Sort an array in descending order based on the sum of its occurrence

Given an unsorted array of integers which may contain repeated elements, sort the elements in descending order of some of its occurrence. If there exists more than one element whose sum of occurrences are the same then, the one which is greater will come first. 
Examples:

Input: arr[] = [2, 4, 1, 2, 4, 2, 10] 
Output: arr[] = [10, 4, 4, 2, 2, 2, 1]
Explanation: 
Here, 2 appears 3 times, 4 appears 2 times, 1 appears 1 time and 10 appears 1 time. 
Thus, 
Sum of all occurrences of 2 in given array = 2 * 3 = 6, 
Sum of all occurrences of 4 = 4 * 2 = 8, 
Sum of all occurrences of 10 = 10 * 1 = 10, 
Sum of all occurrences of 1 = 1 * 1 = 1. 
Therefore sorting the array in descending order based on the above sum = [ 10, 4, 4, 2, 2, 2, 1 ]
Input2: arr[] = [2, 3, 2, 3, 2, 1, 1, 9, 6, 9, 1, 2] 
Output2: [9, 9, 2, 2, 2, 2, 6, 3, 3, 1, 1, 1] 

Approach:

1. Create a map that will map elements to it appearance ie if a occur one time then a will map to a but if a occur m times then a will be map to a*m.
2. After doing the mapping sort the dictionary in descending order based on their values not keys. In case of tie, sort based on values.
3. Finally copy the sorted dictionary keys in the final output array and there frequency is obtain based on their key-value pairs i.e. if a is mapped to a*m it means we need to include a, m times in output array.

Below is the implementation of the above approach:

C++

// c++ code for the above approach

#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>

using namespace std;

vector<int> sort_desc(vector<int>& arr) {

    unordered_map<int, int> d_sum, d_count;

    for (auto x : arr) {
        if (d_sum.find(x) == d_sum.end()) {
            d_sum[x] = x;
            d_count[x] = 1;
        }
        else {
            d_sum[x] += x;
            d_count[x] += 1;
        }
    }

    vector<pair<int, int>> l;
    for (auto& x : d_sum) {
        l.push_back({ x.second, x.first });
    }

    sort(l.rbegin(), l.rend());

    vector<int> ans;
    for (auto& x : l) {
        ans.insert(ans.end(), d_count[x.second], x.second);
    }

    return ans;
}

int main() {
    vector<int> arr{ 3, 5, 2, 2, 3, 1, 3, 1 };

    // Function Call
    vector<int> sorted_arr = sort_desc(arr);

    for (auto x : sorted_arr) {
        cout << x << " ";
    }

    return 0;
}

// This code is contributed by princekumaras

Python

# Python3 program to sort elements of
# arr[] in descending order of sum
# of its occurrence

def sort_desc(arr):

    # to store sum of all
    # occurrences of an elements
    d_sum = {}

    # to store count of
    # occurrence of elements
    d_count ={}

    # to store final result
    ans = []

    # to store maximum sum of occurrence
    mx = 0

    # Traverse and calculate sum
    # of occurrence and count of
    # occurrence of elements of arr[]
    for x in arr:
        if x not in d_sum:
            d_sum[x] = x
            d_count[x] = 1
        else:
            d_sum[x] += x
            d_count[x] += 1

    # sort d_sum in decreasing order of its value
    l = sorted(d_sum.items(),
               reverse = True,
               key = lambda x:(x[1], x[0]))

    # store the final result
    for x in l:
        ans += [x[0]] * d_count[x[0]]

    return ans

# Driver Code
arr = [3, 5, 2, 2, 3, 1, 3, 1]
print(sort_desc(arr))

C#

using System;
using System.Collections.Generic;
using System.Linq;

class GFG
{
  static List<int> SortDesc(int[] arr)
  {
    // to store sum of all
    // occurrences of an elements
    Dictionary<int, int> d_sum = new Dictionary<int, int>();

    // to store count of
    // occurrence of elements
    Dictionary<int, int> d_count = new Dictionary<int, int>();

    // to store final result
    List<int> ans = new List<int>();

    // to store maximum sum of occurrence
    int mx = 0;

