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In order to win the prize for most cookies sold, my friend Alice and I are going to merge our Girl Scout Cookies orders and enter as one unit.

Each order is represented by an "order id" (an integer).

We have our lists of orders sorted numerically already, in arrays. Write a method to merge our arrays of orders into one sorted array.

For example:

int[] myArray = new int[]{3, 4, 6, 10, 11, 15}; int[] alicesArray = new int[]{1, 5, 8, 12, 14, 19}; System.out.println(Arrays.toString(mergeArrays(myArray, alicesArray))); // prints [1, 3, 4, 5, 6, 8, 10, 11, 12, 14, 15, 19]

We can do this in time and space.

If you're running a built-in sorting method, your algorithm probably takes time for that sort.

Think about edge cases! What happens when we've merged in all of the elements from one of our arrays but we still have elements to merge in from our other array?

We could simply concatenate (join together) the two arrays into one, then sort the result:

public static int[] mergeSortedArrays(int[] myArray, int[] alicesArray) { int[] mergedArray = Arrays.copyOf(myArray, myArray.length + alicesArray.length); System.arraycopy(alicesArray, 0, mergedArray, myArray.length, alicesArray.length); Arrays.sort(mergedArray); return mergedArray; }

What would the time cost be?

, where n is the total length of our output array (the sum of the lengths of our inputs).

We can do better. With this algorithm, we're not really taking advantage of the fact that the input arrays are themselves already sorted. How can we save time by using this fact?

A good general strategy for thinking about an algorithm is to try writing out a sample input and performing the operation by hand. If you're stuck, try that!

Since our arrays are sorted, we know they each have their smallest item in the 0th index. So the smallest item overall is in the 0th index of one of our input arrays!

Which 0th element is it? Whichever is smaller!

To start, let's just write a method that chooses the 0th element for our sorted array.

public static int[] mergeArrays(int[] myArray, int[] alicesArray) { // make an array big enough to fit the elements from both arrays int[] mergedArray = new int[myArray.length + alicesArray.length]; int headOfMyArray = myArray[0]; int headOfAlicesArray = alicesArray[0]; // case: 0th comes from my array if (headOfMyArray < headOfAlicesArray) { mergedArray[0] = headOfMyArray; // case: 0th comes from Alice's array } else { mergedArray[0] = headOfAlicesArray; } // eventually we'll want to return the merged array return mergedArray; }

Okay, good start! That works for finding the 0th element. Now how do we choose the next element?

Let's look at a sample input:

[3, 4, 6, 10, 11, 15] // myArray [1, 5, 8, 12, 14, 19] // alicesArray

To start we took the 0th element from alicesArray and put it in the 0th slot in the output array:

[3, 4, 6, 10, 11, 15] // myArray [1, 5, 8, 12, 14, 19] // alicesArray [1, x, x, x, x, x] // mergedArray

We need to make sure we don't try to put that 1 in mergedArray again. We should mark it as "already merged" somehow. For now, we can just cross it out:

[3, 4, 6, 10, 11, 15] // myArray [x, 5, 8, 12, 14, 19] // alicesArray [1, x, x, x, x, x] // mergedArray

Or we could even imagine it's removed from the array:

[3, 4, 6, 10, 11, 15] // myArray [5, 8, 12, 14, 19] // alicesArray [1, x, x, x, x, x] // mergedArray

Now to get our next element we can use the same approach we used to get the 0th element—it's the smallest of the earliest unmerged elements in either array! In other words, it's the smaller of the leftmost elements in either array, assuming we've removed the elements we've already merged in.

So in general we could say something like:

1. We'll start at the beginnings of our input arrays, since the smallest elements will be there.
2. As we put items in our final mergedArray, we'll keep track of the fact that they're "already merged."
3. At each step, each array has a first "not-yet-merged" item.
4. At each step, the next item to put in the mergedArray is the smaller of those two "not-yet-merged" items!

Can you implement this in code?

public static int[] mergeArrays(int[] myArray, int[] alicesArray) { int[] mergedArray = new int[myArray.length + alicesArray.length]; int currentIndexAlices = 0; int currentIndexMine = 0; int currentIndexMerged = 0; while (currentIndexMerged < mergedArray.length) { int firstUnmergedAlices = alicesArray[currentIndexAlices]; int firstUnmergedMine = myArray[currentIndexMine]; // case: next comes from my array if (firstUnmergedMine < firstUnmergedAlices) { mergedArray[currentIndexMerged] = firstUnmergedMine; currentIndexMine++; // case: next comes from Alice's array } else { mergedArray[currentIndexMerged] = firstUnmergedAlices; currentIndexAlices++; } currentIndexMerged++; } return mergedArray; }

Okay, this algorithm makes sense. To wrap up, we should think about edge cases and check for bugs. What edge cases should we worry about?

Here are some edge cases:

1. One or both of our input arrays is 0 elements or 1 element
2. One of our input arrays is longer than the other.
3. One of our arrays runs out of elements before we're done merging.

Actually, (3) will always happen. In the process of merging our arrays, we'll certainly exhaust one before we exhaust the other.

Does our method handle these cases correctly?

