Implementing Counting Sort using map in C++
Last Updated :
23 May, 2023
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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++:
// 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)