    // Traverse and calculate sum
    // of occurrence and count of
    // occurrence of elements of arr[]
    for (int i = 0; i < arr.Length; i++)
    {
      int x = arr[i];
      if (d_sum.ContainsKey(x))
      {
        d_sum[x] += x;
        d_count[x]++;
      }
      else
      {
        d_sum[x] = x;
        d_count[x] = 1;
      }
    }

    // sort d_sum in decreasing order of its value
    var l = d_sum.OrderByDescending(d => d.Value).ThenBy(d => d.Key);

    // store the final result
    foreach (var kvp in l)
    {
      int x = kvp.Key;
      int count = d_count[x];
      while (count > 0)
      {
        ans.Add(x);
        count--;
      }
    }

    return ans;
  }

  // Driver code
  static void Main(string[] args)
  {
    int[] arr = { 3, 5, 2, 2, 3, 1, 3, 1 };
    Console.WriteLine(string.Join(" ", SortDesc(arr)));
  }
}


// This code is contributed by phasing17

Javascript

// JavaScript program to sort elements of
// arr[] in descending order of sum
// of its occurrence
function sort_desc(arr) {

    // to store sum of all
    // occurrences of an elements
    let d_sum = {};

    // to store count of
    // occurrence of elements
    let d_count = {};

    // to store final result
    let ans = [];

    // to store maximum sum of occurrence
    let mx = 0;

    // Traverse and calculate sum
    // of occurrence and count of
    // occurrence of elements of arr[]
    for (let i = 0; i < arr.length; i++) {
        let x = arr[i];
        if (x in d_sum) {
            d_sum[x] += x;
            d_count[x]++;
        } else {
            d_sum[x] = x;
            d_count[x] = 1;
        }
    }

    // sort d_sum in decreasing order of its value
    let l = Object.entries(d_sum).sort(function(a, b) {
        return b[1] - a[1] || a[0] - b[0];
    });

    // store the final result
    for (let i = 0; i < l.length; i++) {
        let x = l[i][0];
        let count = d_count[x];
        while (count > 0) {
            ans.push(parseInt(x));
            count--;
        }
    }

    return ans;
}

// Driver Code
let arr = [3, 5, 2, 2, 3, 1, 3, 1];
console.log(sort_desc(arr).join(' '));

Java

import java.util.*;
import java.util.stream.*;

class GFG {

    static List<Integer> sortDesc(int[] arr)
    {
        // to store sum of all
        // occurrences of an elements
        Map<Integer, Integer> d_sum = new HashMap<>();

        // to store count of
        // occurrence of elements
        Map<Integer, Integer> d_count = new HashMap<>();

        // to store final result
        List<Integer> ans = new ArrayList<>();

        // to store maximum sum of occurrence
        int mx = 0;

        // Traverse and calculate sum
        // of occurrence and count of
        // occurrence of elements of arr[]
        for (int i = 0; i < arr.length; i++) {
            int x = arr[i];
            if (d_sum.containsKey(x)) {
                d_sum.put(x, d_sum.get(x) + x);
                d_count.put(x, d_count.get(x) + 1);
            }
            else {
                d_sum.put(x, x);
                d_count.put(x, 1);
            }
        }

        // sort d_sum in decreasing order of its value
        List<Map.Entry<Integer, Integer> > l
            = new ArrayList<>(d_sum.entrySet());
        Collections.sort(l, (e1, e2) -> {
            int cmp
                = e2.getValue().compareTo(e1.getValue());
            if (cmp == 0) {
                cmp = e1.getKey().compareTo(e2.getKey());
            }
            return cmp;
        });

        // store the final result
        for (Map.Entry<Integer, Integer> e : l) {
            int x = e.getKey();
            int count = d_count.get(x);
            while (count > 0) {
                ans.add(x);
                count--;
            }
        }

        return ans;
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 3, 5, 2, 2, 3, 1, 3, 1 };
        List<Integer> sortedList = sortDesc(arr);
        System.out.println(
            sortedList.stream()
                .map(Object::toString)
                .collect(Collectors.joining(" ")));
    }
}

Output:

[3, 3, 3, 5, 2, 2, 1, 1]

Time Complexity: O(N*log N). 
Space Complexity: O(N).