Merge Sort
Goal: Sort an array from low to high (or high to low)
Invented in 1945 by John von Neumann, merge sort is a fairly efficient sorting algorithm with a best, worst, and average time complexity of O(n log n).
The idea behind merge sort is to divide and conquer: to divide a big problem into smaller problems and solving many small problems instead of solving a big one. I think of merge sort as split first and merge after.
Assume you're given an array of n numbers and you need to put them in the right order. The merge sort algorithm works as follows:
- Put the numbers in a pile. The pile is unsorted.
- Split the pile into 2. Now you have two unsorted piles of numbers.
- Keep splitting the resulting piles until you can't split anymore. In the end, you will have n piles with 1 number in each pile.
- Begin to merge the piles together by sequentially pairing a pile with another pile. During each merge, you put the contents in sorted order. This is fairly easy because each individual pile is already sorted.
An example
Splitting
Let's say the numbers to sort are [2, 1, 5, 4, 9]. This is your unsorted pile. The goal is to keep splitting the pile until you can't split anymore.
First, split the array into two halves: [2, 1] and [5, 4, 9]. Can you keep splitting them? Yes you can!
Focus on the left pile. [2, 1] will split into [2] and [1]. Can you keep splitting them? No. Time to check the other pile.
[5, 4, 9] splits to [5] and [4, 9]. Unsurprisingly, [5] can't be split anymore, but [4, 9] splits into [4] and [9].
The splitting process ends with the following piles: [2] [1] [5] [4] [9]. Notice that each pile consists of just one element.
Merging
Now that you've split the array, you'll merge the piles together while sorting them. Remember, the idea is to solve many small problems rather than a big one. For each merge iteration you'll only be concerned at merging one pile with another.
Given the piles [2] [1] [5] [4] [9], the first pass will result in [1, 2] and [4, 5] and [9]. Since [9] is the odd one out, you can't merge it with anything during this pass.
The next pass will merge [1, 2] and [4, 5] together. This results in [1, 2, 4, 5], with the [9] left out again since it's the odd one out.
You're left with only two piles and [9] finally gets its chance to merge, resulting in the sorted array [1, 2, 4, 5, 9].
Top-down implementation
Here's what merge sort may look like in Swift:
func mergeSort(_ array: [Int]) -> [Int] {
guard array.count > 1 else { return array } // 1
let middleIndex = array.count / 2 // 2
let leftArray = mergeSort(Array(array[0..<middleIndex])) // 3
let rightArray = mergeSort(Array(array[middleIndex..<array.count])) // 4
return merge(leftPile: leftArray, rightPile: rightArray) // 5
}
A step-by-step explanation of how the code works:
If the array is empty or only contains a single element, there's no way to split it into smaller pieces. You'll just return the array.
Find the middle index.
Using the middle index from the previous step, recursively split the left side of the array.
Also recursively split the right side of the array.
Finally, merge all the values together, making sure that it's always sorted.
Here's the merging algorithm:
func merge(leftPile: [Int], rightPile: [Int]) -> [Int] {
// 1
var leftIndex = 0
var rightIndex = 0
// 2
var orderedPile = [Int]()
// 3
while leftIndex < leftPile.count && rightIndex < rightPile.count {
if leftPile[leftIndex] < rightPile[rightIndex] {
orderedPile.append(leftPile[leftIndex])
leftIndex += 1
} else if leftPile[leftIndex] > rightPile[rightIndex] {
orderedPile.append(rightPile[rightIndex])
rightIndex += 1
} else {
orderedPile.append(leftPile[leftIndex])
leftIndex += 1
orderedPile.append(rightPile[rightIndex])
rightIndex += 1
}
}
// 4
while leftIndex < leftPile.count {
orderedPile.append(leftPile[leftIndex])
leftIndex += 1
}
while rightIndex < rightPile.count {
orderedPile.append(rightPile[rightIndex])
rightIndex += 1
}
return orderedPile
}
This method may look scary but it is quite straightforward:
You need two indexes to keep track of your progress for the two arrays while merging.
This is the merged array. It's empty right now, but you'll build it up in subsequent steps by appending elements from the other arrays.
This while loop will compare the elements from the left and right sides, and append them to the
orderedPilewhile making sure that the result stays in order.If control exits from the previous while loop, it means that either
leftPileorrightPilehas its contents completely merged into theorderedPile. At this point, you no longer need to do comparisons. Just append the rest of the contents of the other array until there's no more to append.
