Sorting Array Elements By Frequency (Using STL)

Hello Everyone,

Given an array of integers, sort the array according to frequency of elements. If frequencies of two elements are same, print them in increasing order.

Examples:

Input : arr[] = {2, 3, 2, 4, 5, 12, 2, 3, 3, 3, 12}
Output : 3 3 3 3 2 2 2 12 12 4 5
Explanation :
No. Freq
2 : 3
3 : 4
4 : 1
5 : 1
12 : 2

Initially we create a map such that map[element] = freq. Once we are done building the map, we create an array of pairs. A pair which stores elements and their corresponding frequency will be used for the purpose of sorting. We write a custom compare function which compares two pairs firstly on the basis of frequency and if there is a tie on the basis of values.

Below is its c++ implementation :

// C++ program to sort elements by frequency using

// STL

#include <bits/stdc++.h>

using namespace std;

// function to compare two pairs for inbuilt sort

bool compare(pair< int , int > &p1,

pair< int , int > &p2)

{

// If frequencies are same, compare

// values

if (p1.second == p2.second)

return p1.first < p2.first;

return p1.second > p2.second;

}

// function to print elements sorted by freq

void printSorted( int arr[], int n)

{

// Store items and their frequencies

map< int , int > m;

for ( int i = 0; i < n; i++)

m[arr[i]]++;

// no of distinct values in the array

// is equal to size of map.

int s = m.size();

// an array of pairs

pair< int , int > p[s];

// Fill (val, freq) pairs in an array

// of pairs.

int i = 0;

for ( auto it = m.begin(); it != m.end(); ++it)

p[i++] = make_pair(it->first, it->second);

// sort the array of pairs using above

// compare function.

sort(p, p + s, compare);

cout << "Elements sorted by frequency are: " ;

for ( int i = 0; i < s; i++)

{

int freq = p[i].second;

while (freq--)

cout << p[i].first << " " ;

}

}

// driver program

int main()

{

int arr[] = {2, 3, 2, 4, 5, 12, 2, 3,

3, 3, 12};

int n = sizeof (arr)/ sizeof (arr[0]);

printSorted(arr, n);

return 0;

}

Output:

Elements sorted by frequency are: 3 3 3 3 2 2 2 12 12 4 5

Time Complexity : O(n Log n)