CodingBat: Java. Array-2, Part III

tripleUp:

public boolean tripleUp(int[] nums) {
	for (int i = 0; i <= nums.length - 3; i++)
		if (nums[i + 1] == nums[i] + 1 && nums[i + 2] == nums[i] + 2)
			return true;
	return false;
}

shiftLeft:

public int[] shiftLeft(int[] nums) {
	if (nums.length > 0) {
		int first = nums[0];
		for (int i = 0; i < nums.length - 1; i++)
			nums[i] = nums[i + 1];
		nums[nums.length - 1] = first;
	}
	return nums;
}

tenRun:

public int[] tenRun(int[] nums) {
	boolean replace = false;
	int multiple = 0;

	for (int i = 0; i < nums.length; i++) {
		if (nums[i] % 10 == 0) {
			if (!replace) {
				replace = true;
				multiple = nums[i];
			} else
				multiple = nums[i];
		}
		if (nums[i] % 10 != 0 && replace) nums[i] = multiple;
	}
	return nums;
}

pre4:

public int[] pre4(int[] nums) {
		int count = 0;
		for (int i = 0; i < nums.length; i++) {
			if (nums[i] != 4) count++;
			else break;
		}
		int[] result = new int[count];
		for (int i = 0; i < result.length; i++)
			result[i] = nums[i];
		return result;
}

post4:

public int[] post4(int[] nums) {
	int last4 = 0;
	for (int i = 0; i < nums.length; i++)
		if (nums[i] == 4) last4 = i;
	
	int[] res = new int[nums.length - (last4 + 1)];
	for (int i = last4 + 1, j = 0; i < nums.length; i++, j++)
		res[j] = nums[i];
	
	return res;
}

notAlone:

public int[] notAlone(int[] nums, int val) {
	for (int i = 1; i < nums.length - 1; i++)
		if (nums[i] == val && nums[i - 1] != val
			&& nums[i + 1] != val)
			nums[i] = Math.max(nums[i - 1], nums[i + 1]);
	return nums;
}

zeroFront:

public int[] zeroFront(int[] nums) {
	int[] res      = new int[nums.length];
	int zeroPos    = 0;
	int nonZeroPos = res.length - 1;

	for (int i = 0; i < nums.length; i++)
		if (nums[i] == 0)
			res[zeroPos++]    = 0;
		else
			res[nonZeroPos--] = nums[i];

	return res;
}

Note that the order of the non-zero numbers does not matter. For an alternative solution that modifies the given array, please see the comment by ‘aaaaaaaa’ below.

withoutTen:

public int[] withoutTen(int[] nums) {
	int[] copy = new int[nums.length];
	int index = 0;

	for (int i = 0; i < nums.length; i++)
		if (nums[i] != 10) {
			copy[index] = nums[i];
			index++;
		}
	return copy;
}

zeroMax:

public int[] zeroMax(int[] nums) {
	int largestOdd = 0;
	for (int i = nums.length - 1; i >= 0; i--) {
		if (nums[i] % 2 == 1 && nums[i] > largestOdd)
			largestOdd = nums[i];
		if (nums[i] == 0)
			nums[i] = largestOdd;
	}
	return nums;
}

evenOdd:

public int[] evenOdd(int[] nums) {
	int[] res   = new int[nums.length];
	int evenPos = 0;
	int oddPos  = res.length - 1;

	for (int i = 0; i < nums.length; i++)
		if (nums[i] % 2 == 0)
			res[evenPos++] = nums[i];
		else
			res[oddPos--]  = nums[i];
	return res;
}

The solution is similar to “zeroFront”, which is given above.

22 thoughts on “CodingBat: Java. Array-2, Part III

  1. aaaaaaaa

    your zero front solution is n^2 and others too….

    public int[] zeroFront(int[] nums) {
    int count = 0;
    for (int i = 0; i < nums.length; i++) {
    if (nums[i] == 0) {
    nums[i] = nums[count];
    nums[count] = 0;
    count++;
    }
    }
    return nums;
    }

    Reply
    1. Gregor Ulm Post author

      As I mentioned in the post, efficiency is hardly a concern, given these input sizes. But thanks for pointing this out.

      Reply
        1. Gregor Ulm Post author

          You shouldn’t get too hung up on that. The instructions on the Array-2 page mention “2 loops”, even though you can solve some of them with just one loop. I’ll have a look later, though, and might rewrite the three solutions in which I used nested for-loops.

          Reply
      1. Joseph

        I don’t understand the solution to without ten. In particular, how are the elements of the initial array that ARE 10 replaced by zeros? What part of your code specifies that operation?

