Count smaller elements on right side | GeeksforGeeks

Write a function to count number of smaller elements on right of each element in an array. Given an unsorted array arr[] of distinct integers, construct another array countSmaller[] such that countSmaller[i] contains count of smaller elements on right side of each element arr[i] in array.
Examples:

Input:   arr[] =  {12, 1, 2, 3, 0, 11, 4}  

Output:  countSmaller[]  =  {6, 1, 1, 1, 0, 1, 0} 

Method 2 (Use Self Balancing BST)
A Self Balancing Binary Search Tree (AVL, Red Black,.. etc) can be used to get the solution in O(nLogn) time complexity. We can augment these trees so that every node N contains size the subtree rooted with N. We have used AVL tree in the following implementation.

We traverse the array from right to left and insert all elements one by one in an AVL tree. While inserting a new key in an AVL tree, we first compare the key with root. If key is greater than root, then it is greater than all the nodes in left subtree of root. So we add the size of left subtree to the count of smaller element for the key being inserted. We recursively follow the same approach for all nodes down the root.

// Inserts a new key to the tree rotted with node. Also, updates *count

// to contain count of smaller elements for the new key

struct node* insert(struct node* node, int key, int *count)

{

    /* 1.  Perform the normal BST rotation */

    if (node == NULL)

        return(newNode(key));

    if (key < node->key)

        node->left  = insert(node->left, key, count);

    else

    {

        node->right = insert(node->right, key, count);

        // UPDATE COUNT OF SMALLER ELEMENTS FOR KEY

        *count = *count + size(node->left) + 1;

    }

    /* 2. Update height and size of this ancestor node */

    node->height = max(height(node->left), height(node->right)) + 1;

    node->size   = size(node->left) + size(node->right) + 1;

    /* 3. Get the balance factor of this ancestor node to check whether

       this node became unbalanced */

    int balance = getBalance(node);

    // If this node becomes unbalanced, then there are 4 cases

    // Left Left Case

    if (balance > 1 && key < node->left->key)

        return rightRotate(node);

    // Right Right Case

    if (balance < -1 && key > node->right->key)

        return leftRotate(node);

    // Left Right Case

    if (balance > 1 && key > node->left->key)

    {

        node->left =  leftRotate(node->left);

        return rightRotate(node);

    }

    // Right Left Case

    if (balance < -1 && key < node->right->key)

    {

        node->right = rightRotate(node->right);

        return leftRotate(node);

    }

    /* return the (unchanged) node pointer */

    return node;

}

// The following function updates the countSmaller array to contain count of

// smaller elements on right side.

void constructLowerArray (int arr[], int countSmaller[], int n)

{

  int i, j;

  struct node *root = NULL;

  // initialize all the counts in countSmaller array as 0

  for  (i = 0; i < n; i++)

     countSmaller[i] = 0;

  // Starting from rightmost element, insert all elements one by one in

  // an AVL tree and get the count of smaller elements

  for (i = n-1; i >= 0; i--)

  {

     root = insert(root, arr[i], &countSmaller[i]);

  }

}

http://ideone.com/QtAjCa
  static class Node{
    int data;
    Node left;
    Node right;
    int count;
    int smallerCount;
    public Node(int data){
      this.data = data;
    }
  }

  public static void main(String[] args){
    int[] arr = new int[]{12, 1, 2, 3, 0, 11, 4};
    System.out.println("Input array");
    for(int i: arr)
      System.out.print(i+",");

    Node dummyNode = new Node(-Integer.MAX_VALUE);
    int[] countSmaller = new int[arr.length];
    for(int i=arr.length-1;i>=0;i--){
      Node node = new Node(arr[i]);
      push(node,dummyNode);
      countSmaller[i] = node.smallerCount;
    }

    System.out.println("\nOutput array");
    for(int i: countSmaller)
      System.out.print(i+",");
    System.out.println("");
  }

  private static void push(Node node, Node head){
    if(head != null){
      if(head.data < node.data){
        head.count++;
        //System.out.println("head.data="+head.data);
        node.smallerCount += (head.left == null)?1:(head.left.count+2);
        if(head.right == null){
          node.smallerCount--;  //to account for dummy node;
          //System.out.println(node.data+" - number of smaller elements on right >>"+node.smallerCount);
          head.right = node;
        } else{
          push(node, head.right);
        }
      } else if(head.data > node.data){
        head.count++;
        if(head.left == null){
          head.left = node;
          node.smallerCount--;  //to account for dummy node;
          //System.out.println(node.data+" - number of smaller elements on right >>"+node.smallerCount);
        } else{
          push(node,head.left);
        }
      }
    }
  }
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