Sort elements by frequency | Set 4 (Efficient approach using hash)
Print the elements of an array in the decreasing frequency if 2 numbers have the same frequency then print the one which came first.
Examples:
Input : arr[] = {2, 5, 2, 8, 5, 6, 8, 8} Output : arr[] = {8, 8, 8, 2, 2, 5, 5, 6} Input : arr[] = {2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8} Output : arr[] = {8, 8, 8, 2, 2, 5, 5, 6, -1, 9999999}
We have discussed different approaches in below posts :
Sort elements by frequency | Set 1
Sort elements by frequency | Set 2
Sorting Array Elements By Frequency | Set 3 (Using STL)
All of the above approaches work in O(n Log n) time where n is total number of elements. In this post, a new approach is discussed that works in O(n + m Log m) time where n is total number of elements and m is total number of distinct elements.
The idea is to use hashing.
- We insert all elements and their counts into a hash. This step takes O(n) time where n is number of elements.
- We copy the contents of hash to an array (or vector) and sort them by counts. This step takes O(m Log m) time where m is total number of distinct elements.
- For maintaining the order of elements if the frequency is the same, we use another hash which has the key as elements of the array and value as the index. If the frequency is the same for two elements then sort elements according to the index.
The below image is a dry run of the above approach:
We do not need to declare another map m2, as it does not provide the proper expected result for the problem.
instead, we need to just check for the first values of the pairs sent as parameters in the sortByVal function.
Below is the implementation of the above approach:
C++
// CPP program for the above approach #include <bits/stdc++.h> using namespace std; // Used for sorting by frequency. And if frequency is same, // then by appearance bool sortByVal( const pair< int , int >& a, const pair< int , int >& b) { // If frequency is same then sort by index if (a.second == b.second) return a.first < b.first; return a.second > b.second; } // function to sort elements by frequency vector< int >sortByFreq( int a[], int n) { vector< int >res; unordered_map< int , int > m; vector<pair< int , int > > v; for ( int i = 0; i < n; ++i) { // Map m is used to keep track of count // of elements in array m[a[i]]++; } // Copy map to vector copy(m.begin(), m.end(), back_inserter(v)); // Sort the element of array by frequency sort(v.begin(), v.end(), sortByVal); for ( int i = 0; i < v.size(); ++i) while (v[i].second--) { res.push_back(v[i].first); } return res; } // Driver program int main() { int a[] = { 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 }; int n = sizeof (a) / sizeof (a[0]); vector< int >res; res = sortByFreq(a, n); for ( int i = 0;i < res.size(); i++) cout<<res[i]<< " " ; return 0; } |
Python3
# Used for sorting by frequency. And if frequency is same, # then by appearance from functools import cmp_to_key def sortByVal(a,b): # If frequency is same then sort by index if (a[ 1 ] = = b[ 1 ]): return a[ 0 ] - b[ 0 ] return b[ 1 ] - a[ 1 ] # function to sort elements by frequency def sortByFreq(a, n): res = [] m = {} v = [] for i in range (n): # Map m is used to keep track of count # of elements in array if (a[i] in m): m[a[i]] = m[a[i]] + 1 else : m[a[i]] = 1 for key,value in m.items(): v.append([key,value]) # Sort the element of array by frequency v.sort(key = cmp_to_key(sortByVal)) for i in range ( len (v)): while (v[i][ 1 ]): res.append(v[i][ 0 ]) v[i][ 1 ] - = 1 return res # Driver program a = [ 2 , 5 , 2 , 6 , - 1 , 9999999 , 5 , 8 , 8 , 8 ] n = len (a) res = [] res = sortByFreq(a, n) for i in range ( len (res)): print (res[i],end = " " ) # This code is contributed by shinjanpatra |
Java
