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Count pairs in BST with sum greater than K

  • Last Updated : 28 Sep, 2021
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Given a binary search tree containing N distinct nodes and a value K. The task is to count pairs in the given binary search tree whose sum is greater than the given value K.

Examples:  

Input:  
        5      
       / \      
      3   7      
     / \ / \  
    2  4 6  8   

     k = 11
Output: 6
Explanation:
There are 6 pairs which are (4, 8), (5, 7), (5, 8), (6, 7), (6, 8) and (7, 8).

Input: 
 
         8      
        / \      
       3   9      
       \   / \  
        5 6  18   

       k = 23
Output: 3
Explanation:
There are 3 pairs which are (6, 18), (8, 18) and (9, 18).

Naive Approach:
To solve the problem mentioned above we have to store inorder traversal of BST in an array then run two loops to generate all pairs and one by one check if the current pair’s sum is greater than k or not.

Efficient Approach: 
The above method can be optimized if we store the inorder traversal of BST in an array and take the initial and last index of the array in l and r variable to find the total pair in the inorder array. Initially assign l as 0 and r as n-1. Consider a variable and initialize it to zero. This variable result will be our final answer. Now iterate until l < r and if the current left and current right have a sum greater than K, all elements from l+1 to r form a pair with it otherwise it doesn’t, therefore, increment current left. Finally, return the result.

Below is the implementation of the above approach:  



C++




// C++ program to Count
// pair in BST whose Sum
// is greater than K
 
#include <bits/stdc++.h>
using namespace std;
 
// Structure of each node of BST
struct node {
    int key;
    struct node *left, *right;
};
 
// Function to create a new BST node
node* newNode(int item)
{
    node* temp = new node();
 
    temp->key = item;
    temp->left = temp->right = NULL;
 
    return temp;
}
 
/* Function to insert a new
node with given key in BST */
struct node* insert(struct node* node, int key)
{
 
    // check if the tree is empty
    if (node == NULL)
        return newNode(key);
 
    if (key < node->key)
 
        node->left = insert(node->left, key);
 
    else if (key > node->key)
 
        node->right = insert(node->right, key);
 
    /* return the (unchanged) node pointer */
    return node;
}
 
// Function to return the size of the tree
int sizeOfTree(node* root)
{
    if (root == NULL) {
        return 0;
    }
 
    // Calculate left size recursively
    int left = sizeOfTree(root->left);
 
    // Calculate right size recursively
    int right = sizeOfTree(root->right);
 
    // Return total size recursively
    return (left + right + 1);
}
 
// Function to store inorder traversal of BST
void storeInorder(node* root, int inOrder[],
                  int& index)
{
 
    // Base condition
    if (root == NULL) {
        return;
    }
 
    // Left recursive call
    storeInorder(root->left, inOrder, index);
 
    // Store elements in inorder array
    inOrder[index++] = root->key;
 
    // Right recursive call
    storeInorder(root->right, inOrder, index);
}
 
// function to count the pair of BST
// whose sum is greater than k
int countPairUtil(int inOrder[], int j, int k)
{
    int i = 0;
    int pair = 0;
    while (i < j) {
 
        // check if sum of value at index
        // i and j is greater than k
        if (inOrder[i] + inOrder[j] > k) {
            pair += j - i;
 
            j--;
        }
        else {
            i++;
        }
    }
 
    // Return number of total pair
    return pair;
}
 
// Function to count the
// pair of BST whose sum is
// greater than k
int countPair(node* root, int k)
{
 
    // Store the size of BST
    int numNode = sizeOfTree(root);
 
    // Auxiliary array for storing
    // the inorder traversal of BST
    int inOrder[numNode + 1];
 
    int index = 0;
 
    storeInorder(root, inOrder, index);
 
    // Function call to count the pair
    return countPairUtil(inOrder, index - 1, k);
}
 
// Driver code
int main()
{
 
    // create tree
    struct node* root = NULL;
    root = insert(root, 5);
    insert(root, 3);
    insert(root, 2);
    insert(root, 4);
    insert(root, 7);
    insert(root, 6);
    insert(root, 8);
 
    int k = 11;
 
    // Print the number of pair
    cout << countPair(root, k);
 
    return 0;
}

Java




// Java program to Count
// pair in BST whose Sum
// is greater than K
class GFG{
  
// Structure of each node of BST
static class node {
    int key;
    node left, right;
};
static int index;
 
// Function to create a new BST node
static node newNode(int item)
{
    node temp = new node();
  
    temp.key = item;
    temp.left = temp.right = null;
  
    return temp;
}
  
/* Function to insert a new
node with given key in BST */
static node insert(node node, int key)
{
  
    // check if the tree is empty
    if (node == null)
        return newNode(key);
  
    if (key < node.key)
  
        node.left = insert(node.left, key);
  
    else if (key > node.key)
  
        node.right = insert(node.right, key);
  
    /* return the (unchanged) node pointer */
    return node;
}
  
// Function to return the size of the tree
static int sizeOfTree(node root)
{
    if (root == null) {
        return 0;
    }
  
