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; ` `} ` |

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Output:

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

Time Complexity : O(n Log n)

This article is contributed by **Aditi Sharma**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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