// Java program for the above approach import java.util.*; // Used for sorting by frequency. And if frequency is same, // then by appearance class SortByValue implements Comparator<Map.Entry<Integer, Integer> > { // Used for sorting in descending order of values public int compare(Map.Entry<Integer, Integer> o1, Map.Entry<Integer, Integer> o2) { // If frequency is same then sort by index if (o1.getValue() == o2.getValue()) return o1.getKey() - o2.getKey(); return o2.getValue() - o1.getValue(); } } class GFG { // Function to sort elements by frequency static Vector<Integer> sortByFreq( int a[], int n) { // Map to store the frequency of the elements HashMap<Integer, Integer> m = new HashMap<>(); // Vector to store the sorted elements Vector<Integer> v = new Vector<>(); // Insert elements and their frequency in the map for ( int i = 0 ; i < n; i++) { int x = a[i]; if (m.containsKey(x)) m.put(x, m.get(x) + 1 ); else m.put(x, 1 ); } // Copy map to vector Vector<Map.Entry<Integer, Integer> > v1 = new Vector<>(m.entrySet()); // Sort the vector elements by frequency Collections.sort(v1, new SortByValue()); // Traverse the vector and insert elements // in the vector v for ( int i = 0 ; i < v1.size(); i++) for ( int j = 0 ; j < v1.get(i).getValue(); j++) v.add(v1.get(i).getKey()); return v; } // Driver program public static void main(String[] args) { int a[] = { 2 , 5 , 2 , 6 , - 1 , 9999999 , 5 , 8 , 8 , 8 }; int n = a.length; Vector<Integer> v = sortByFreq(a, n); // Print the elements of vector for ( int i = 0 ; i < v.size(); i++) System.out.print(v.get(i) + " " ); } } |
Javascript
<script> // JavaScript program for the above approach // Used for sorting by frequency. And if frequency is same, // then by appearance function sortByVal(a,b) { // If frequency is same then sort by index if (a[1] == b[1]) return a[0] - b[0]; return b[1] - a[1]; } // function to sort elements by frequency function sortByFreq(a, n) { let res = []; let m = new Map(); let v = []; for (let i = 0; i < n; ++i) { // Map m is used to keep track of count // of elements in array if (m.has(a[i])) m.set(a[i],m.get(a[i])+1); else m.set(a[i],1); } for (let [key,value] of m){ v.push([key,value]); } // Sort the element of array by frequency v.sort(sortByVal) for (let i = 0; i < v.length; ++i) while (v[i][1]--) { res.push(v[i][0]); } return res; } // Driver program let a = [ 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 ]; let n = a.length; let res = []; res = sortByFreq(a, n); for (let i = 0;i < res.length; i++) document.write(res[i], " " ); // This code is contributed by shinjanpatra </script> |
C#
using System; using System.Collections.Generic; using System.Linq; // Used for sorting by frequency. And if frequency is same, // then by appearance class SortByValue : IComparer<KeyValuePair< int , int >> { // Used for sorting in descending order of values public int Compare(KeyValuePair< int , int > o1, KeyValuePair< int , int > o2) { // If frequency is same then sort by index if (o1.Value == o2.Value) return o1.Key - o2.Key; return o2.Value - o1.Value; } } class GFG { // Function to sort elements by frequency static List< int > sortByFreq( int [] a, int n) { // Dictionary to store the frequency of the elements Dictionary< int , int > m = new Dictionary< int , int >(); // List to store the sorted elements List< int > res = new List< int >(); // Insert elements and their frequency in the dictionary for ( int i = 0; i < n; i++) { int x = a[i]; if (m.ContainsKey(x)) m[x]++; else m.Add(x, 1); } // Copy dictionary to list List<KeyValuePair< int , int >> v = new List<KeyValuePair< int , int >>(m); // Sort the list elements by frequency v.Sort( new SortByValue()); // Traverse the list and insert elements // in the list v foreach (KeyValuePair< int , int > kvp in v) { for ( int i = 0; i < kvp.Value; i++) res.Add(kvp.Key); } return res; } // Driver program public static void Main( string [] args) { int [] a = { 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 }; int n = a.Length; List< int > res = sortByFreq(a, n); // Print the elements of list foreach ( int i in res) Console.Write(i + " " ); } } |
8 8 8 2 2 5 5 -1 6 9999999
Time Complexity: O(n) + O(m Log m) where n is total number of elements and m is total number of distinct elements
Auxiliary Space: O(n)
This article is contributed by Aarti_Rathi and Ankur Singh and improved by Ankur Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Simple way to sort by frequency.
The Approach:
Here In This approach we first we store the element by there frequency in vector_pair format(Using Mapping stl map) then sort it according to frequency then reverse it and apply bubble sort to make the condition true decreasing frequency if 2 numbers have the same frequency then print the one which came first. then print the vector.
