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Data Structures & Algorithms in Dart

Second Edition · Flutter · Dart 3.0 · VS Code 1.78

Section VI: Challenge Solutions

Section 6: 21 chapters
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18. Chapter 18 Solutions
Written by Jonathan Sande

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Solution to Challenge 1

You’re starting with the following list:

var list = [46, 500, 459, 1345, 13, 999];

LSD-Radix Sort

Modify your radixSort extension by adding the following print statement at the bottom of the while loop:

print(buckets);
[[500], [], [], [13], [], [1345], [46], [], [], [459, 999]]
[[500], [13], [], [], [1345, 46], [459], [], [], [], [999]]
[[13, 46], [], [], [1345], [459], [500], [], [], [], [999]]
[[13, 46, 459, 500, 999], [1345], [], [], [], [], [], [], [], []]

MSD-Radix Sort

To see the items in your buckets on each recursion, add the following line to _msdRadixSorted right above where you set bucketOrder:

print(buckets);
// final bucketOrder = ...
[[], [1345, 13], [], [], [46, 459], [500], [], [], [], [999]]
[[], [], [], [1345, 13], [], [], [], [], [], []]
[[], [], [], [], [1345], [], [], [], [], []]
[[], [], [], [], [], [459], [46], [], [], []]

Solution to Challenge 2

You can find the number of unique UTF-16 code units in a list of words by adding each code unit to a set:

int uniqueCharacters(List<String> words) {
  final uniqueChars = <int>{};
  for (final word in words) {
    for (final codeUnit in word.codeUnits) {
      uniqueChars.add(codeUnit);
    }
  }
  return uniqueChars.length;
}
final words = ['done', 'ad', 'eel', 'zoo', 'adept', 'do'];
print(uniqueCharacters(words)); // 9

Solution to Challenge 3

Rather than just checking if you’ve reached the most significant digit of the largest number, you can count how many numbers are left to sort. If there’s only one, then you’re finished.

extension RadixSort on List<int> {
  void radixSort() {
    const base = 10;
    var place = 1;
    // 1
    var unsorted = length;
    // 2
    while (unsorted > 1) {
      // 3
      unsorted = 0;
      final buckets = List.generate(base, (_) => <int>[]);
      forEach((number) {
        final remainingPart = number ~/ place;
        final digit = remainingPart % base;
        buckets[digit].add(number);
        // 4
        if (remainingPart ~/ base > 0) {
          unsorted++;
        }
      });
      place *= base;
      clear();
      addAll(buckets.expand((element) => element));
    }
  }
}
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