Find distance between two nodes of a Binary Tree

Difficulty Level : Hard
Last Updated : 18 Jan, 2021

Find the distance between two keys in a binary tree, no parent pointers are given. The distance between two nodes is the minimum number of edges to be traversed to reach one node from another.

The distance between two nodes can be obtained in terms of lowest common ancestor. Following is the formula.

Dist(n1, n2) = Dist(root, n1) + Dist(root, n2) - 2*Dist(root, lca) 
'n1' and 'n2' are the two given keys
'root' is root of given Binary Tree.
'lca' is lowest common ancestor of n1 and n2
Dist(n1, n2) is the distance between n1 and n2.

Following is the implementation of the above approach. The implementation is adopted from the last code provided in Lowest Common Ancestor Post.

C++

/* C++ program to find distance between n1 and n2 using 
   one traversal */
#include <iostream>
using namespace std;
 
// A Binary Tree Node
struct Node
{
    struct Node *left, *right;
    int key;
};
 
// Utility function to create a new tree Node
Node* newNode(int key)
{
    Node *temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Returns level of key k if it is present in tree, 
// otherwise returns -1
int findLevel(Node *root, int k, int level)
{
    // Base Case
    if (root == NULL)
        return -1;
 
    // If key is present at root, or in left subtree
    // or right subtree, return true;
    if (root->key == k)
        return level;
 
    int l = findLevel(root->left, k, level+1);
    return (l != -1)? l : findLevel(root->right, k, level+1);
}
 
// This function returns pointer to LCA of two given
// values n1 and n2. It also sets d1, d2 and dist if 
// one key is not ancestor of other
// d1 --> To store distance of n1 from root
// d2 --> To store distance of n2 from root
// lvl --> Level (or distance from root) of current node
// dist --> To store distance between n1 and n2
Node *findDistUtil(Node* root, int n1, int n2, int &d1, 
                            int &d2, int &dist, int lvl)
{
    // Base case
    if (root == NULL) return NULL;
 
    // If either n1 or n2 matches with root's key, report
    // the presence by returning root (Note that if a key is
    // ancestor of other, then the ancestor key becomes LCA
    if (root->key == n1)
    {
         d1 = lvl;
         return root;
    }
    if (root->key == n2)
    {
         d2 = lvl;
         return root;
    }
 
    // Look for n1 and n2 in left and right subtrees
    Node *left_lca  = findDistUtil(root->left, n1, n2, 
                                   d1, d2, dist, lvl+1);
    Node *right_lca = findDistUtil(root->right, n1, n2,
                                   d1, d2, dist, lvl+1);
 
    // If both of the above calls return Non-NULL, then
    // one key is present in once subtree and other is 
    // present in other. So this node is the LCA
    if (left_lca && right_lca)
    {
        dist = d1 + d2 - 2*lvl;
        return root;
    }
 
    // Otherwise check if left subtree or right subtree 
    // is LCA
    return (left_lca != NULL)? left_lca: right_lca;
}
 
// The main function that returns distance between n1
// and n2. This function returns -1 if either n1 or n2
// is not present in Binary Tree.
int findDistance(Node *root, int n1, int n2)
{
    // Initialize d1 (distance of n1 from root), d2 
    // (distance of n2 from root) and dist(distance 
    // between n1 and n2)
    int d1 = -1, d2 = -1, dist;
    Node *lca = findDistUtil(root, n1, n2, d1, d2,
                                          dist, 1);
 
    // If both n1 and n2 were present in Binary 
    // Tree, return dist
    if (d1 != -1 && d2 != -1)
        return dist;
 
    // If n1 is ancestor of n2, consider n1 as root 
    // and find level of n2 in subtree rooted with n1
    if (d1 != -1)
    {
        dist = findLevel(lca, n2, 0);
        return dist;
    }
 
    // If n2 is ancestor of n1, consider n2 as root 
    // and find level of n1 in subtree rooted with n2
    if (d2 != -1)
    {
        dist = findLevel(lca, n1, 0);
        return dist;
    }
 
    return -1;
}
 
// Driver program to test above functions
int main()
{
    // Let us create binary tree given in the
    // above example
    Node * root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);
    cout << "Dist(4, 5) = " << findDistance(root, 4, 5);
    cout << "nDist(4, 6) = " << findDistance(root, 4, 6);
    cout << "nDist(3, 4) = " << findDistance(root, 3, 4);
    cout << "nDist(2, 4) = " << findDistance(root, 2, 4);
    cout << "nDist(8, 5) = " << findDistance(root, 8, 5);
    return 0;
}