        Reply
        1. William

          When you make a new array, it’s all 0’s til you change the number.
          So in short, the code simply goes through, upping the index for the original array every time, and only upping the index on the new array when you’re adding something that’s not 10, thus it would add each number not 10 one by one to the new array, then the rest of the new array is left at the default 0.

          Reply
    2. Renil Babu

      for tenRun small code which changes its elements itself..
      for(int i=0;i<nums.length;i++)
      {
      int a = nums [i];
      if (a%10==0)
      {
      if ( (i+1) < nums.length && nums [i+1] % 10! =0) nums [i+1] = a;
      }
      }
      return nums;
      }

      Reply
  2. Justin

    Shift left can use modulus

    public int[] shiftLeft(int[] nums) {
    int len = nums.length;
    int[] result = new int[len];

    for (int i = 0; i < len; i++) {
    result[i] = nums[(i + 1) % len];
    }

    return result;

    }

    Reply
    1. William

      Wow nice code, just had to say that was brilliant to use the modulus to sort of reset it at the end, it would count up like normal til it got to i+1 == len, so of course the remainder would be 0 to grab the first. So neat ^_^
      Here was mine, same basic idea only long winded:

      public int[] shiftLeft(int[] nums) {
      int res[] = new int[nums.length];
      for(int i=0,j=1;i<res.length;i++,j++){//i is for res,j for nums
      if(j==res.length)j=0;//resets j to get the first nums to put at end
      res[i]=nums[j];
      }
      return res;
      }

      Reply
  3. Joseph

    Why is my code not working? The only difference between my code and the solution give is instead of writing.

    efirst[e++] I wrote

    efirst[e];
    e++.

    When I changed the nation as below, it worked. Shouldn’t both work?

    public int[] evenOdd(int[] nums) {
    int [] efirst= new int [nums.length];
    int e=0;
    int o=efirst.length-1;
    for (int i=0;i<nums.length;i++){

    if (nums[i]%2==0)
    efirst[e++]=nums[i];
    e++;
    if (nums[i]%2!=0)
    efirst[o–]=nums[i];
    o–;
    }
    return efirst;
    }

    Reply
  4. Stephan

    My solution for tenRun. Much simpler and it seems to work for all cases.

    private static String tenRun(int[] nums) {
    for (int i = 0; i < nums.length – 1; i++) {

    if (nums[i] % 10 == 0) {
    while (nums[i + 1] % 10 != 0) {
    nums[i + 1] = nums[i];
    }

    }

    }

    return nums;

    Reply
      1. Joseph

        My solution is almost identical:

        Actually, I do not really understand Gregor’s solution. Perhaps someone can explain the logic.

        public int[] tenRun(int[] nums) {
        for (int i= 0; i<nums.length; i++){
        if(nums[i]%10==0)
        for (int j=i+1; j<nums.length; j++){
        if(nums[j]%10!=0) nums[j]=nums[i];
        else break;}
        }
        return nums;

        }

        Reply
  5. John Doe

    I’m new to this kind of For loop

    for (int i = last4 + 1, j = 0; i < nums.length; i++, j++)

    I only discovered this because of codingBat and Gregors solutions.

    Basically i'm asking how the f. i train my brain to think like this because of 5 exercises where this kind of for loop was used i only managed to figure out one.

    Reply
      1. Gregor Ulm Post author

        This just means that you initialize two variables, i and j, and increment both until a certain condition is no longer fulfilled (here: i < nums.length).

        Reply
  6. Barto

    For TripleUp:

    private static boolean trippleUp( int… nums )
    {
    for( int i = 0; i < nums.length – 2; i++ )
    {
    if( nums[i] == nums[i + 1] – 1 && nums[i + 1] == nums[i + 2] – 1 )
    {
    return true;
    }
    }
    return false;
    }

    Reply
  7. Coding God

    Cleaner solution:

    public int[] evenOdd(int[] nums) {
    int [] nums1 = new int [nums.length];
    int evens = 0;
    int odds = 0;
    int even = 0;
    int odd = 0;
    for(int i = 0; i < nums.length; i++)
    {
    if(nums[i] % 2 == 0)
    {
    evens++;
    even = nums[i];
    }
    else
    {
    odds++;
    odd = nums[i];
    }
    }
    for(int j = 0; j < evens; j++)
    {
    nums1[j] = even;
    }
    for(int k = evens; k < nums.length; k++)
    {
    nums1[k] = odd;
    }
    return nums1;
    }

    Reply
  8. Rickyg

    Wouldn’t this be much simpler solution to TenRun? It passed the test, hope I didn’t miss anything.

    public int[] tenRun(int[] nums) {
    for (int i=0; i<nums.length-1; i++){
    if (nums[i] % 10 == 0 && nums[i+1] % 10 != 0) nums[i+1] = nums[i];
    }
    return nums;
    }

    Reply

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