C++
#include <bits/stdc++.h> #include<iostream> using namespace std; //map all the number and sort by frequency. void the_helper( int a[],vector<pair< int , int >>&res, int n){ map< int , int >mp; for ( int i=0;i<n;i++)mp[a[i]]++; for ( auto it:mp)res.push_back({it.second,it.first}); sort(res.begin(),res.end()); } int main() { int a[] = {2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8}; vector<pair< int , int >>res; the_helper(a,res,10); reverse(res.begin(),res.end()); for ( int i=0;i<res.size();i++){ if (res[i].first==res[i+1].first){ for ( int j=i;j<res.size();j++){ if (res[i].second>res[j].second&&res[i].first==res[j].first){ swap(res[i],res[j]); } } } } for ( int i=0;i<res.size();i++){ for ( int j=0;j<res[i].first;j++)cout<<res[i].second<< " " ; // cout<<endl; } return 0; } |
Java
// Java program for the above approach import java.util.*; // pair class class Pair{ int first; int second; public Pair( int first, int second){ this .first = first; this .second = second; } } public class Main{ // map all the number and sort by frequency public static void the_helper( int [] a, List<Pair> res, int n){ // create a new map to store the frequency of each number in the array Map<Integer, Integer> mp = new HashMap<>(); for ( int i = 0 ; i < n; i++){ // check if the map already has the number, // if yes, increase its count by 1, else add it to the map with count 1 if (mp.containsKey(a[i])) mp.put(a[i], mp.get(a[i])+ 1 ); else mp.put(a[i], 1 ); } // loop through the map entries and add them to the result list as pairs for (Map.Entry<Integer, Integer> entry : mp.entrySet()){ res.add( new Pair(entry.getValue(), entry.getKey())); } // sort the result list in ascending order of // frequency using a lambda expression res.sort((x, y) -> x.first - y.first); } // driver program public static void main(String[] args) { int [] a = { 2 , 5 , 2 , 6 , - 1 , 9999999 , 5 , 8 , 8 , 8 }; List<Pair> res = new ArrayList<>(); the_helper(a, res, 10 ); // reverse the result list to get it in descending order of frequency Collections.reverse(res); // loop through the result list and swap pairs with // equal frequencies if the second element of the earlier // pair is greater than the second element of the later pair for ( int i = 0 ; i < res.size()- 1 ; i++){ if (res.get(i).first == res.get(i+ 1 ).first){ for ( int j = i; j < res.size(); j++){ if (res.get(i).second > res.get(j).second && res.get(i).first == res.get(j).first){ Pair temp = res.get(j); res.set(j, res.get(i)); res.set(i, temp); } } } } System.out.println(); // loop through the result list and print each pair's //second element the number of times indicated by its first element for (Pair p : res){ for ( int i = 0 ; i < p.first; i++){ System.out.print(p.second + " " ); } } } } // this code is contributed by bhardwajji |
Python3
# python3 program for the above approach import collections # map all the number and sort by frequency def the_helper(a, res, n): mp = collections.defaultdict( int ) for i in range (n): mp[a[i]] + = 1 for key, val in mp.items(): res.append((val, key)) res.sort() # main function if __name__ = = '__main__' : a = [ 2 , 5 , 2 , 6 , - 1 , 9999999 , 5 , 8 , 8 , 8 ] res = [] the_helper(a, res, len (a)) res.reverse() for i in range ( len (res) - 1 ): if res[i][ 0 ] = = res[i + 1 ][ 0 ]: for j in range (i + 1 , len (res)): if res[i][ 0 ] = = res[j][ 0 ] and res[i][ 1 ] > res[j][ 1 ]: res[i], res[j] = res[j], res[i] for i in range ( len (res)): for j in range (res[i][ 0 ]): print (res[i][ 1 ], end = ' ' ) # print() # uncomment to print each frequency on a new line |
C#
using System; using System.Collections.Generic; using System.Linq; public class Program { //map all the number and sort by frequency. public static void the_helper( int [] a, List<Tuple< int , int >> res, int n) { Dictionary< int , int > mp = new Dictionary< int , int >(); for ( int i = 0; i < n; i++) { if (mp.ContainsKey(a[i])) { mp[a[i]]++; } else { mp[a[i]] = 1; } } foreach ( var it in mp) { res.Add( new Tuple< int , int >(it.Value, it.Key)); } res.Sort(); } public static void Main() { int [] a = { 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 }; List<Tuple< int , int >> res = new List<Tuple< int , int >>(); the_helper(a, res, 10); res.Reverse(); for ( int i = 0; i < res.Count; i++) { if (i < res.Count - 1 && res[i].Item1 == res[i+1].Item1) { for ( int j = i; j < res.Count; j++) { if (res[i].Item2 > res[j].Item2 && res[i].Item1 == res[j].Item1) { var temp = res[i]; res[i] = res[j]; res[j] = temp; } } } } for ( int i=0;i<res.Count;i++){ for ( int j = 0; j < res[i].Item1; j++) { Console.Write(res[i].Item2 + " " ); } } } } |
Javascript
// JavaScript program for the above approach // pair class class pair{ constructor(first, second){ this .first = first; this .second = second; } } // map all the number and sort by frequency function the_helper(a, res, n){ mp = new Map(); for (let i = 0; i<n; i++){ if (mp.has(a[i])) mp.set(a[i], mp.get(a[i])+1); else mp.set(a[i], 1); } mp.forEach( function (value, key){ res.push( new pair(value, key)); }) res.sort( function (a, b){ return a.first - b.first; }); } // driver program let a = [2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8]; let res = []; the_helper(a, res, 10); res.reverse(); for (let i = 0; i < res.length-1; i++){ if (res[i].first == res[i+1].first){ for (let j = i; j < res.length; j++){ if (res[i].second > res[j].second && res[i].first == res[j].first){ let temp = res[j]; res[j] = res[i]; res[i] = temp; } } } } console.log( "\n" ); for (let i = 0; i < res.length; i++){ for (let j = 0; j < res[i].first; j++){ console.log(res[i].second + " " ); } } // this code is contributed by Yash Agarwal(yashagarwal2852002) |
8 8 8 2 2 5 5 -1 6 9999999
Time Complexity: O(n^2) I.e it take O(n) for getting the frequency sorted vector but for sorting in decreasing frequency if 2 numbers have the same frequency then print the one which came first we use bubble sort so it take O(n^2).
Auxiliary Space: O(n),for vector.
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