Java

// A Java Program to find distance between 
// n1 and n2 using one traversal 
class GFG
{
   
// (To the moderator) in Java solution these
// variables are declared as pointers hence 
//changes made to them reflects in the whole program 
 
// Global static variable 
 static int d1 = -1;
 static int d2 = -1;
 static int dist = 0;
 
// A Binary Tree Node 
static class Node
{
     Node left, right;
     int key;
 
    // constructor 
     Node(int key)
    {
        this.key = key;
        left = null;
        right = null;
    }
}
 
// Returns level of key k if it is present 
// in tree, otherwise returns -1 
 static int findLevel(Node root, int k,
                                   int level)
{
    
    // Base Case 
    if (root == null)
    {
        return -1;
    }
 
    // If key is present at root, or in left 
    // subtree or right subtree, return true; 
    if (root.key == k)
    {
        return level;
    }
 
    int l = findLevel(root.left, k, level + 1);
    return (l != -1)? l : findLevel(root.right, k, 
                                        level + 1);
}
 
// This function returns pointer to LCA of 
// two given values n1 and n2. It also sets
// d1, d2 and dist if one key is not ancestor of other 
// d1 -. To store distance of n1 from root 
// d2 -. To store distance of n2 from root 
// lvl -. Level (or distance from root) of current node 
// dist -. To store distance between n1 and n2 
public static Node findDistUtil(Node root, int n1, 
                                   int n2, int lvl)
{
 
    // Base case 
    if (root == null)
    {
        return null;
    }
 
    // If either n1 or n2 matches with root's 
    // key, report the presence by returning 
    // root (Note that if a key is ancestor of 
    // other, then the ancestor key becomes LCA 
    if (root.key == n1)
    {
        d1 = lvl;
        return root;
    }
    if (root.key == n2)
    {
        d2 = lvl;
        return root;
    }
 
    // Look for n1 and n2 in left and right subtrees 
    Node left_lca = findDistUtil(root.left, n1, 
                                   n2, lvl + 1);
    Node right_lca = findDistUtil(root.right, n1, 
                                     n2, lvl + 1);
 
    // If both of the above calls return Non-null, 
    // then one key is present in once subtree and 
    // other is present in other, So this node is the LCA 
    if (left_lca != null && right_lca != null)
    {
        dist = (d1 + d2) - 2 * lvl;
        return root;
    }
 
    // Otherwise check if left subtree 
    // or right subtree is LCA 
    return (left_lca != null)? left_lca : right_lca;
}
 
// The main function that returns distance 
// between n1 and n2. This function returns -1 
// if either n1 or n2 is not present in 
// Binary Tree. 
public static int findDistance(Node root, int n1, int n2)
{
    d1 = -1;
    d2 = -1;
    dist = 0;
    Node lca = findDistUtil(root, n1, n2, 1);
 
    // If both n1 and n2 were present
    // in Binary Tree, return dist 
    if (d1 != -1 && d2 != -1)
    {
        return dist;
    }
 
    // If n1 is ancestor of n2, consider
    // n1 as root and find level 
    // of n2 in subtree rooted with n1 
    if (d1 != -1)
    {
        dist = findLevel(lca, n2, 0);
        return dist;
    }
 
    // If n2 is ancestor of n1, consider 
    // n2 as root and find level 
    // of n1 in subtree rooted with n2 
    if (d2 != -1)
    {
        dist = findLevel(lca, n1, 0);
        return dist;
    }
    return -1;
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Let us create binary tree given
    // in the above example 
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);
    root.right.left = new Node(6);
    root.right.right = new Node(7);
    root.right.left.right = new Node(8);
    System.out.println("Dist(4, 5) = " + 
                       findDistance(root, 4, 5));
    System.out.println("Dist(4, 6) = " + 
                       findDistance(root, 4, 6));
    System.out.println("Dist(3, 4) = " + 
                       findDistance(root, 3, 4));
    System.out.println("Dist(2, 4) = " + 
                       findDistance(root, 2, 4));
    System.out.println("Dist(8, 5) = " + 
                       findDistance(root, 8, 5));
}
}
 
// This code is contributed by gauravrajput1

Python

# Python Program to find distance between 
# n1 and n2 using one traversal
 
class Node:
    def __init__(self, data):
        self.data = data
        self.right = None
        self.left = None
 
def pathToNode(root, path, k):
 
    # base case handling
    if root is None:
        return False
 
     # append the node value in path
    path.append(root.data)
  
    # See if the k is same as root's data
    if root.data == k :
        return True
  
    # Check if k is found in left or right 
    # sub-tree
    if ((root.left != None and pathToNode(root.left, path, k)) or
            (root.right!= None and pathToNode(root.right, path, k))):
        return True
  
