function merge_sort!(
arr::Vector{T},
l::Int = 1,
r::Int = length(arr),
temp::Vector{T} = Vector{T}(undef, r - l + 1),
) where {T}
if l >= r
return
end
# split
mid = (l + r) >> 1
merge_sort!(arr, l, mid)
merge_sort!(arr, mid + 1, r)
# merge
l_pos = l # pos of the left part
r_pos = mid + 1 # pos of the right part
for t_pos in 1:r-l+1
if l_pos <= mid && (r_pos > r || arr[l_pos] < arr[r_pos])
temp[t_pos] = arr[l_pos]
l_pos += 1
else
temp[t_pos] = arr[r_pos]
r_pos += 1
end
end
for i in l:r
arr[i] = temp[i-l+1]
end
end
Given an array of n elements, write a function to sort the array
Best case - O(n log n)
Average - O(n log n)
Worst case - O(n log n)
O(n)
arr = [1, 3, 9, 5, 0, 2]
Divide the array in two halves [1, 3, 9] and [5, 0, 2]
Recursively call merge sort function for both these halves which will provide sorted halves
=> [1, 3, 9] & [0, 2, 5]
Now merge both these halves to get the sorted array [0, 1, 2, 3, 5, 9]
arr = [1, 9, 2, 5, 7, 3, 6, 4]
Divide the array into two halves [1, 9, 2, 5] and [7, 3, 6, 4]
As you can see that the above two halves are not yet sorted, so divide both of them into two halves again.
This time we get four arrays as [1, 9], [2, 5], [7, 3] and [6, 4].
We see that the last two arrays are again not sorted, so we divide them again into two halves and we will get [7], [3], [6], and [4].
Since an array of a single element is sorted, we now have all the arrays sorted, now we only need to merge them appropriately.
First, the arrays of one element will be merged as they were divided in last, and are at top of the recursion stack, so we get [3,7] and [4,6].
Now the merge will occur accordingly to the recursion stack, [1, 9] and [2, 5] will be merged and will make [1, 2, 5, 9].
Similarly [3, 7] and [4, 6] will be merged and made [3, 4, 6, 7].
At the next stack level [1, 2, 5, 9] and [3, 4, 6, 7] will be merged and we will get the final sorted array as [1, 2, 3, 4, 5, 6, 7, 9].