Implementing Counting Sort using map in C++

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Counting Sort is one of the best sorting algorithms which can sort in O(n) time complexity but the disadvantage with the counting sort is it’s space complexity, for a small collection of values, it will also require a huge amount of unused space. So, we need two things to overcome this:

A data structure which occupies the space for input elements only and not for all the elements other than inputs.
The stored elements must be in sorted order because if it’s unsorted then storing them will be of no use.

So Map in C++ satisfies both the condition. Thus we can achieve this through a map.

Examples:

Input: arr[] = {1, 4, 3, 5, 1}

Output: 1 1 3 4 5

Input: arr[] = {1, -1, -3, 8, -3}

Output: -3 -3 -1 1 8

Below is the implementation of Counting Sort using map in C++:

CPP

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
 
// Function to sort the array using counting sort 
void countingSort(vector<int> arr, int n) 
{ 
 
    // Map to store the frequency 
    // of the array elements 
    map<int, int> freqMap; 
    for (auto i = arr.begin(); i != arr.end(); i++) { 
        freqMap[*i]++; 
    } 
 
    int i = 0; 
 
    // For every element of the map 
    for (auto it : freqMap) { 
 
        // Value of the element 
        int val = it.first; 
 
        // Its frequency 
        int freq = it.second; 
        for (int j = 0; j < freq; j++) 
            arr[i++] = val; 
    } 
 
    // Print the sorted array 
    for (auto i = arr.begin(); i != arr.end(); i++) { 
        cout << *i << " "; 
    } 
} 
 
// Driver code 
int main() 
{ 
    vector<int> arr = { 1, 4, 3, 5, 1 }; 
    int n = arr.size(); 
 
    countingSort(arr, n); 
 
    return 0; 
} 

Output:

1 1 3 4 5

Time Complexity: O(n log(n))

Auxiliary Space: O(n)

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