# Sorting element of an array by frequency in decreasing order

Given an array arr[] of N integers. The task is to sort the array arr[] according to the frequency of elements in decreasing order.
Note: If the frequencies of the two elements are the same, then the smaller element should come first.

Examples:

Input: arr[] = { 4, 4, 5, 6, 4, 2, 2, 8, 5 }
Output: 4 4 4 2 2 5 5 6 8

Input: arr[] = { 9, 9, 5, 8, 5 }
Output: 5 5 9 9 8

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. Store the frequency of all element in array arr[].
```For arr[] = { 4, 4, 5, 6, 4, 2, 2, 8, 5 }
The frequency array of the above array is:
freq[] = { 0, 0, 2, 0, 3, 2, 0, 1, 0, 1}
```
2. Traverse the frequency array and for all the element having frequency greater than 1, update the value in the array arr[] as:

freq = 2
arr = 100000*freq + (100000 – 2) = 299998

freq = 3
arr = 100000*freq + (100000 – 4) = 399996

freq = 2
arr = 100000*freq + (100000 – 5) = 299995

freq = 2
arr = 100000*freq + (100000 – 6) = 199994

freq = 2
arr = 100000*freq + (100000 – 2) = 199994

Now array becomes:
arr[] = {299998, 399996, 299995, 199994, 199992, 2, 2, 8, 5}

3. Sort the array arr[] in decreasing order.
4. Traverse the array arr[] and to get the element and frequency of that element as per the updation of array element in step 2, do the following:

To get the frequency of current element:
frequency = arr[i]/100000;
To get the value:
value = 100000 – arr[i]%100000

For Example:

if arr[i] = 399996
frequency = arr[i]/100000 = 399996/100000 = 3
value = 100000 – arr[i]%100000 = 100000 – 99996 = 4
The element 4 is having frequency 3.

5. For each element in arr[] find the value and frequency(say f) at each index and print the value f number of times.

Below is the implementation of the above approach:

## C++

 `// C++ program to sort an array in ` `// decreasing order of their frequnecy ` `#include "bits/stdc++.h" ` `using` `namespace` `std; ` ` `  `// Function that return the index ` `// upto all the array elements are ` `// updated. ` `int` `sortByFreq(``int``* arr, ``int` `n) ` `{ ` ` `  `    ``// Initialise maxE = -1 ` `    ``int` `maxE = -1; ` ` `  `    ``// Find the maximum element of ` `    ``// arr[] ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``maxE = max(maxE, arr[i]); ` `    ``} ` ` `  `    ``// Create frequency array freq[] ` `    ``int` `freq[maxE + 1] = { 0 }; ` ` `  `    ``// Update the frequency array as ` `    ``// per the occurence of element in ` `    ``// arr[] ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``freq[arr[i]]++; ` `    ``} ` ` `  `    ``// Initialise cnt to 0 ` `    ``int` `cnt = 0; ` ` `  `    ``// Traversing freq[] ` `    ``for` `(``int` `i = 0; i <= maxE; i++) { ` ` `  `        ``// If freq of an element is ` `        ``// greater than 0 update the ` `        ``// value of arr[] at index cnt ` `        ``// & increment cnt ` `        ``if` `(freq[i] > 0) { ` ` `  `            ``int` `value = 100000 - i; ` `            ``arr[cnt] = 100000 * freq[i] + value; ` `            ``cnt++; ` `        ``} ` `    ``} ` ` `  `    ``// Return cnt ` `    ``return` `cnt; ` `} ` ` `  `// Function that print array arr[] ` `// elements in sorted order ` `void` `printSortedArray(``int``* arr, ``int` `cnt) ` `{ ` ` `  `    ``// Traversing arr[] till index cnt ` `    ``for` `(``int` `i = 0; i < cnt; i++) { ` ` `  `        ``// Find frequency of elements ` `        ``int` `frequency = arr[i] / 100000; ` ` `  `        ``// Find value at index i ` `        ``int` `value = 100000 - (arr[i] % 100000); ` ` `  `        ``// Traversing till frequency ` `        ``// to print value at index i ` `        ``for` `(``int` `j = 0; j < frequency; j++) { ` `            ``cout << value << ``' '``; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `arr[] = { 4, 4, 5, 6, 4, 2, 2, 8, 5 }; ` ` `  `    ``// Size of array arr[] ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// Function call to get cnt ` `    ``int` `cnt = sortByFreq(arr, n); ` ` `  `    ``// Sort the arr[] in decreasing order ` `    ``sort(arr, arr + cnt, greater<``int``>()); ` ` `  `    ``// Function that prints elements ` `    ``// in decreasing order ` `    ``printSortedArray(arr, cnt); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to sort an array in ` `// decreasing order of their frequnecy ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function that return the index ` `// upto all the array elements are ` `// updated. ` `static` `int` `sortByFreq(Integer []arr, ``int` `n) ` `{ ` `  `  `    ``// Initialise maxE = -1 ` `    ``int` `maxE = -``1``; ` `  `  `    ``// Find the maximum element of ` `    ``// arr[] ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``maxE = Math.max(maxE, arr[i]); ` `    ``} ` `  `  `    ``// Create frequency array freq[] ` `    ``int` `freq[] = ``new` `int``[maxE + ``1``]; ` `  `  `    ``// Update the frequency array as ` `    ``// per the occurence of element in ` `    ``// arr[] ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``freq[arr[i]]++; ` `    ``} ` `  `  `    ``// Initialise cnt to 0 ` `    ``int` `cnt = ``0``; ` `  `  `    ``// Traversing freq[] ` `    ``for` `(``int` `i = ``0``; i <= maxE; i++) { ` `  `  `        ``// If freq of an element is ` `        ``// greater than 0 update the ` `        ``// value of arr[] at index cnt ` `        ``// & increment cnt ` `        ``if` `(freq[i] > ``0``) { ` `  `  `            ``int` `value = ``100000` `- i; ` `            ``arr[cnt] = ``100000` `* freq[i] + value; ` `            ``cnt++; ` `        ``} ` `    ``} ` `  `  `    ``// Return cnt ` `    ``return` `cnt; ` `} ` `  `  `// Function that print array arr[] ` `// elements in sorted order ` `static` `void` `printSortedArray(Integer []arr, ``int` `cnt) ` `{ ` `  `  `    ``// Traversing arr[] till index cnt ` `    ``for` `(``int` `i = ``0``; i < cnt; i++) { ` `  `  `        ``// Find frequency of elements ` `        ``int` `frequency = arr[i] / ``100000``; ` `  `  `        ``// Find value at index i ` `        ``int` `value = ``100000` `- (arr[i] % ``100000``); ` `  `  `        ``// Traversing till frequency ` `        ``// to print value at index i ` `        ``for` `(``int` `j = ``0``; j < frequency; j++) { ` `            ``System.out.print(value + ``" "``); ` `        ``} ` `    ``} ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `  `  `    ``Integer arr[] = { ``4``, ``4``, ``5``, ``6``, ``4``, ``2``, ``2``, ``8``, ``5` `}; ` `  `  `    ``// Size of array arr[] ` `    ``int` `n = arr.length; ` `  `  `    ``// Function call to get cnt ` `    ``int` `cnt = sortByFreq(arr, n); ` `  `  `    ``// Sort the arr[] in decreasing order ` `    ``Arrays.sort(arr, Collections.reverseOrder()); ` `  `  `    ``// Function that prints elements ` `    ``// in decreasing order ` `    ``printSortedArray(arr, cnt); ` `  `  `} ` `} ` ` `  `// This code is contributed by sapnasingh4991 `

