Sort an array in descending order based on the sum of its occurrence
Last Updated :
27 Mar, 2023
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:
- 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.
- After doing the mapping sort the dictionary in descending order based on their values not keys. In case of tie, sort based on values.
- 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++
#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 };
vector< int > sorted_arr = sort_desc(arr);
for ( auto x : sorted_arr) {
cout << x << " " ;
}
return 0;
}
|
Python
def sort_desc(arr):
d_sum = {}
d_count = {}
ans = []
mx = 0
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
l = sorted (d_sum.items(),
reverse = True ,
key = lambda x:(x[ 1 ], x[ 0 ]))
for x in l:
ans + = [x[ 0 ]] * d_count[x[ 0 ]]
return ans
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)
{
Dictionary< int , int > d_sum = new Dictionary< int , int >();
Dictionary< int , int > d_count = new Dictionary< int , int >();
List< int > ans = new List< int >();
int mx = 0;
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;
}
}
var l = d_sum.OrderByDescending(d => d.Value).ThenBy(d => d.Key);
foreach ( var kvp in l)
{
int x = kvp.Key;
int count = d_count[x];
while (count > 0)
{
ans.Add(x);
count--;
}
}
return ans;
}
static void Main( string [] args)
{
int [] arr = { 3, 5, 2, 2, 3, 1, 3, 1 };
Console.WriteLine( string .Join( " " , SortDesc(arr)));
}
}
|
Javascript
function sort_desc(arr) {
let d_sum = {};
let d_count = {};
let ans = [];
let mx = 0;
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;
}
}
let l = Object.entries(d_sum).sort( function (a, b) {
return b[1] - a[1] || a[0] - b[0];
});
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;
}
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)
{
Map<Integer, Integer> d_sum = new HashMap<>();
Map<Integer, Integer> d_count = new HashMap<>();
List<Integer> ans = new ArrayList<>();
int mx = 0 ;
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 );
}
}
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;
});
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;
}
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).
Share your thoughts in the comments
Please Login to comment...