Sort elements by frequency | Set 2 | GeeksforGeeks

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}.

1) 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.
2) 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.
3) Sort ‘count[]‘ according to frequency of the elements. This step takes O(nLohn) time if a O(nLogn) sorting algorithm is used.
4) 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.

struct BSTNode

{

    struct BSTNode *left;

    int data;

    int freq;

    struct BSTNode *right;

};

struct dataFreq

{

    int data;

    int freq;

};

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 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 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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