Given an array of integers, sort the array according to frequency of elements. For example, if the input array is {2, 3, 2, 4, 5, 12, 2, 3, 3, 3, 12}, then modify the array to {3, 3, 3, 3, 2, 2, 2, 12, 12, 4, 5}.

In the previous post, we have discussed all methods for sorting according to frequency. In this post, method 2 is discussed in detail and C++ implementation for the same is provided.

Following is detailed algorithm.

- Create a BST and while creating BST maintain the count i,e frequency of each coming element in same BST. This step may take O(nLogn) time if a self balancing BST is used.
- Do Inorder traversal of BST and store every element and count of each element in an auxiliary array. Let us call the auxiliary array as ‘count[]’. Note that every element of this array is element and frequency pair. This step takes O(n) time.
- Sort ‘count[]’ according to frequency of the elements. This step takes O(nLohn) time if a O(nLogn) sorting algorithm is used.
- Traverse through the sorted array ‘count[]’. For each element x, print it ‘freq’ times where ‘freq’ is frequency of x. This step takes O(n) time.

The overall time complexity of the algorithm can be minimum **O(nLogn)** if we use a **O(nLogn)** sorting algorithm and use a self-balancing BST with **O(Logn)** insert operation.

Following is the implementation of the above algorithm.

`// Implementation of above algorithm in C++.` `#include <iostream>` `#include <stdlib.h>` `using` `namespace` `std;` ` ` `/* A BST node has data, freq, left and right pointers */` `struct` `BSTNode` `{` ` ` `struct` `BSTNode *left;` ` ` `int` `data;` ` ` `int` `freq;` ` ` `struct` `BSTNode *right;` `};` ` ` `// A structure to store data and its frequency` `struct` `dataFreq` `{` ` ` `int` `data;` ` ` `int` `freq;` `};` ` ` `/* Function for qsort() implementation. Compare frequencies to` ` ` `sort the array according to decreasing order of frequency */` `int` `compare(` `const` `void` `*a, ` `const` `void` `*b)` `{` ` ` `return` `( (*(` `const` `dataFreq*)b).freq - (*(` `const` `dataFreq*)a).freq );` `}` ` ` `/* Helper function that allocates a new node with the given data,` ` ` `frequency as 1 and NULL left and right pointers.*/` `BSTNode* newNode(` `int` `data)` `{` ` ` `struct` `BSTNode* node = ` `new` `BSTNode;` ` ` `node->data = data;` ` ` `node->left = NULL;` ` ` `node->right = NULL;` ` ` `node->freq = 1;` ` ` `return` `(node);` `}` ` ` `// A utility function to insert a given key to BST. If element` `// is already present, then increases frequency` `BSTNode *insert(BSTNode *root, ` `int` `data)` `{` ` ` `if` `(root == NULL)` ` ` `return` `newNode(data);` ` ` `if` `(data == root->data) ` `// If already present` ` ` `root->freq += 1;` ` ` `else` `if` `(data < root->data)` ` ` `root->left = insert(root->left, data);` ` ` `else` ` ` `root->right = insert(root->right, data);` ` ` `return` `root;` `}` ` ` `// Function to copy elements and their frequencies to count[].` `void` `store(BSTNode *root, dataFreq count[], ` `int` `*index)` `{` ` ` `// Base Case` ` ` `if` `(root == NULL) ` `return` `;` ` ` ` ` `// Recur for left substree` ` ` `store(root->left, count, index);` ` ` ` ` `// Store item from root and increment index` ` ` `count[(*index)].freq = root->freq;` ` ` `count[(*index)].data = root->data;` ` ` `(*index)++;` ` ` ` ` `// Recur for right subtree` ` ` `store(root->right, count, index);` `}` ` ` `// The main function that takes an input array as an argument` `// and sorts the array items according to frequency` `void` `sortByFrequency(` `int` `arr[], ` `int` `n)` `{` ` ` `// Create an empty BST and insert all array items in BST` ` ` `struct` `BSTNode *root = NULL;` ` ` `for` `(` `int` `i = 0; i < n; ++i)` ` ` `root = insert(root, arr[i]);` ` ` ` ` `// Create an auxiliary array 'count[]' to store data and` ` ` `// frequency pairs. The maximum size of this array would` ` ` `// be n when all elements are different` ` ` `dataFreq count[n];` ` ` `int` `index = 0;` ` ` `store(root, count, &index);` ` ` ` ` `// Sort the count[] array according to frequency (or count)` ` ` `qsort` `(count, index, ` `sizeof` `(count[0]), compare);` ` ` ` ` `// Finally, traverse the sorted count[] array and copy the` ` ` `// i'th item 'freq' times to original array 'arr[]'` ` ` `int` `j = 0;` ` ` `for` `(` `int` `i = 0; i < index; i++)` ` ` `{` ` ` `for` `(` `int` `freq = count[i].freq; freq > 0; freq--)` ` ` `arr[j++] = count[i].data;` ` ` `}` `}` ` ` `// A utility function to print an array of size n` `void` `printArray(` `int` `arr[], ` `int` `n)` `{` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `cout << arr[i] << ` `" "` `;` ` ` `cout << endl;` `}` ` ` `/* Driver program to test above functions */` `int` `main()` `{` ` ` `int` `arr[] = {2, 3, 2, 4, 5, 12, 2, 3, 3, 3, 12};` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `sortByFrequency(arr, n);` ` ` `printArray(arr, n);` ` ` `return` `0;` `}` |

**Output**:

3 3 3 3 2 2 2 12 12 5 4

**Exercise:**

The above implementation doesn’t guarantee original order of elements with same frequency (for example, 4 comes before 5 in input, but 4 comes after 5 in output). Extend the implementation to maintain original order. For example, if two elements have same frequency then print the one which came 1st in input array.

This article is compiled by **Chandra Prakash**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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