    # If not present in subtree rooted with root, 
    # remove root from path and return False 
    path.pop()
    return False
 
def distance(root, data1, data2):
    if root:
        # store path corresponding to node: data1
        path1 = []
        pathToNode(root, path1, data1)
 
        # store path corresponding to node: data2
        path2 = []
        pathToNode(root, path2, data2)
 
        # iterate through the paths to find the 
        # common path length
        i=0
        while i<len(path1) and i<len(path2):
            # get out as soon as the path differs 
            # or any path's length get exhausted
            if path1[i] != path2[i]:
                break
            i = i+1
 
        # get the path length by deducting the 
        # intersecting path length (or till LCA)
        return (len(path1)+len(path2)-2*i)
    else:
        return 0
 
# Driver Code to test above functions
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.right.right= Node(7)
root.right.left = Node(6)
root.left.right = Node(5)
root.right.left.right = Node(8)
 
dist = distance(root, 4, 5)
print "Distance between node {} & {}: {}".format(4, 5, dist) 
 
dist = distance(root, 4, 6)
print "Distance between node {} & {}: {}".format(4, 6, dist) 
 
dist = distance(root, 3, 4)
print "Distance between node {} & {}: {}".format(3, 4, dist) 
 
dist = distance(root, 2, 4)
print "Distance between node {} & {}: {}".format(2, 4, dist) 
 
dist = distance(root, 8, 5)
print "Distance between node {} & {}: {}".format(8, 5, dist) 
 
# This program is contributed by Aartee

C#

// A C# Program to find distance between 
// n1 and n2 using one traversal 
using System;
 
class GFG
{
// (To the moderator) in c++ solution these
// variables are declared as pointers hence 
//changes made to them reflects in the whole program 
 
// Global static variable 
public static int d1 = -1;
public static int d2 = -1;
public static int dist = 0;
 
// A Binary Tree Node 
public class Node
{
    public Node left, right;
    public int key;
 
    // constructor 
    public Node(int key)
    {
        this.key = key;
        left = null;
        right = null;
    }
}
 
// Returns level of key k if it is present 
// in tree, otherwise returns -1 
public static int findLevel(Node root, int k,
                                   int level)
{
    // Base Case 
    if (root == null)
    {
        return -1;
    }
 
    // If key is present at root, or in left 
    // subtree or right subtree, return true; 
    if (root.key == k)
    {
        return level;
    }
 
    int l = findLevel(root.left, k, level + 1);
    return (l != -1)? l : findLevel(root.right, k, 
                                        level + 1);
}
 
// This function returns pointer to LCA of 
// two given values n1 and n2. It also sets
// d1, d2 and dist if one key is not ancestor of other 
// d1 --> To store distance of n1 from root 
// d2 --> To store distance of n2 from root 
// lvl --> Level (or distance from root) of current node 
// dist --> To store distance between n1 and n2 
public static Node findDistUtil(Node root, int n1, 
                                   int n2, int lvl)
{
 
    // Base case 
    if (root == null)
    {
        return null;
    }
 
    // If either n1 or n2 matches with root's 
    // key, report the presence by returning 
    // root (Note that if a key is ancestor of 
    // other, then the ancestor key becomes LCA 
    if (root.key == n1)
    {
        d1 = lvl;
        return root;
    }
    if (root.key == n2)
    {
        d2 = lvl;
        return root;
    }
 
    // Look for n1 and n2 in left and right subtrees 
    Node left_lca = findDistUtil(root.left, n1, 
                                   n2, lvl + 1);
    Node right_lca = findDistUtil(root.right, n1, 
                                     n2, lvl + 1);
 
    // If both of the above calls return Non-NULL, 
    // then one key is present in once subtree and 
    // other is present in other, So this node is the LCA 
    if (left_lca != null && right_lca != null)
    {
        dist = (d1 + d2) - 2 * lvl;
        return root;
    }
 
    // Otherwise check if left subtree 
    // or right subtree is LCA 
    return (left_lca != null)? left_lca : right_lca;
}
 
// The main function that returns distance 
// between n1 and n2. This function returns -1 
// if either n1 or n2 is not present in 
// Binary Tree. 
public static int findDistance(Node root, int n1, int n2)
{
    d1 = -1;
    d2 = -1;
    dist = 0;
    Node lca = findDistUtil(root, n1, n2, 1);
 
    // If both n1 and n2 were present
    // in Binary Tree, return dist 
    if (d1 != -1 && d2 != -1)
    {
        return dist;
    }
 
