0) {
					if (CurrentCounter > Counter) {
						Counter = CurrentCounter;
					}
					EvenNeeded = false;
					CurrentCounter = 0;
				}
			} else {
				if (iArray[i] == 0) {
					CurrentCounter++;
					EvenNeeded = false;
				} else if (iArray[i] % 2 == 0) {
					CurrentCounter++;
					EvenNeeded = false;
				} else if (iArray[i] % 2 == 1) {
					if (CurrentCounter > Counter) {
						Counter = CurrentCounter;
					}
					EvenNeeded = true;
					CurrentCounter = 0;
				}
			}
		}

		if (CurrentCounter > Counter) {
			Counter = CurrentCounter;
		}
		CounterB = Counter;

		if (CounterA > CounterB) {
			System.out.println(CounterA + 1);
		} else {
			System.out.println(CounterB + 1);

		}

	}
}

Quite a long one, isn’t it?

The original code of the solution looks a little better and shorter:

import java.util.Scanner;

public class LongestOddEvenSequence {

	public static void main(String[] args) {

		// Read and parse the input numbers
		Scanner scanner = new Scanner(System.in);
		String inputLine = scanner.nextLine();
		String[] numbers = inputLine.split("[ ()]+");
		int[] nums = new int[numbers.length-1];
		for (int i = 0; i < nums.length; i++) {
			nums[i] = Integer.parseInt(numbers[i+1]);
		}

		// Find the longest alternating sequence
		int bestLen = 0;
		int len = 0;
		boolean shouldBeOdd = (nums[0] % 2 != 0);
		for (int num : nums) {
			boolean isOdd = num % 2 != 0;
			if (isOdd == shouldBeOdd || num == 0) {
				len++;
			} else {
				shouldBeOdd = isOdd;
				len = 1;
			}
			shouldBeOdd = !shouldBeOdd;
			if (len > bestLen) {
				bestLen = len;
			}
		}

		System.out.println(bestLen);
	}

}

What is worthy in the original code:

• Usage of a regular expression for the splitting of the inputLine into array;
• A boolean is declared with the evaluation of an equation! Quite a good approach (line 19)!
• The for each loop – quite a good idea, indeed, I have used the standard for;
• In the for each loop we have no usage of TRUE and FALSE (check my code for the mess). Indeed, this way the code looks much better and understandable!

Takeaway  message – Use the following boolean declaration, whenever possible:

boolean isOdd = num % 2 != 0;
boolean shouldBeOdd = (nums[0] % 2 != 0);			
shouldBeOdd = !shouldBeOdd;

Enjoy it!