## Python3

 `# Python program to sort an array in ` `# decreasing order of their frequnecy ` ` `  `# Function that return the index ` `# upto all the array elements are ` `# updated. ` `def` `sortByFreq(arr, n): ` ` `  `    ``# Initialise maxE = -1 ` `    ``maxE ``=` `-``1``; ` ` `  `    ``# Find the maximum element of ` `    ``# arr[] ` `    ``for` `i ``in` `range``(n): ` `        ``maxE ``=` `max``(maxE, arr[i]) ` ` `  `    ``# Create frequency array freq[] ` `    ``freq ``=` `[``0``]``*``(maxE ``+` `1``); ` ` `  `    ``# Update the frequency array as ` `    ``# per the occurence of element in ` `    ``# arr[] ` `    ``for` `i ``in` `range``(n): ` `        ``freq[arr[i]] ``+``=` `1``; ` `     `  `    ``# Initialise cnt to 0 ` `    ``cnt ``=` `0``; ` ` `  `    ``# Traversing freq[] ` `    ``for` `i ``in` `range``(maxE``+``1``): ` ` `  `        ``# If freq of an element is ` `        ``# greater than 0 update the ` `        ``# value of arr[] at index cnt ` `        ``# & increment cnt ` `        ``if` `(freq[i] > ``0``): ` ` `  `            ``value ``=` `100000` `-` `i; ` `            ``arr[cnt] ``=` `100000` `*` `freq[i] ``+` `value; ` `            ``cnt ``+``=` `1``; ` `         `  `    ``# Return cnt ` `    ``return` `cnt; ` ` `  `# Function that print array arr[] ` `# elements in sorted order ` `def` `printSortedArray(arr, cnt): ` ` `  `    ``# Traversing arr[] till index cnt ` `    ``for` `i ``in` `range``(cnt): ` ` `  `        ``# Find frequency of elements ` `        ``frequency ``=` `arr[i] ``/` `100000``; ` ` `  `        ``# Find value at index i ` `        ``value ``=` `100000` `-` `(arr[i] ``%` `100000``); ` ` `  `        ``# Traversing till frequency ` `        ``# to print value at index i ` `        ``for` `j ``in` `range``(``int``(frequency)): ` `            ``print``(value, end``=``" "``) ` `             `  `# Driver code ` `if` `__name__``=``=``'__main__'``:  ` ` `  `    ``arr ``=` `[ ``4``, ``4``, ``5``, ``6``, ``4``, ``2``, ``2``, ``8``, ``5` `] ` ` `  `    ``# Size of array arr[] ` `    ``n ``=` `len``(arr) ` ` `  `    ``# Function call to get cnt ` `    ``cnt ``=` `sortByFreq(arr, n); ` ` `  `    ``# Sort the arr[] in decreasing order ` `    ``arr.sort(reverse ``=` `True``) ` ` `  `    ``# Function that prints elements ` `    ``# in decreasing order ` `    ``printSortedArray(arr, cnt); ` ` `  `# This code is contributed by Princi Singh `

Output:

```4 4 4 2 2 5 5 6 8
```

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.