We discuss three sorting algorithm that have O(n) time, improving over the O(n log(n)) algorithms by imposing extra requirements on the input elements.
The input is an array of integers and the integers fall in the half-open range [0,k). We can sort them using a technique called counting sort that is similar to the one we used for checking anagrams.
We’ll start by giving some intuition for why the algorithm works. Suppose we want to sort the following array.
A = [2, 8, 7, 1], k = 10
Here’s the output of sorting:
B = [1, 2, 7, 8]
0 1 2 3
Let’s focus on just one element of the input, say 8, and think about
which position it should move to. It belongs at position 3 because
there are three elements in the array that are less-than 8.
The main idea behind counting sort is to count the number of elements that are less-than (or equal) to each element.
Here’s the algorithm:
C[i] is the count for integer i
C = [0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0]
0 1 2 3 4 5 6 7 8 9 10
L stores the cumulative sum (aka. prefix sum) of the counts. In other words, L[i] says how many elements in the input are less-than or equal to element i.
L = [0, 1, 2, 2, 2, 2, 2, 3, 4, 4, 4]
0 1 2 3 4 5 6 7 8 9 10
Sorted output:
B = [1, 2, 7, 8]
0 1 2 3
The following array maps each element to its location in B:
[-, 0, 1, -, -, -, -, 2, 3, -, -]
0 1 2 3 4 5 6 7 8 9 10
How does this relate to L? Just subtract one from L[x] to compute the location for element x in the output.
However, if there are duplicates, going from the cummulative sum in L to the output gets a bit more complicated.
Another example with duplicate elements:
A = [3, 5, 2, 2, 8, 3]
0 1 2 3 4 5
Here are the counts
0 1 2 3 4 5 6 7 8
C = [0, 0, 2, 2, 0, 1, 0, 0, 1]
and the cummulative sum
L = [0, 0, 2, 4, 4, 5, 5, 5, 6]
0 1 2 3 4 5 6 7 8
Working back-to-front through A.
Where does A[5]=3 go? L[3] = 4, 4 - 1 = 3.
B = [0, 0, 0, 3, 0, 0]
0 1 2 3 4 5
Update the cummulative sum L to reflect that we’ve dealt with A[5]=3 by subtracting one from L[3].
L = [0, 0, 2, 3, 4, 5, 5, 5, 6]
0 1 2 3 4 5 6 7 8
Where does A[4]=8 go? L[8]=6, 6-1=5.
B = [0, 0, 0, 3, 0, 8]
Subtract one from L[8].
L = [0, 0, 2, 3, 4, 5, 5, 5, 5]
0 1 2 3 4 5 6 7 8
Where does A[3]=2 go? L[2]=2, 2-1=1.
B = [0, 2, 0, 3, 0, 8]
Subtract one from L[2].
L = [0, 0, 1, 3, 4, 5, 5, 5, 5]
0 1 2 3 4 5 6 7 8
where does 2 go? 1-1=0
B = [2, 2, 0, 3, 0, 8]
L = [0, 0, 0, 3, 4, 5, 5, 5, 5]
where does 5 go? 5-1=4
B = [2, 2, 0, 3, 5, 8]
L = [0, 0, 0, 3, 4, 4, 5, 5, 5]
where does 3 go? 3-1=2
B = [2, 2, 3, 3, 5, 8]
L = [0, 0, 0, 2, 4, 4, 5, 5, 5]
Counting sort in Java:
static void counting_sort(int[] A, int[] B, int k) {
int[] C = new int[k+1]; // counts of each element of A
int[] L = new int[k+1]; // L[j] = number of elements less or equal j.
// stage 1: counting
for (int i = 0; i != A.length; ++i) { // O(n)
++C[A[i]];
}
// stage 2: cummulative sum
L[0] = C[0];
for (int j = 1; j != k+1; ++j) { // O(k)
L[j] = C[j] + L[j-1];
}
// stage 3: produce output
for (int j = A.length - 1; j != -1; --j) { // O(n)
int elt = A[j];
int num_le = L[elt];
B[num_le - 1] = elt;
L[elt] = num_le - 1;
}
// total time complexity: O(n + k)
// if k is a constant, O(n)
// space complexity: O(k)
}
Among equal elements, they appear in the output in the same order that they appeared in the input. If the elements are merely integers, this doesn’t matter. But if the elements are something like personel records sorted by unique ID’s, then this might matter.
Radix sort also works on integers, and it sorts them by one digit at a time, starting with the least significant digit.
It’s important to use a stable sort for the sorting of each digit.
Example:
V V V
329 720 720 329
457 355 329 355
657 436 436 436
839 -> 457 -> 839 -> 457
436 657 355 657
720 329 457 720
355 839 657 839
static void radix_sort(int[] A, int d) {
int[] B = new int[A.length];
for (int i = 0; i != d; ++i) { // O(n*d)
counting_sort(A, B, 10, extract_digit(i,d)); // k=10, O(n+10) = O(n)
// swap A and B
for (int j = 0; j != A.length; ++j) { // O(n)
int tmp = A[j];
A[j] = B[j];
B[j] = tmp;
}
}
}
Had to update counting_sort to extract key from element, using function f.
static void counting_sort<E>(E[] A, int[] B, int k,
Function<E,Integer> f)
{
int[] C = new int[k+1]; // counts of each element of A
int[] L = new int[k+1]; // L[j] = number of elements less or equal j.
// Compute C
for (int i = 0; i != A.length; ++i) {
++C[f.apply(A[i])];
}
// Compute L
L[0] = C[0];
for (int j = 1; j != k+1; ++j) {
L[j] = C[j] + L[j-1];
}
// Generate output
for (int j = A.length - 1; j != -1; --j) {
int key = f.apply(A[j]);
int num_le = L[key];
B[num_le - 1] = A[j];
L[key] = num_le - 1;
}
}
Time complexity of radix_sort: O(d (n + k))
Bucket Sort assumes that the input is drawn from a uniform distribution. It then partitions the space into buckets and puts the input elements into their buckets.
Let’s fix the space to be [0,1). Then if we make the bucket array B the same size as A, we can just multiply the element number by the length of A to get the bucket number.
static void bucket_sort(double[] A) {
// Allocate the buckets
ArrayList<ArrayList<Double>> B = new ArrayList<>();
for (int i = 0; i != A.length; ++i) { // O(n)
B.add(new ArrayList<Double>());
}
// Distribute the elements of A to the buckets
for (int i = 0; i != A.length; ++i) { // O(n)
int bucket = (int)Math.floor(A[i] * A.length); // O(1)
B.get(bucket).add(A[i]); // O(1)
}
// Sort each bucket
for (int i = 0; i != B.size(); ++i) { // n iter,
B.get(i).sort((Double x, Double y) -> x < y ? -1 : (x > y) ? 1 : 0); // worst:O(n log n)
// average: O(1)
}
// Put the results back in A
int k = 0;
for (int i = 0; i != B.size(); ++i) { // n iters, O(n^2)? really O(n)
for (int j = 0; j != B.get(i).size(); ++j) { // n iters
A[k] = B.get(i).get(j); // O(1) , really once per input element
++k;
}
}
// total: average O(n)
// worst O(n^2 log n)
}
Time complexity of bucket_sort: