Counting Sort
Algorithm · Core
Step through it: Sort First pattern
Step 1 of 4
Goal: does the list contain a duplicate? Checking every pair works but is slow. Idea: sort first, so equal numbers end up side by side.
Step 2 of 4
After sorting: 2, 2, 5, 7, 9. Equal numbers are now neighbours.
Step 3 of 4
Now one pass comparing each number with the next: 2 and 2 match. Duplicate found. Sorting often turns a hard search into a simple neighbour check, and makes tricks like two pointers and interval merging possible.
Step 4 of 4
Your turn.
Complexity
| Name | Best | Average | Worst | Memory | Stable | Comments |
|---|---|---|---|---|---|---|
| Counting sort | n + r | n + r | n + r | n + r | Yes | r - biggest number in array |
In computer science, counting sort is an algorithm for sorting a collection of objects according to keys that are small integers; that is, it is an integer sorting algorithm. It operates by counting the number of objects that have each distinct key value, and using arithmetic on those counts to determine the positions of each key value in the output sequence. Its running time is linear in the number of items and the difference between the maximum and minimum key values, so it is only suitable for direct use in situations where the variation in keys is not significantly greater than the number of items. However, it is often used as a subroutine in another sorting algorithm, radix sort, that can handle larger keys more efficiently.
Because counting sort uses key values as indexes into an array,
it is not a comparison sort, and the Ω(n log n) lower bound for
comparison sorting does not apply to it. Bucket sort may be used
for many of the same tasks as counting sort, with a similar time
analysis; however, compared to counting sort, bucket sort requires
linked lists, dynamic arrays or a large amount of preallocated
memory to hold the sets of items within each bucket, whereas
counting sort instead stores a single number (the count of items)
per bucket.
Counting sorting works best when the range of numbers for each array element is very small.
Algorithm
Step I
In first step we calculate the count of all the elements of the
input array A. Then Store the result in the count array C.
The way we count is depicted below.
Step II
In second step we calculate how many elements exist in the input
array A which are less than or equals for the given index.
Ci = numbers of elements less than or equals to i in input array.
Step III
In this step we place the input array A element at sorted
position by taking help of constructed count array C ,i.e what
we constructed in step two. We used the result array B to store
the sorted elements. Here we handled the index of B start from
zero.
Code
import Sort from '../Sort';
export default class CountingSort extends Sort {
/**
* @param {number[]} originalArray
* @param {number} [smallestElement]
* @param {number} [biggestElement]
*/
sort(originalArray, smallestElement = undefined, biggestElement = undefined) {
// Init biggest and smallest elements in array in order to build number bucket array later.
let detectedSmallestElement = smallestElement || 0;
let detectedBiggestElement = biggestElement || 0;
if (smallestElement === undefined || biggestElement === undefined) {
originalArray.forEach((element) => {
// Visit element.
this.callbacks.visitingCallback(element);
// Detect biggest element.
if (this.comparator.greaterThan(element, detectedBiggestElement)) {
detectedBiggestElement = element;
}
// Detect smallest element.
if (this.comparator.lessThan(element, detectedSmallestElement)) {
detectedSmallestElement = element;
}
});
}
// Init buckets array.
// This array will hold frequency of each number from originalArray.
const buckets = Array(detectedBiggestElement - detectedSmallestElement + 1).fill(0);
originalArray.forEach((element) => {
// Visit element.
this.callbacks.visitingCallback(element);
buckets[element - detectedSmallestElement] += 1;
});
// Add previous frequencies to the current one for each number in bucket
// to detect how many numbers less then current one should be standing to
// the left of current one.
for (let bucketIndex = 1; bucketIndex < buckets.length; bucketIndex += 1) {
buckets[bucketIndex] += buckets[bucketIndex - 1];
}
// Now let's shift frequencies to the right so that they show correct numbers.
// I.e. if we won't shift right than the value of buckets[5] will display how many
// elements less than 5 should be placed to the left of 5 in sorted array
// INCLUDING 5th. After shifting though this number will not include 5th anymore.
buckets.pop();
buckets.unshift(0);
// Now let's assemble sorted array.
const sortedArray = Array(originalArray.length).fill(null);
for (let elementIndex = 0; elementIndex < originalArray.length; elementIndex += 1) {
// Get the element that we want to put into correct sorted position.
const element = originalArray[elementIndex];
// Visit element.
this.callbacks.visitingCallback(element);
// Get correct position of this element in sorted array.
const elementSortedPosition = buckets[element - detectedSmallestElement];
// Put element into correct position in sorted array.
sortedArray[elementSortedPosition] = element;
// Increase position of current element in the bucket for future correct placements.
buckets[element - detectedSmallestElement] += 1;
}
// Return sorted array.
return sortedArray;
}
}References
Practice
Sort First
- Contains DuplicateEasy
- Valid AnagramEasy
- Meeting RoomsEasy
- Merge Sorted ArrayEasy
- Sort ColorsMedium
- Largest NumberMedium
- H-IndexMedium
- Wiggle Sort IIMedium
- Sort an ArrayMedium
- Custom Sort StringMedium
- Insertion Sort ListMedium
- Maximum GapHard
- Count of Smaller Numbers After SelfHard
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