Find a pair with given sum in BST
Last Updated :
21 Feb, 2023
Given a BST and a sum, find if there is a pair with the given sum.
Example:
Input: sum = 28, given BST
Output: Pair is found (16, 12)
Pair with given sum using Hashing
The idea is based on Hashing. We traverse binary search tree by inorder way and insert node’s value into a set. Also check for any node, difference between given sum and node’s value in set, if it is found then pair exists otherwise it doesn’t exist.
Follow the steps mentioned below to implement the idea:
- Traverse the tree, while traversing store the value of a node in the set
- If for a current node with value x, there exists a y for which x + y = sum then check it using set and return the pair.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
struct Node *left, *right;
};
Node* NewNode(int data)
{
Node* temp = (Node*)malloc(sizeof(Node));
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}
Node* insert(Node* root, int key)
{
if (root == NULL)
return NewNode(key);
if (key < root->data)
root->left = insert(root->left, key);
else
root->right = insert(root->right, key);
return root;
}
bool findpairUtil(Node* root, int sum,
unordered_set<int>& set)
{
if (root == NULL)
return false;
if (findpairUtil(root->left, sum, set))
return true;
if (set.find(sum - root->data) != set.end()) {
cout << "Pair is found (" << sum - root->data
<< ", " << root->data << ")" << endl;
return true;
}
else
set.insert(root->data);
return findpairUtil(root->right, sum, set);
}
void findPair(Node* root, int sum)
{
unordered_set<int> set;
if (!findpairUtil(root, sum, set))
cout << "Pairs do not exit" << endl;
}
int main()
{
Node* root = NULL;
root = insert(root, 15);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 12);
root = insert(root, 16);
root = insert(root, 25);
root = insert(root, 10);
int sum = 28;
findPair(root, sum);
return 0;
}
|
Java
import java.util.*;
public class GFG {
static class Node {
int data;
Node left, right;
};
static Node NewNode(int data){
Node temp = new Node();
temp.data = data;
temp.left = null;
temp.right = null;
return temp;
}
static Node insert(Node root, int key){
if (root == null) return NewNode(key);
if (key < root.data) root.left = insert(root.left, key);
else root.right = insert(root.right, key);
return root;
}
static boolean findpairUtil(Node root, int sum, HashSet<Integer> set){
if (root == null) return false;
if (findpairUtil(root.left, sum, set)) return true;
if (set.contains(sum - root.data)){
System.out.println("Pair is found (" + (sum - root.data) + ", " + root.data + ")");
return true;
}
else set.add(root.data);
return findpairUtil(root.right, sum, set);
}
static void findPair(Node root, int sum){
HashSet<Integer> set = new HashSet<Integer>();
if (!findpairUtil(root, sum, set))
System.out.print("Pairs do not exit \n");
}
public static void main(String[] args){
Node root = null;
root = insert(root, 15);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 12);
root = insert(root, 16);
root = insert(root, 25);
root = insert(root, 10);
int sum = 28;
findPair(root, sum);
}
}
|
Python3
import sys
import math
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
def insert(root, data):
if root is None:
return Node(data)
if(data < root.data):
root.left = insert(root.left, data)
if(data > root.data):
root.right = insert(root.right, data)
return root
def findPairUtil(root, summ, unsorted_set):
if root is None:
return False
if findPairUtil(root.left, summ, unsorted_set):
return True
if unsorted_set and (summ-root.data) in unsorted_set:
print("Pair is Found ({},{})".format(summ-root.data, root.data))
return True
else:
unsorted_set.add(root.data)
return findPairUtil(root.right, summ, unsorted_set)
def findPair(root, summ):
unsorted_set = set()
if(not findPairUtil(root, summ, unsorted_set)):
print("Pair do not exist!")
