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DSA in JavaScript
Searching & Sorting

Counting Sort

Algorithm · Core

Level: CoreSort First

Step through it: Sort First pattern

nums
5
2
9
2
7

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.

Complexity

NameBestAverageWorstMemoryStableComments
Counting sortn + rn + rn + rn + rYesr - 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

CountingSort.js
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

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