Comparison Sorting
Quicksort usually has a running time of n x log(n), but is there an algorithm that can sort even faster? In general, this is not possible. Most sorting algorithms are comparison sorts, i.e. they sort a list just by comparing the elements to one another. A comparison sort algorithm cannot beat n x log(n) (worst-case) running time, since n x log(n) represents the minimum number of comparisons needed to know where to place each element. For more details, you can see these notes (PDF).
Alternative Sorting
Another sorting method, the counting sort, does not require comparison. Instead, you create an integer array whose index range covers the entire range of values in your array to sort. Each time a value occurs in the original array, you increment the counter at that index. At the end, run through your counting array, printing the value of each non-zero valued index that number of times.
Example
arr = [1,1,1,3,2,1]
All of the values are in the range [0...3]
, so create an array of zeros result = [0,0,0,0]
. The results of each iteration follow:
i arr[i] result
0 1 [0, 1, 0, 0]
1 1 [0, 2, 0, 0]
2 3 [0, 2, 0, 1]
3 2 [0, 2, 1, 1]
4 1 [0, 3, 1, 1]
The frequency array is [0,3,1,1]
. These values can be used to create the sorted array as well: sorted = [1,1,1,2,3]
.
Note
For this exercise, always return a frequency array with 100 elements. The example above shows only the first 4 elements, the remainder being zeros.
Challenge
Given a list of integers, count and return the number of times each value appears as an array of integers.
Function Description
Complete the countingSort
function in the editor below.
countingSort
has the following parameter(s):
arr[n]
: an array of integers
Returns
int[100]
: a frequency array
Input Format
The first line contains an integer , the number of items in . Each of the next lines contains an integer where .
Constraints
100 <= n <- 10^6
0 <-= arr[i] < 100
Sample Input
100
63 25 73 1 98 73 56 84 86 57 16 83 8 25 81 56 9 53 98 67 99 12 83 89 80 91 39 86 76 85 74 39 25 90 59 10 94 32 44 3 89 30 27 79 46 96 27 32 18 21 92 69 81 40 40 34 68 78 24 87 42 69 23 41 78 22 6 90 99 89 50 30 20 1 43 3 70 95 33 46 44 9 69 48 33 60 65 16 82 67 61 32 21 79 75 75 13 87 70 33
Sample Output
0 2 0 2 0 0 1 0 1 2 1 0 1 1 0 0 2 0 1 0 1 2 1 1 1 3 0 2 0 0 2 0 3 3 1 0 0 0 0 2 2 1 1 1 2 0 2 0 1 0 1 0 0 1 0 0 2 1 0 1 1 1 0 1 0 1 0 2 1 3 2 0 0 2 1 2 1 0 2 2 1 2 1 2 1 1 2 2 0 3 2 1 1 0 1 1 1 0 2 2
Explanation
Each of the resulting values represents the number of times result[i] appeared in arr.
Solution
// Convert sample input string into a usable List<int>
string s = "63 25 73 1 98 73 56 84 86 57 16 83 8 25 81 56 9 53 98 67 99 12 83 89 80 91 39 86 76 85 74 39 25 90 59 10 94 32 44 3 89 30 27 79 46 96 27 32 18 21 92 69 81 40 40 34 68 78 24 87 42 69 23 41 78 22 6 90 99 89 50 30 20 1 43 3 70 95 33 46 44 9 69 48 33 60 65 16 82 67 61 32 21 79 75 75 13 87 70 33";
string[] inputStrings = s.Split(' ');
int[] inputArray = Array.ConvertAll<string, int>(inputStrings, s => int.Parse(s));
List<int> arrList = inputArray.ToList();
foreach (var item in arrList)
{
Console.Write(item + " ");
Console.WriteLine();
}
List<int> output = countingSort1(arrList);
foreach (var item in output)
{
Console.Write(item + " ");
}
static List<int> countingSort1(List<int> arr)
{
int[] result = new int[100];
// take the input List<int> and break it into chunks of 100
for (int i = 0; i < arr.Count; i += 100)
{
for (int j = i; j < i + 100; j++)
{
// the int value is the result array index
// increment the count at that index
result[arr[j]]++;
}
}
return result.ToList();
}
```
result[arr[i]]++;
on your solution? I find it hard to comprehend. I appreciate it your time. Thank you so much \$\endgroup\$