    // Calculate left size recursively
    int left = sizeOfTree(root.left);
  
    // Calculate right size recursively
    int right = sizeOfTree(root.right);
  
    // Return total size recursively
    return (left + right + 1);
}
  
// Function to store inorder traversal of BST
static void storeInorder(node root, int inOrder[])
{
  
    // Base condition
    if (root == null) {
        return;
    }
  
    // Left recursive call
    storeInorder(root.left, inOrder);
  
    // Store elements in inorder array
    inOrder[index++] = root.key;
  
    // Right recursive call
    storeInorder(root.right, inOrder);
}
  
// function to count the pair of BST
// whose sum is greater than k
static int countPairUtil(int inOrder[], int j, int k)
{
    int i = 0;
    int pair = 0;
    while (i < j) {
  
        // check if sum of value at index
        // i and j is greater than k
        if (inOrder[i] + inOrder[j] > k) {
            pair += j - i;
  
            j--;
        }
        else {
            i++;
        }
    }
  
    // Return number of total pair
    return pair;
}
  
// Function to count the
// pair of BST whose sum is
// greater than k
static int countPair(node root, int k)
{
  
    // Store the size of BST
    int numNode = sizeOfTree(root);
  
    // Auxiliary array for storing
    // the inorder traversal of BST
    int []inOrder = new int[numNode + 1];
  
    index = 0;
  
    storeInorder(root, inOrder);
  
    // Function call to count the pair
    return countPairUtil(inOrder, index - 1, k);
}
  
// Driver code
public static void main(String[] args)
{
  
    // create tree
    node root = null;
    root = insert(root, 5);
    insert(root, 3);
    insert(root, 2);
    insert(root, 4);
    insert(root, 7);
    insert(root, 6);
    insert(root, 8);
  
    int k = 11;
  
    // Print the number of pair
    System.out.print(countPair(root, k));
  
}
}
 
// This code is contributed by Princi Singh

Python3




# Python3 program to count pair in
# BST whose sum is greater than K
index = 0
 
# Structure of each node of BST
class newNode:
     
    # Function to create a new BST node
    def __init__(self, item):
         
        self.key = item
        self.left = None
        self.right = None
 
# Function to insert a new
# node with given key in BST
def insert(node, key):
     
    # Check if the tree is empty
    if (node == None):
        return newNode(key)
 
    if (key < node.key):
        node.left = insert(node.left, key)
 
    elif(key > node.key):
        node.right = insert(node.right, key)
 
    # Return the (unchanged) node pointer
    return node
 
# Function to return the size of the tree
def sizeOfTree(root):
     
    if (root == None):
        return 0
 
    # Calculate left size recursively
    left = sizeOfTree(root.left)
 
    # Calculate right size recursively
    right = sizeOfTree(root.right)
 
    # Return total size recursively
    return (left + right + 1)
 
# Function to store inorder traversal of BST
def storeInorder(root, inOrder):
     
    global index
     
    # Base condition
    if (root == None):
        return
 
    # Left recursive call
    storeInorder(root.left, inOrder)
 
    # Store elements in inorder array
    inOrder[index] = root.key
    index += 1
 
    # Right recursive call
    storeInorder(root.right, inOrder)
 
# Function to count the pair of BST
# whose sum is greater than k
def countPairUtil(inOrder, j, k):
     
    i = 0
    pair = 0
     
    while (i < j):
         
        # Check if sum of value at index
        # i and j is greater than k
        if (inOrder[i] + inOrder[j] > k):
            pair += j - i
            j -= 1
        else:
            i += 1
 
    # Return number of total pair
    return pair
 
# Function to count the
# pair of BST whose sum is
# greater than k
def countPair(root, k):
     
    global index
     
    # Store the size of BST
    numNode = sizeOfTree(root)
 
    # Auxiliary array for storing
    # the inorder traversal of BST
    inOrder = [0 for i in range(numNode + 1)]
 
    storeInorder(root, inOrder)
 
    # Function call to count the pair
    return countPairUtil(inOrder, index - 1, k)
 
# Driver code
if __name__ == '__main__':
     
    # Create tree
    root = None
    root = insert(root, 5)
    insert(root, 3)
    insert(root, 2)
    insert(root, 4)
    insert(root, 7)
    insert(root, 6)
    insert(root, 8)
 
    k = 11
 
    # Print the number of pair
    print(countPair(root, k))
 
# This code is contributed by ipg2016107

C#




// C# program to Count
// pair in BST whose Sum
// is greater than K
using System;
 
class GFG{
   
// Structure of each node of BST
class node {
    public int key;
    public node left, right;
};
static int index;
  
// Function to create a new BST node
static node newNode(int item)
{
    node temp = new node();
   
    temp.key = item;
    temp.left = temp.right = null;
   
    return temp;
}
   
/* Function to insert a new
node with given key in BST */
static node insert(node node, int key)
{
   
