We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. The Full Counting Sort
  2. Discussions

The Full Counting Sort

Sort by

recency

|

81 Discussions

|

  • benjuniour26
     + 0 comments

    Python solution:

    def countSort(arr):
    mid_point = len(arr)//2
    mid_point_arr = [[elem[0], "-"] for elem in arr[:mid_point]]
    dashed_arr = mid_point_arr + arr[mid_point:]
    dashed_arr.sort(key=lambda elem: int(elem[0]))
    [print(elem[1], end=" ") for elem in dashed_arr]

  • victor_reial
     + 0 comments

    Rust best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    fn full_countint_sort(arr: &[Vec<String>]) {
    //Time complexity: O(n)
    //Space complexity (ignoring input): O(1)
    let mut sorted_array: Vec<String> = vec!["".to_string(); 101];
    for index in 0..arr.len() {
        let sorted_index = arr[index][0]
            .parse::<usize>()
            .expect("To be able to parse correctly");
        if index < arr.len() / 2 {
            sorted_array[sorted_index].push_str(" -");
        } else {
            sorted_array[sorted_index].push(' ');
            sorted_array[sorted_index].push_str(&arr[index][1]);
        }
    }
    let mut result_array = String::new();
    for string in sorted_array {
        if !string.is_empty() {
            if result_array.is_empty() {
                result_array = string[1..].to_string();
            } else {
                result_array.push_str(&string);
            }
        }
    }
    println!("{}", result_array);
}

  • victor_reial
     + 0 comments

    Python best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    def full_counting_sort(arr):
    #Time complexity: O(n)
    #Space complexity (ignoring input): O(1)
    sorted_array = [""]*101
    for index in range(0, len(arr)):
        sorted_index = int(arr[index][0])
        if index < len(arr)/2:
            sorted_array[sorted_index] += " -"
        else:
            sorted_array[sorted_index] += " "
            for letter in arr[index][1]:
                sorted_array[sorted_index] += letter

    result_string = ""
    for string in sorted_array:
        result_string += string

    print(result_string[1:])

  • michael_chen_0
     + 0 comments

    Python 3 solution:

    def countSort(arr: list[list[str]]) -> None:
    for i in range(len(arr) // 2):
        arr[i][1] = "-"
    arr.sort(key=lambda x: int(x[0]))
    print(*(x[1] for x in arr))

  • shadowinc
     + 0 comments
    1. create empty array of 100 arrays
    2. add elements acc to index mentioned
    3. remove empty elements
    4. convert to string

    def countSort(arr):

    arr_sort = [[] for _ in range(100)]
arr_len = len(arr)
for j in range(arr_len):
    f = int(arr[j][0])
    s = arr[j][1]
    if j < arr_len // 2:
        arr_sort[f].append('-')
    else:
        arr_sort[f].append(s)

res = [y for x in arr_sort for y in x]
print(" ".join(res))