If both arrays are empty, we're fine. But for all other edge cases, we'll get an ArrayIndexOutOfBoundsException.

How can we fix this?

We can probably solve these cases at the same time. They're not so different—they just have to do with indexing past the end of arrays.

To start, we could treat each of our arrays being out of elements as a separate case to handle, in addition to the 2 cases we already have. So we have 4 cases total. Can you code that up?

Be sure you check the cases in the right order!

public static int[] mergeArrays(int[] myArray, int[] alicesArray) { int[] mergedArray = new int[myArray.length + alicesArray.length]; int currentIndexAlices = 0; int currentIndexMine = 0; int currentIndexMerged = 0; while (currentIndexMerged < mergedArray.length) { // case: my array is exhausted if (currentIndexMine >= myArray.length) { mergedArray[currentIndexMerged] = alicesArray[currentIndexAlices]; currentIndexAlices++; // case: Alice's array is exhausted } else if (currentIndexAlices >= alicesArray.length) { mergedArray[currentIndexMerged] = myArray[currentIndexMine]; currentIndexMine++; // case: my item is next } else if (myArray[currentIndexMine] < alicesArray[currentIndexAlices]) { mergedArray[currentIndexMerged] = myArray[currentIndexMine]; currentIndexMine++; // case: Alice's item is next } else { mergedArray[currentIndexMerged] = alicesArray[currentIndexAlices]; currentIndexAlices++; } currentIndexMerged++; } return mergedArray; }

Cool. This'll work, but it's a bit repetitive. We have these two lines twice:

mergedArray[currentIndexMerged] = myArray[currentIndexMine]; currentIndexMine++;

Same for these two lines:

mergedArray[currentIndexMerged] = alicesArray[currentIndexAlices]; currentIndexAlices++;

That's not DRY. Maybe we can avoid repeating ourselves by bringing our code back down to just 2 cases.

See if you can do this in just one "if else" by combining the conditionals.

You might try to simply squish the middle cases together:

if (isAlicesArrayExhausted || (myArray[currentIndexMine] < alicesArray[currentIndexAlices])) { mergedArray[currentIndexMerged] = myArray[currentIndexMine]; currentIndexMine++;

But what happens when myArray is exhausted?

We'll get an ArrayIndexOutOfBoundsException when we try to access myArray[currentIndexMine]!

How can we fix this?

First, we allocate our answer array, getting its size by adding the size of myArray and alicesArray.

We keep track of a current index in myArray, a current index in alicesArray, and a current index in mergedArray. So at each step, there's a "current item" in alicesArray and in myArray. The smaller of those is the next one we add to the mergedArray!

But careful: we also need to account for the case where we exhaust one of our arrays and there are still elements in the other. To handle this, we say that the current item in myArray is the next item to add to mergedArray only if myArray is not exhausted AND, either:

1. alicesArray is exhausted, or
2. the current item in myArray is less than the current item in alicesArray
public static int[] mergeArrays(int[] myArray, int[] alicesArray) { // set up our mergedArray int[] mergedArray = new int[myArray.length + alicesArray.length]; int currentIndexAlices = 0; int currentIndexMine = 0; int currentIndexMerged = 0; while (currentIndexMerged < mergedArray.length) { boolean isMyArrayExhausted = currentIndexMine >= myArray.length; boolean isAlicesArrayExhausted = currentIndexAlices >= alicesArray.length; // case: next comes from my array // my array must not be exhausted, and EITHER: // 1) Alice's array IS exhausted, or // 2) the current element in my array is less // than the current element in Alice's array if (!isMyArrayExhausted && (isAlicesArrayExhausted || (myArray[currentIndexMine] < alicesArray[currentIndexAlices]))) { mergedArray[currentIndexMerged] = myArray[currentIndexMine]; currentIndexMine++; // case: next comes from Alice's array } else { mergedArray[currentIndexMerged] = alicesArray[currentIndexAlices]; currentIndexAlices++; } currentIndexMerged++; } return mergedArray; }

The if statement is carefully constructed to avoid an ArrayIndexOutOfBoundsException from indexing past the end of an array. We take advantage of Java's short circuit evaluation and check first if the arrays are exhausted.

time and additional space, where n is the number of items in the merged array.

The added space comes from allocating the mergedArray. There's no way to do this "in place" because neither of our input arrays are necessarily big enough to hold the merged array.

But if our inputs were linked lists, we could avoid allocating a new structure and do the merge by simply adjusting the next pointers in the list nodes!

In our implementation above, we could avoid tracking currentIndexMerged and just compute it on the fly by adding currentIndexMine and currentIndexAlices. This would only save us one integer of space though, which is hardly anything. It's probably not worth the added code complexity.

What if we wanted to merge several sorted arrays? Write a method that takes as an input an array of sorted arrays and outputs a single sorted array with all the items from each array.

Do we absolutely have to allocate a new array to use for the merged output? Where else could we store our merged array? How would our method need to change?

We spent a lot of time figuring out how to cleanly handle edge cases.

Sometimes it's easy to lose steam at the end of a coding interview when you're debugging. But keep sprinting through to the finish! Think about edge cases. Look for off-by-one errors.

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