Sort Array according to modulus of their values with their frequencies
Given an array arr containing N positive integers, sort them according to the increasing modulus of their values with their frequencies.
Example:
Input: arr[]={1, 1, 5, 3, 2, 3, 3, 3, 4, 5, 4, 5}
Output: 2 4 4 1 1 5 5 5 3 3 3 3
Explanation:
The elements are sorted in the following order:
2 % frequency(2) = 2 % 1 = 0
4 % frequency(4) = 4 % 2 = 0
1 % frequency(1) = 1 % 2 = 1
5 % frequency(5) = 5 % 3 = 2
3 % frequency(4) = 3 % 4 = 3Input: arr[]={2, 9, 8, 2, 8, 9, 2, 8, 5}
Output: 5 9 9 2 2 2 8 8 8
Approach: To solve this question, store the frequencies of each number in a map and then create a custom comparator, which will do the sorting. That compare function will accept two integer values, say A and B as parameters and the condition in that comparator to sort the array is:
(A % frequency(A)) > (B % frequency(B))
Algorithm:
- Create an unordered map to store the frequencies of each number in the array.
- Sort the array using a custom comparator that compares the modulus of the values of two elements in the array with their respective frequencies.
- If the modulus is the same, compare the values of the elements and return true if the first element is smaller than the second element.
- If the modulus is not the same, compare the modulus of the two elements and return true if the modulus of the first element is smaller than the modulus of the second element.
- Print the sorted array.
Pseudocode:
function customSort(array): freq_map = empty unordered map for element in array: increment the frequency of element in freq_map sort array using a custom comparator: if modulus of A with freq[A] is equal to modulus of B with freq[B]: return A < B else: return modulus of A with freq[A] < modulus of B with freq[B] print the sorted array
Below is the implementation of the above approach.
C++
// C++ code for the above approach #include <bits/stdc++.h> using namespace std; // Function to sort the numbers in an array // according to the modulus of their values // with their frequencies void customSort(vector< int >& arr) { // Map to store // the frequencies of each number unordered_map< int , int > freq; for ( auto x : arr) { freq[x]++; } // Sorting them using // a custom comparator sort(arr.begin(), arr.end(), [&]( int A, int B) { if (A % freq[A] == B % freq[B]) { return A < B; } return A % freq[A] < B % freq[B]; }); // Printing the numbers for ( auto x : arr) { cout << x << ' ' ; } } // Driver Code int main() { vector< int > arr = { 2, 9, 8, 2, 8, 9, 2, 8, 5 }; customSort(arr); } |
Java
// Java code for the above approach import java.util.*; class GFG{ // Function to sort the numbers in an array // according to the modulus of their values // with their frequencies static void customSort(Integer []arr) { // Map to store // the frequencies of each number HashMap<Integer,Integer> freq = new HashMap<Integer,Integer>(); for ( int x : arr) { if (freq.containsKey(x)){ freq.put(x, freq.get(x)+ 1 ); } else { freq.put(x, 1 ); } } // Sorting them using // a custom comparator Arrays.sort(arr, new Comparator<Integer>(){ @Override public int compare(Integer a, Integer b) { // If both are odd or even // then sorting in increasing order if ((a % freq.get(a)) == (b % freq.get(b))) { return a < b?- 1 : 1 ; } // Sorting on the basis of last bit if // if one is odd and the other one is even return ((a % freq.get(a)) < (b % freq.get(b)))?- 1 : 1 ; } }); // Printing the numbers for ( int x : arr) { System.out.print(x + " " ); } } // Driver Code public static void main(String[] args) { Integer []arr = { 2 , 9 , 8 , 2 , 8 , 9 , 2 , 8 , 5 }; customSort(arr); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 code for the above approach # Function to sort the numbers in an array # according to the modulus of their values # with their frequencies from collections import OrderedDict from filecmp import cmp from functools import cmp_to_key freq = {} def custom_sort(A, B): global freq if (A % freq[A] = = B % freq[B]): return A - B return ((A % freq[A]) - (B % freq[B])) def arr_sort(a,b): return a[ 0 ] - b[ 0 ] def customSort(arr): global freq # Map to store # the frequencies of each number for x in arr : if x in freq: freq[x] = freq[x] + 1 else : freq[x] = 1 freq = OrderedDict( sorted (freq.items(),key = cmp_to_key(arr_sort))) # Sorting them using # a custom comparator arr.sort(key = cmp_to_key(custom_sort)) # Printing the numbers for x in arr: print (x,end = ' ' ) # Driver Code arr = [ 2 , 9 , 8 , 2 , 8 , 9 , 2 , 8 , 5 ] customSort(arr) # This code is contributed by shinjanpatra |
C#
// C# code for the above approach using System; using System.Collections.Generic; class GFG { // Function to sort the numbers in an array // according to the modulus of their values // with their frequencies static void customSort( int [] arr) { // Map to store // the frequencies of each number Dictionary< int , int > freq = new Dictionary< int , int >(); foreach ( int x in arr) { if (freq.ContainsKey(x)) { freq[x] += 1; } else { freq[x] = 1; } } // Sorting them using // a custom comparator Array.Sort(arr, (a, b) => { // If both are odd or even // then sorting in increasing order if ((a % freq[a]) == (b % freq[b])) { return a < b ? -1 : 1; } // Sorting on the basis of last bit if // if one is odd and the other one is even return ((a % freq[a]) < (b % freq[b])) ? -1 : 1; }); // Printing the numbers foreach ( int x in arr) { Console.Write(x + " " ); } } // Driver Code public static void Main() { int [] arr = { 2, 9, 8, 2, 8, 9, 2, 8, 5 }; customSort(arr); } } // This code is contributed by Saurabh Jaiswal |
Javascript
<script> // Javascript code for the above approach // Function to sort the numbers in an array // according to the modulus of their values // with their frequencies function customSort(arr) { // Map to store // the frequencies of each number let freq = new Map(); for (x of arr) { if (freq.has(x)) { freq.set(x, freq.get(x) + 1) } else { freq.set(x, 1) } } freq = new Map([...freq].sort((a, b) => a[0] - b[0])); // Sorting them using // a custom comparator arr.sort((A, B) => { if (A % freq.get(A) == B % freq.get(B)) { return A - B; } return ((A % freq.get(A)) - (B % freq.get(B))); }); // Printing the numbers for (x of arr) { document.write(x + ' ' ); } } // Driver Code let arr = [2, 9, 8, 2, 8, 9, 2, 8, 5]; customSort(arr); // This code is contributed by gfgking. </script> |
5 9 9 2 2 2 8 8 8
Time Complexity: O(N*log2N) as sorting takes N*log2N time.
Auxiliary Space: O(N)
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