As an example of how merge() works, suppose that we have the following piles: leftPile = [1, 7, 8] and rightPile = [3, 6, 9]. Note that each of these piles is individually sorted already -- that is always true with merge sort. These are merged into one larger sorted pile in the following steps:
leftPile rightPile orderedPile
[ 1, 7, 8 ] [ 3, 6, 9 ] [ ]
l r
The left index, here represented as l, points at the first item from the left pile, 1. The right index, r, points at 3. Therefore, the first item we add to orderedPile is 1. We also move the left index l to the next item.
leftPile rightPile orderedPile
[ 1, 7, 8 ] [ 3, 6, 9 ] [ 1 ]
-->l r
Now l points at 7 but r is still at 3. We add the smallest item to the ordered pile, so that's 3. The situation is now:
leftPile rightPile orderedPile
[ 1, 7, 8 ] [ 3, 6, 9 ] [ 1, 3 ]
l -->r
This process repeats. At each step we pick the smallest item from either leftPile or rightPile and add it to orderedPile:
leftPile rightPile orderedPile
[ 1, 7, 8 ] [ 3, 6, 9 ] [ 1, 3, 6 ]
l -->r
leftPile rightPile orderedPile
[ 1, 7, 8 ] [ 3, 6, 9 ] [ 1, 3, 6, 7 ]
-->l r
leftPile rightPile orderedPile
[ 1, 7, 8 ] [ 3, 6, 9 ] [ 1, 3, 6, 7, 8 ]
-->l r
Now there are no more items in the left pile. We simply add the remaining items from the right pile, and we're done. The merged pile is [ 1, 3, 6, 7, 8, 9 ].
Notice that this algorithm is very simple: it moves from left-to-right through the two piles and at every step picks the smallest item. This works because we guarantee that each of the piles is already sorted.
Bottom-up implementation
The implementation of merge sort you've seen so far is called "top-down" because it first splits the array into smaller piles and then merges them. When sorting an array (as opposed to, say, a linked list) you can actually skip the splitting step and immediately start merging the individual array elements. This is called the "bottom-up" approach.
Time to step up the game a little. :-) Here is a complete bottom-up implementation in Swift:
func mergeSortBottomUp<T>(_ a: [T], _ isOrderedBefore: (T, T) -> Bool) -> [T] {
let n = a.count
var z = [a, a] // 1
var d = 0
var width = 1
while width < n { // 2
var i = 0
while i < n { // 3
var j = i
var l = i
var r = i + width
let lmax = min(l + width, n)
let rmax = min(r + width, n)
while l < lmax && r < rmax { // 4
if isOrderedBefore(z[d][l], z[d][r]) {
z[1 - d][j] = z[d][l]
l += 1
} else {
z[1 - d][j] = z[d][r]
r += 1
}
j += 1
}
while l < lmax {
z[1 - d][j] = z[d][l]
j += 1
l += 1
}
while r < rmax {
z[1 - d][j] = z[d][r]
j += 1
r += 1
}
i += width*2
}
width *= 2
d = 1 - d // 5
}
return z[d]
}
It looks a lot more intimidating than the top-down version but notice that the main body includes the same three while loops from merge().
Notable points:
Merge sort needs a temporary working array because you can't merge the left and right piles and at the same time overwrite their contents. But allocating a new array for each merge is wasteful. Therefore, we're using two working arrays and we'll switch between them using the value of
d, which is either 0 or 1. The arrayz[d]is used for reading,z[1 - d]is used for writing. This is called double-buffering.Conceptually, the bottom-up version works the same way as the top-down version. First, it merges small piles of 1 element each, then it merges piles of 2 elements each, then piles of 4 elements each, and so on. The size of the pile is given by
width. Initially,widthis1but at the end of each loop iteration we multiply it by 2. So this outer loop determines the size of the piles being merged. And in each step, the subarrays to merge become larger.The inner loop steps through the piles and merges each pair of piles into a larger one. The result is written in the array given by
z[1 - d].This is the same logic as in the top-down version. The main difference is that we're using double-buffering, so values are read from
z[d]and written intoz[1 - d]. It also uses anisOrderedBeforefunction to compare the elements rather than just<, so this merge sort is generic and you can use it to sort any kind of object you want.At this point, the piles of size
widthfrom arrayz[d]have been merged into larger piles of sizewidth * 2in arrayz[1 - d]. Here we swap the active array, so that in the next step we'll read from the new piles we've just created.
This function is generic, so you can use it to sort any type you desire, as long as you provide a proper isOrderedBefore closure to compare the elements.
Example of how to use it:
let array = [2, 1, 5, 4, 9]
mergeSortBottomUp(array, <) // [1, 2, 4, 5, 9]
Performance
The speed of merge sort is dependent on the size of the array it needs to sort. The larger the array, the more work it needs to do.
Whether or not the initial array is sorted already doesn't affect the speed of merge sort since you'll be doing the same amount splits and comparisons regardless of the initial order of the elements.
Therefore, the time complexity for the best, worst, and average case will always be O(n log n).
A disadvantage of merge sort is that it needs a temporary "working" array equal in size to the array being sorted. It is not an in-place sort, unlike for example quicksort.
Most implementations of merge sort produce a stable sort. This means that array elements that have identical sort keys will stay in the same order relative to each other after sorting. This is not important for simple values such as numbers or strings, but it can be an issue when sorting more complex objects.
See also
Written by Kelvin Lau. Additions by Matthijs Hollemans.