Sort elements by frequency 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 :
CPP
// 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; } |
Java
// Java program to sort elements by frequency using import java.util.*; public class Main { // function to compare two pairs for inbuilt sort static boolean compare(Map.Entry<Integer, Integer> p1, Map.Entry<Integer, Integer> p2) { // If frequencies are same, compare values if (p1.getValue().equals(p2.getValue())) return p1.getKey() < p2.getKey(); return p1.getValue() > p2.getValue(); } // function to print elements sorted by freq static void printSorted( int [] arr, int n) { // Store items and their frequencies Map<Integer, Integer> m = new HashMap<Integer, Integer>(); for ( int i = 0 ; i < n; i++) m.put(arr[i], m.getOrDefault(arr[i], 0 ) + 1 ); // no of distinct values in the array // is equal to size of map. int s = m.size(); // an array of Map.Entry pairs List<Map.Entry<Integer, Integer> > list = new ArrayList<Map.Entry<Integer, Integer> >( m.entrySet()); // sort the list of Map.Entry pairs using above // compare function. Collections.sort( list, (p1, p2) -> compare(p1, p2) ? - 1 : 1 ); System.out.print( "Elements sorted by frequency are: " ); for (Map.Entry<Integer, Integer> entry : list) { int freq = entry.getValue(); while (freq-- > 0 ) System.out.print(entry.getKey() + " " ); } } // driver program public static void main(String[] args) { int [] arr = { 2 , 3 , 2 , 4 , 5 , 12 , 2 , 3 , 3 , 3 , 12 }; int n = arr.length; printSorted(arr, n); } } |
C#
using System; using System.Collections.Generic; using System.Linq; public class MainClass { // function to compare two pairs for inbuilt sort static bool compare(KeyValuePair< int , int > p1, KeyValuePair< int , int > p2) { // If frequencies are same, compare values if (p1.Value == p2.Value) return p1.Key < p2.Key; return p1.Value > p2.Value; } // function to print elements sorted by freq static void printSorted( int [] arr, int n) { // Store items and their frequencies Dictionary< int , int > m = new Dictionary< int , int >(); for ( int i = 0; i < n; i++) { if (m.ContainsKey(arr[i])) m[arr[i]]++; else m[arr[i]] = 1; } // no of distinct values in the array // is equal to size of dictionary. int s = m.Count; // an array of KeyValuePair pairs List<KeyValuePair< int , int >> list = new List<KeyValuePair< int , int >>(m); // sort the list of KeyValuePair pairs using above // compare function. list.Sort((p1, p2) => compare(p1, p2) ? -1 : 1); Console.Write( "Elements sorted by frequency are: " ); foreach (KeyValuePair< int , int > entry in list) { int freq = entry.Value; while (freq-- > 0) Console.Write(entry.Key + " " ); } } // driver program public static void Main( string [] args) { int [] arr = { 2, 3, 2, 4, 5, 12, 2, 3, 3, 3, 12 }; int n = arr.Length; printSorted(arr, n); } } |
Elements sorted by frequency are: 3 3 3 3 2 2 2 12 12 4 5
Time Complexity : O(n Log n)
Space Complexity: O(n)
The above algorithm requires O(n) space for the hash map and the array of pairs.
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