# Sort elements by frequency | Set 2

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}.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 ` `#include ` `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), 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); ` `    ``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.

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