    // check if the tree is empty
    if (node == null)
        return newNode(key);
   
    if (key < node.key)
   
        node.left = insert(node.left, key);
   
    else if (key > node.key)
   
        node.right = insert(node.right, key);
   
    /* return the (unchanged) node pointer */
    return node;
}
   
// Function to return the size of the tree
static int sizeOfTree(node root)
{
    if (root == null) {
        return 0;
    }
   
    // Calculate left size recursively
    int left = sizeOfTree(root.left);
   
    // Calculate right size recursively
    int right = sizeOfTree(root.right);
   
    // Return total size recursively
    return (left + right + 1);
}
   
// Function to store inorder traversal of BST
static void storeInorder(node root, int []inOrder)
{
   
    // Base condition
    if (root == null) {
        return;
    }
   
    // Left recursive call
    storeInorder(root.left, inOrder);
   
    // Store elements in inorder array
    inOrder[index++] = root.key;
   
    // Right recursive call
    storeInorder(root.right, inOrder);
}
   
// function to count the pair of BST
// whose sum is greater than k
static int countPairUtil(int []
                          
                          
                         inOrder, int j, int k)
{
    int i = 0;
    int pair = 0;
    while (i < j) {
   
        // check if sum of value at index
        // i and j is greater than k
        if (inOrder[i] + inOrder[j] > k) {
            pair += j - i;
   
            j--;
        }
        else {
            i++;
        }
    }
   
    // Return number of total pair
    return pair;
}
   
// Function to count the
// pair of BST whose sum is
// greater than k
static int countPair(node root, int k)
{
   
    // Store the size of BST
    int numNode = sizeOfTree(root);
   
    // Auxiliary array for storing
    // the inorder traversal of BST
    int []inOrder = new int[numNode + 1];
   
    index = 0;
   
    storeInorder(root, inOrder);
   
    // Function call to count the pair
    return countPairUtil(inOrder, index - 1, k);
}
   
// Driver code
public static void Main(String[] args)
{
   
    // create tree
    node root = null;
    root = insert(root, 5);
    insert(root, 3);
    insert(root, 2);
    insert(root, 4);
    insert(root, 7);
    insert(root, 6);
    insert(root, 8);
   
    int k = 11;
   
    // Print the number of pair
    Console.Write(countPair(root, k));
   
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
    // JavaScript program to Count
    // pair in BST whose Sum
    // is greater than K
     
    // Structure of each node of BST
    class node {
        constructor(item) {
           this.left = null;
           this.right = null;
           this.key = item;
        }
    }
    let index;
 
    // Function to create a new BST node
    function newNode(item)
    {
        let temp = new node(item);
        return temp;
    }
 
    /* Function to insert a new
    node with given key in BST */
    function insert(node, key)
    {
 
        // check if the tree is empty
        if (node == null)
            return newNode(key);
 
        if (key < node.key)
 
            node.left = insert(node.left, key);
 
        else if (key > node.key)
 
            node.right = insert(node.right, key);
 
        /* return the (unchanged) node pointer */
        return node;
    }
 
    // Function to return the size of the tree
    function sizeOfTree(root)
    {
        if (root == null) {
            return 0;
        }
 
        // Calculate left size recursively
        let left = sizeOfTree(root.left);
 
        // Calculate right size recursively
        let right = sizeOfTree(root.right);
 
        // Return total size recursively
        return (left + right + 1);
    }
 
    // Function to store inorder traversal of BST
    function storeInorder(root, inOrder)
    {
 
        // Base condition
        if (root == null) {
            return;
        }
 
        // Left recursive call
        storeInorder(root.left, inOrder);
 
        // Store elements in inorder array
        inOrder[index++] = root.key;
 
        // Right recursive call
        storeInorder(root.right, inOrder);
    }
 
    // function to count the pair of BST
    // whose sum is greater than k
    function countPairUtil(inOrder, j, k)
    {
        let i = 0;
        let pair = 0;
        while (i < j) {
 
            // check if sum of value at index
            // i and j is greater than k
            if (inOrder[i] + inOrder[j] > k) {
                pair += j - i;
 
                j--;
            }
            else {
                i++;
            }
        }
 
        // Return number of total pair
        return pair;
    }
 
    // Function to count the
    // pair of BST whose sum is
    // greater than k
    function countPair(root, k)
    {
 
        // Store the size of BST
        let numNode = sizeOfTree(root);
 
        // Auxiliary array for storing
        // the inorder traversal of BST
        let inOrder = new Array(numNode + 1);
 
        index = 0;
 
        storeInorder(root, inOrder);
 
        // Function call to count the pair
        return countPairUtil(inOrder, index - 1, k);
    }
     
    // create tree
    let root = null;
    root = insert(root, 5);
    insert(root, 3);
    insert(root, 2);
    insert(root, 4);
    insert(root, 7);
    insert(root, 6);
    insert(root, 8);
   
    let k = 11;
   
    // Print the number of pair
    document.write(countPair(root, k));
   
</script>
Output: 
6

 

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