if __name__ == '__main__':
root = None
root = insert(root, 15)
root = insert(root, 10)
root = insert(root, 20)
root = insert(root, 8)
root = insert(root, 12)
root = insert(root, 16)
root = insert(root, 25)
root = insert(root, 10)
summ = 28
findPair(root, summ)
|
C#
using System;
using System.Collections.Generic;
class GFG {
class Node {
public int data;
public Node left, right;
};
static Node NewNode(int data)
{
Node temp = new Node();
temp.data = data;
temp.left = null;
temp.right = null;
return temp;
}
static Node insert(Node root, int key)
{
if (root == null)
return NewNode(key);
if (key < root.data)
root.left = insert(root.left, key);
else
root.right = insert(root.right, key);
return root;
}
static bool findpairUtil(Node root, int sum,
HashSet<int> set)
{
if (root == null)
return false;
if (findpairUtil(root.left, sum, set))
return true;
if (set.Contains(sum - root.data)) {
Console.WriteLine("Pair is found ("
+ (sum - root.data) + ", "
+ root.data + ")");
return true;
}
else
set.Add(root.data);
return findpairUtil(root.right, sum, set);
}
static void findPair(Node root, int sum)
{
HashSet<int> set = new HashSet<int>();
if (!findpairUtil(root, sum, set))
Console.Write("Pairs do not exit"
+ "\n");
}
public static void Main(String[] args)
{
Node root = null;
root = insert(root, 15);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 12);
root = insert(root, 16);
root = insert(root, 25);
root = insert(root, 10);
int sum = 28;
findPair(root, sum);
}
}
|
Javascript
class Node {
constructor()
{
this.data = 0;
this.left = null;
this.right = null;
}
};
function NewNode(data)
{
var temp = new Node();
temp.data = data;
temp.left = null;
temp.right = null;
return temp;
}
function insert(root, key)
{
if (root == null)
return NewNode(key);
if (key < root.data)
root.left = insert(root.left, key);
else
root.right = insert(root.right, key);
return root;
}
function findpairUtil(root, sum, set)
{
if (root == null)
return false;
if (findpairUtil(root.left, sum, set))
return true;
if (set.has(sum - root.data)) {
console.log("Pair is found ("
+ (sum - root.data) + ", "
+ root.data + ")<br>");
return true;
}
else
set.add(root.data);
return findpairUtil(root.right, sum, set);
}
function findPair(root, sum)
{
var set = new Set();
if (!findpairUtil(root, sum, set))
console.log("Pairs do not exit"
+ "\n");
}
var root = null;
root = insert(root, 15);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 12);
root = insert(root, 16);
root = insert(root, 25);
root = insert(root, 10);
var sum = 28;
findPair(root, sum);
|
Output
Pair is found (12, 16)
Time Complexity: O(N)
Auxiliary Space: O(N)
Pair with a given sum using Two pointers
Key Idea:
The main idea is to find the Inorder Traversal of the BST and store it in a vector. We know that Inorder traversal of BST will be in sorted order.Now we will Apply Two pointers Technique to find that whether there exist two elements in the vector that sums up to the given value.
Algorithm:
1. First find the Inorder traversal of the Given BST and store it in a vector (Let v).
2. Take two pointers i and j. Keep i at the start of v and j at the end of the v.
Now, if sum of elements at the ith index and jth index is greater that the given element then decrement j,
if sum of elements at the ith index and jth index is less that the given element then increment i,
else, these two elements are our required answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
struct TreeNode {
int val;
struct TreeNode *left, *right;
};
TreeNode* NewNode(int data)
{
TreeNode* temp = (TreeNode*)malloc(sizeof(TreeNode));
temp->val = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}
TreeNode* insert(TreeNode* root, int key)
{
if (root == NULL)
return NewNode(key);
if (key < root->val)
root->left = insert(root->left, key);
else
root->right = insert(root->right, key);
return root;
}
void inorder(TreeNode* root, vector<int>& v)
{
if (root == NULL) {
return;
}
inorder(root->left, v);
v.push_back(root->val);
inorder(root->right, v);
}
pair<int, int> findTarget(TreeNode* root, int k)
{
vector<int> v;
inorder(root, v);
int n = v.size();
int i = 0;
int j = n - 1;
while (j > i) {
if (v[i] + v[j] == k) {
return { v[i], v[j] };
}
else if (v[i] + v[j] > k) {
j--;
}
else {
i++;
}
}
return { -1, -1 };
}
int main()
{
TreeNode* root = NULL;
root = insert(root, 15);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 12);
root = insert(root, 16);
root = insert(root, 25);
root = insert(root, 10);
int k = 28;
auto a = findTarget(root, k);
cout << a.first << " " << a.second << endl;
