# Sorting Array Elements By Frequency | Set 3 (Using STL)

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

We have discussed different approaches in below posts :

Sort elements by frequency | Set 1

Sort elements by frequency | Set 2

We can solve this problem using map and pairs. 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 freq 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)

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