Thread: Need help with an Array problem!

Hey guys, really having some trouble solving this Array problem.

I have been trying to use a for loop to loop through the array to find an odd or even number, then a second loop checks whether there are another 2 odd or even numbers following.

Really having trouble wrapping my head around the logic behind this one, very frustrating

The question is this:

////////////////////////////// PROBLEM STATEMENT //////////////////////////////
// Given an array of ints, print true if the array contains either 3 even or
// 3 odd values all next to each other.
// {2, 1, 3, 5} -> true
// {2, 1, 2, 5} -> false
// {2, 4, 2, 5} -> true

You mean like this?

  Scanner keyboard = new Scanner(System.in);
  int length = keyboard.nextInt();
  int[] mainArray = new int[length];
  int evenCount = 0, oddCount = 0;
 
  for (int i = 0; i < length; i++){
    mainArray[i] = keyboard.nextInt();
 
  if (mainArray[i] % 2 == 0)
    evenCount++;
  else 
    oddCount++;
  }
 
  System.out.print (evenCount == 3 || oddCount == 3);

The only problem with this is the Array can be any length (more than just 4).
In the example it only gives 4 length Arrays, but the actual test provides arrays up to 9 long.

This is what I originally tried to do, which works, but I know is terrible code.

 
  Scanner keyboard = new Scanner(System.in);
  int length = keyboard.nextInt();
  int[] mainArray = new int[length];
  int countA = 0, countB = 0;
 
 
  for (int i = 0; i < length; i++){
    mainArray[i] = keyboard.nextInt();
 
    if (mainArray[i] % 2 == 0){
      for (int k = i + 1; k < length; k++){
      if (mainArray[k] % 2 == 0)
      countA++;
    else
      countA = 0;
      break;
      }
    }
  }
 
  for (int i = 0; i < length; i++){
 
    if (mainArray[i] % 2 == 1){
      for (int k = i + 1; k < length; k++){
      if (mainArray[k] % 2 == 1)
      countB++;
    else
      countB = 0;
      break;
      }
    }
  }
 
 System.out.print( countA == 3 || countB >= 2);

Hello eyesackery!
If I were you, I would go with the following logic

loop through the array{
     if (element i is even){
            if (element i+1 and element i+2 are even)
                return true;
                break;
     }
     if (element i is odd){
            if (element i+1 and element i+2 are odds)
                return true;
                break;
    }
    return false;
}

Note: you don't need to check the last two elements since they cannot be three consecutive even or odd numbers starting from them. (means loop until array.lenght-2)
Hope it helps!

Awesome! Thanks for the replies guys, I managed to figure it out thanks to your help. Now I'm stuck on this problem, if anyone could explain to me the logic I would be very thankful

Given a non-empty array, return true if there is a place to split the array so that the sum of the numbers on one side is equal to the sum of the numbers on the other side.

canBalance({1, 1, 1, 2, 1}) → true
canBalance({2, 1, 1, 2, 1}) → false
canBalance({10, 10}) → true

Originally Posted by eyesackery

Awesome! Thanks for the replies guys, I managed to figure it out thanks to your help. Now I'm stuck on this problem, if anyone could explain to me the logic I would be very thankful

What have you tried so far? Before you start implementing the code you should write down the logic in a piece of paper.