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++ 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 |
# 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))
|
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 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( ' ' ));
|
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( " " )));
}
} |
[3, 3, 3, 5, 2, 2, 1, 1]
Time Complexity: O(N*log N).
Space Complexity: O(N).