Sorting algorithms are a must-know for any software engineer preparing for a technical interview. They will help you crack many questions during your coding round.
In this article, we’ll give you a refresher on the counting algorithm:
Counting sort is an algorithm used to sort the elements of an array by counting and storing the frequency of each distinct element in an auxiliary array. Sorting is done by mapping the value of each element as an index of the auxiliary array.
Counting sort is handy while sorting values whose range is not very large.
Let’s assume that array Arr[] = {4, 1, 3, 4, 6, 6} with the maximum element M (here it is 6) is to be sorted.
First, we take an auxiliary array Aux[] of size M+1 (here it is 6+1 = 7) and initialize it with 0.
Next, we store the frequency of each unique element of array Arr[] in Aux[].
Then, we calculate the prefix sum at every index of Aux[] array.
We then create an array sortedArr[] of size equal to the Arr[] array.
Next, traverse array Arr[] from right to left and update sortedArr[] as
sortedArr[ Aux[ Arr[i] ] – 1] = Arr[i] and Aux[] as Aux[ Arr[i] ]–. The reason behind traversing Arr[] from right to left is to preserve the order of equal elements, which eventually makes this sorting algorithm stable.
Consider an array Arr[] of size N that we want to sort:
Step 1: Declare an auxiliary array Aux[] of size max(Arr[])+1 and initialize it with 0.
Step 2: Traverse array Arr[] and map each element of Arr[] as an index of Aux[] array, i.e., execute Aux[Arr[i]]++ for 0 <= i < N.
Step 3: Calculate the prefix sum at every index of array Arr[].
Step 4: Create an array sortedArr[] of size N.
Step 5: Traverse array Arr[] from right to left and update sortedArr[] as
sortedArr[ Aux[ Arr[i] ] – 1] – Arr[i]. Also, update Aux[] as Aux[ Arr[i] ]–.
As we are using the index of an array to map elements of array Arr[], if the difference between the maximum element and minimum element is huge, counting sort will not be feasible.
Counting sort is helpful when we have to sort many numbers that lie in a comparatively small range.
For example, take an array Arr[] = {1000000005,1001000000,1000000007}
Here, maximum element of Arr[] is 1001000000
Can we apply the counting sort algorithm to this type of array?
Yes, we can! Here’s how:
Assumptions for input:
Arr[] =Â array to be sorted
Aux[] = auxiliary array for mapping
sortedArr[] = sorted version of Arr[]
M = maximum element of Arr[]
N = size of Arr[]
// initialising array Aux[] with 0
for i = 0 to M + 1 do
  Aux[ i ] = 0
// Storing count of each element
// in array Aux[]
for i in Arr[] do
  Aux[ Arr[ i ] ] ++
for i from N-1 to 0 in Arr[] do
  sortedArr[ Aux[ Arr[i] ] - 1] = Arr[i]
  Aux[ Arr[i] ] --
return sortedArr[]
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We’ve used C++ to demonstrate how counting sort works. Use this as a reference to code in C, Java, Python, or any programming language of your choice.
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int N = 8;
    int Arr[N] = {4, 3, 12, 1, 5, 5, 3, 9};
    // Finding the maximum element of array Arr[].
    int M = 0;
    for(int i=0;i<N;i++)
        M=max(M,Arr[i]);
    // Initializing Aux[] with 0
    int Aux[M+1];
    for(int i=0;i<=M;i++)
        Aux[i]=0;
    // Mapping each element of Arr[] as an index
    // of Aux[] array
    for(int i=0;i<N;i++)
        Aux[Arr[i]]++;
    // Calculating prefix sum at every index
    // of array Aux[]
    for(int i=1;i<=M;i++)
        Aux[i]+=Aux[i-1];
    // Creating sortedArr[] from Aux[] array
    int sortedArr[N];
    for(int i=N-1;i>=0;i–)
    {
        sortedArr[Aux[Arr[i]] – 1]=Arr[i];
        Aux[Arr[i]]–;
    }
    for(int i=0;i<N;i++)
        cout<<sortedArr[i]<<” “;
    return 0;
}
Consider Arr[] = {4, 3, 12, 1, 5, 5, 3, 9}
Here, M = 12, so we will make an Aux[] array of size 13.
After mapping all elements of array Arr[] as the index of Aux[] array:
Aux[] = {0, 1, 0, 2, 1, 2, 0, 0, 0, 1, 0, 0, 1}
Calculate prefix sum of Aux[] array:
Aux[] = {0, 1, 1, 3, 4, 6, 6, 6, 6, 7, 7, 7, 8}
Now, traverse in array Arr[] from right to left and place each Arr[i] at its correct place in sortedArr[] with the help Aux[] array:
Finally, sortedArr[] = {1, 3, 3, 4, 5, 5, 9, 12}
In this algorithm:
Therefore, the overall time complexity is O(N+M).
In this algorithm:
Therefore, the space complexity is O(N+M).
Question 1: Give an example of a non-comparison-based sorting algorithm. Is it faster than comparison-based sorting algorithms?
Answer:
Question 2: Where should counting sort be used?
Answer: Counting sort works better than the other comparison-based sorting algorithms when we have to sort many numbers that lie in a comparatively smaller range on the number line.
Counting sort has a time complexity of O(N+M), where M is max(arr[])-min(arr[]) and N is equal to size(arr[]). Comparison-based sorting algorithms take O(N log N) time.
When (N+M) << N log N, we can definitely use counting sort, given that we can feasibly allocate memory of O(N+M). On the other hand, if N log N << (N + M), we should use a comparison-based sorting algorithm.
Question 3: Is counting sort stable?
Answer: Yes, counting sort is an example of a stable sorting algorithm, as it does not change the relative order of elements of the same value in the input.
Question 4. Is counting sort an in-place sorting algorithm?
Answer: No, counting sort is not an in-place sorting algorithm. In in-place sorting algorithms, only a small/constant auxiliary space is used; in counting sort, we use an auxiliary array of the size of the range of data.
Sorting algorithms interview questions feature in almost every coding interview for software developers. If you’re looking for guidance and help to nail these questions and more, sign up for our free webinar.
As pioneers in the field of technical interview prep, we have trained thousands of software engineers to crack the toughest coding interviews and land jobs at their dream companies, such as Google, Facebook, Apple, Netflix, Amazon, and more!
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Article contributed by Abhinav Tiwari
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