Sort elements by frequency | Set 2

• Difficulty Level : Hard
• Last Updated : 01 Jul, 2019

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 frequencystruct 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 frequencyBSTNode *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 frequencyvoid 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 nvoid 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.