Counting sort

This is a very fast algorithm, but can only be used with arrays with integer values.

This algorithm can be explained as follows:

  • find minimum (min) and maximum (max) value in the integer array
  • create a new array with length = max – min + 1 to store statistics
  • iterate through the initial array and count the array values. Put the value statistics into the previously created array at indexes equal to the values

For example, the initial array:

[1, 2, 3, 2, 1, 5, 0], min = 0, max = 5

Statistics array:

[1, 2, 1, 1, 0, 1]

  • iterate through a statistic array and restore the initial array:

[0, 1, 1, 2, 3, 5]

  • time complexity: O(n+m), where n – the length of the initial array, m – the length of the statistical array
  • space complexity: O(m)
import java.util.Arrays;

public class ImplementCountingSort {

  private static void sort(int[] input) {
    var min = Integer.MAX_VALUE;
    var max = Integer.MIN_VALUE;
    for (int j : input) {
      if (j < min) {
        min = j;
      }
      if (j > max) {
        max = j;
      }
    }
    var arrayLength = max - min + 1;
    var counts = new int[arrayLength];
    for (var i = 0; i < input.length; i++) {
      counts[input[i] - min]++;
    }
    var i = 0;
    for (var j = 0; j < counts.length; j++) {
      while (counts[j] > 0) {
        input[i] = min + j;
        i++;
        counts[j]--;
      }
    }
  }

  public static void main(String[] args) {
    int[] array = {-6, -6, -6, 3, 2, 1};
    System.out.println(Arrays.toString(array));
    sort(array);
    System.out.println(Arrays.toString(array));
  }
}

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top