}
|
Java
import java.util.*;
public class GFG {
static class pair {
int first, second;
pair(int f, int s)
{
first = f;
second = s;
}
}
static class NewNode {
int val;
NewNode left, right;
NewNode(int data)
{
val = data;
left = null;
right = null;
}
};
static NewNode insert(NewNode root, int key)
{
if (root == null)
return new NewNode(key);
if (key < root.val)
root.left = insert(root.left, key);
else
root.right = insert(root.right, key);
return root;
}
static void inorder(NewNode root, ArrayList<Integer> v)
{
if (root == null) {
return;
}
inorder(root.left, v);
v.add(root.val);
inorder(root.right, v);
}
static pair findTarget(NewNode root, int k)
{
ArrayList<Integer> v = new ArrayList<>();
inorder(root, v);
int n = v.size();
int i = 0;
int j = n - 1;
while (j > i) {
if (v.get(i) + v.get(j) == k) {
return new pair(v.get(i), v.get(j));
}
else if (v.get(i) + v.get(j) > k) {
j--;
}
else {
i++;
}
}
return new pair(-1, -1);
}
public static void main(String[] args)
{
NewNode root = null;
root = insert(root, 15);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 12);
root = insert(root, 16);
root = insert(root, 25);
root = insert(root, 10);
int k = 28;
pair a = findTarget(root, k);
System.out.println(a.first + " " + a.second);
}
}
|
Python3
class TreeNode:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
def NewNode(data):
temp = TreeNode(data)
return temp
def insert(root, key):
if root is None:
return NewNode(key)
if key < root.val:
root.left = insert(root.left, key)
else:
root.right = insert(root.right, key)
return root
def inorder(root, v):
if root is None:
return
inorder(root.left, v)
v.append(root.val)
inorder(root.right, v)
def findTarget(root, k):
v = []
inorder(root, v)
n = len(v)
i = 0
j = n - 1
while j > i:
if v[i] + v[j] == k:
return (v[i], v[j
])
elif v[i] + v[j] > k:
j -= 1
else:
i += 1
return (-1, -1)
if __name__ == '__main__':
root = None
root = insert(root, 15)
root = insert(root, 10)
root = insert(root, 20)
root = insert(root, 8)
root = insert(root, 12)
root = insert(root, 16)
root = insert(root, 25)
root = insert(root, 10)
k = 28
a = findTarget(root, k)
print(a[0], a[1])
|
C#
using System;
using System.Collections.Generic;
public class GFG
{
public class pair
{
public int first, second;
public pair(int f, int s)
{
first = f;
second = s;
}
}
public class NewNode
{
public int val;
public NewNode left, right;
public NewNode(int data)
{
val = data;
left = null;
right = null;
}
};
public static NewNode insert(NewNode root, int key)
{
if (root == null)
return new NewNode(key);
if (key < root.val)
root.left = insert(root.left, key);
else
root.right = insert(root.right, key);
return root;
}
public static void inorder(NewNode root, List<int> v)
{
if (root == null)
{
return;
}
inorder(root.left, v);
v.Add(root.val);
inorder(root.right, v);
}
public static pair findTarget(NewNode root, int k)
{
List<int> v = new List<int>();
inorder(root, v);
int n = v.Count;
int i = 0;
int j = n - 1;
while (j > i)
{
if (v[i] + v[j] == k)
{
return new pair(v[i], v[j]);
}
else if (v[i] + v[j] > k)
{
j--;
}
else
{
i++;
}
}
return new pair(-1, -1);
}
public static void Main (String[] args)
{
NewNode root = null;
root = insert(root, 15);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 12);
root = insert(root, 16);
root = insert(root, 25);
root = insert(root, 10);
int k = 28;
pair a = findTarget(root, k);
Console.WriteLine(a.first + " " + a.second);
}
}
|
Javascript
class TreeNode {
constructor(val) {
this.val = val;
this.left = null;
this.right = null;
}
}
function NewNode(data) {
let temp = new TreeNode(data);
return temp;
}
function insert(root, key) {
if (root === null) {
return NewNode(key);
}
if (key < root.val) {
root.left = insert(root.left, key);
} else {
root.right = insert(root.right, key);
}
return root;
}
function inorder(root, v) {
if (root === null) {
return;
}
inorder(root.left, v);
v.push(root.val);
inorder(root.right, v);
}
function findTarget(root, k) {
let v = [];
inorder(root, v);
let n = v.length;
let i = 0;
let j = n - 1;
while (j > i) {
if (v[i] + v[j] === k) {
return [v[i], v[j]];
} else if (v[i] + v[j] > k) {
j -= 1;
} else {
i += 1;
}
}
return [-1, -1];
}
let root = null;
root = insert(root, 15);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 12);
root = insert(root, 16);
root = insert(root, 25);
root = insert(root, 10);
let k = 28;
let a = findTarget(root, k);
console.log(a[0], a[1]);
|
Time Complexity: O(n) (n = number of nodes)
Auxiliary Space: O(n)
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