countingSortOf<E> function

void countingSortOf<E>(List<E> list, KeyOf<E, int> keyOf, { int? begin, int? end, bool reversed = false, })

Sorts any list of items using counting sort algorithm.

Parameters

• list is any list of integers to be sorted.
• keyOf is an mapping function to get the integer value of the item.
• To perform partial sorting, you can specify the begin or end.
• begin is the start index of the range to be sorted.
• If begin is negative, range starts at the 0
• If begin is not below the length of the list, range will be empty.
• end is the final index if the range to be sorted. It is exclusive.
• If end is above the length of the list, it will be ignored.
• If end is negative, the absolute value of it will be subtracted from the length of the list to determine where the range ends.
• If end is not greater than the begin, the range will be empty.
• Set reversed to true if you want to sort in descending order.

Details

This function maps the list of items with the keyOf to generate a list of numbers. countingSort is applied to this list of numbers, and a sorted list is reconstructed with the mapped values from these numbers.

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Complexity: Time O(n + k) | Space O(n + k) (k is the range of the keys)

Implementation

void countingSortOf<E>(
  List<E> list,
  KeyOf<E, int> keyOf, {
  int? begin,
  int? end,
  bool reversed = false,
}) {
  int b, e;
  int n = list.length;

  // Find the range given the parameters.
  b = 0;
  e = n;
  if (begin != null && b < begin) {
    b = begin;
  }
  if (end != null && end < e) {
    e = end;
    if (e < 0) e += n;
  }

  if (b + 1 >= e) return;

  /// sorts range `[b, e)`
  int l, h, m, i, j, k;
  List<int> keys =
      List<int>.generate(e, (i) => keyOf(list[i]), growable: false);

  // Find the minimum and maximum numbers in range [b, e)
  l = keys[b];
  h = keys[b];
  for (i = b + 1; i < e; ++i) {
    if (keys[i] < l) {
      l = keys[i];
    }
    if (keys[i] > h) {
      h = keys[i];
    }
  }

  m = h - l + 1;
  List<int> count = List<int>.filled(m, 0, growable: false);
  List<E> values = List<E>.filled(m, list[b], growable: false);

  // Count frequencies of the numbers
  for (i = b; i < e; ++i) {
    count[keys[i] - l]++;
    values[keys[i] - l] = list[i];
  }

  // Reconstruct a sorted list
  k = b;
  for (i = 0; i < m; ++i) {
    for (j = count[i]; j > 0; --j) {
      list[k++] = values[i];
    }
  }

  if (reversed) {
    // reverse the order of the list
    reverseList(list, b, e);
  }
}