    // If n1 is ancestor of n2, consider
    // n1 as root and find level 
    // of n2 in subtree rooted with n1 
    if (d1 != -1)
    {
        dist = findLevel(lca, n2, 0);
        return dist;
    }
 
    // If n2 is ancestor of n1, consider 
    // n2 as root and find level 
    // of n1 in subtree rooted with n2 
    if (d2 != -1)
    {
        dist = findLevel(lca, n1, 0);
        return dist;
    }
 
    return -1;
}
 
// Driver Code
public static void Main(string[] args)
{
 
    // Let us create binary tree given
    // in the above example 
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);
    root.right.left = new Node(6);
    root.right.right = new Node(7);
    root.right.left.right = new Node(8);
 
    Console.WriteLine("Dist(4, 5) = " + 
                       findDistance(root, 4, 5));
    Console.WriteLine("Dist(4, 6) = " + 
                       findDistance(root, 4, 6));
    Console.WriteLine("Dist(3, 4) = " + 
                       findDistance(root, 3, 4));
    Console.WriteLine("Dist(2, 4) = " + 
                       findDistance(root, 2, 4));
    Console.WriteLine("Dist(8, 5) = " + 
                       findDistance(root, 8, 5));
}
}
 
// This code is contributed by Shrikant13

Output:

Dist(4, 5) = 2
Dist(4, 6) = 4
Dist(3, 4) = 3
Dist(2, 4) = 1
Dist(8, 5) = 5

Time Complexity: Time complexity of the above solution is O(n) as the method does a single tree traversal.
Thanks to Atul Singh for providing the initial solution for this post.
Better Solution :
We first find the LCA of two nodes. Then we find the distance from LCA to two nodes.

C++

/* C++ Program to find distance between n1 and n2
   using one traversal */
#include <iostream>
using namespace std;
 
// A Binary Tree Node
struct Node
{
    struct Node *left, *right;
    int key;
};
 
// Utility function to create a new tree Node
Node* newNode(int key)
{
    Node *temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return temp;
}
Node* LCA(Node * root, int n1,int n2)
{
    // Your code here
    if (root == NULL)
       return root;
    if (root->key == n1 || root->key == n2)
       return root;
 
    Node* left = LCA(root->left, n1, n2);
    Node* right = LCA(root->right, n1, n2);
 
    if (left != NULL && right != NULL)
         return root;
    if (left == NULL && right == NULL)
           return NULL;
    if (left != NULL)
        return LCA(root->left, n1, n2);
 
    return LCA(root->right, n1, n2);
}
 
// Returns level of key k if it is present in
// tree, otherwise returns -1
int findLevel(Node *root, int k, int level)
{
    if(root == NULL) return -1;
    if(root->key == k) return level;
 
    int left = findLevel(root->left, k, level+1);
   if (left == -1)
       return findLevel(root->right, k, level+1);
    return left;
}
 
int findDistance(Node* root, int a, int b)
{
    // Your code here
    Node* lca = LCA(root, a , b);
 
    int d1 = findLevel(lca, a, 0);
    int d2 = findLevel(lca, b, 0);
 
    return d1 + d2;
}
 
// Driver program to test above functions
int main()
{
    // Let us create binary tree given in
    // the above example
    Node * root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);
    cout << "Dist(4, 5) = " << findDistance(root, 4, 5);
    cout << "\nDist(4, 6) = " << findDistance(root, 4, 6);
    cout << "\nDist(3, 4) = " << findDistance(root, 3, 4);
    cout << "\nDist(2, 4) = " << findDistance(root, 2, 4);
    cout << "\nDist(8, 5) = " << findDistance(root, 8, 5);
    return 0;
}

Java

/* Java Program to find distance between n1 and n2
   using one traversal */
public class GFG {
 
    public static class Node {
        int value;
        Node left;
        Node right;
 
        public Node(int value) {
            this.value = value;
        }
    }
 
    public static Node LCA(Node root, int n1, int n2) 
    {
        if (root == null)
            return root;
        if (root.value == n1 || root.value == n2)
            return root;
 
        Node left = LCA(root.left, n1, n2);
        Node right = LCA(root.right, n1, n2);
 
        if (left != null && right != null)
            return root;
          if (left == null && right == null)
              return null;
        if (left != null)
            return LCA(root.left, n1, n2);
        else
            return LCA(root.right, n1, n2);
    }
 
    // Returns level of key k if it is present in
    // tree, otherwise returns -1
    public static int findLevel(Node root, int a, int level)
    {
        if (root == null)
            return -1;
        if (root.value == a)
            return level;
        int left = findLevel(root.left, a, level + 1);
        if (left == -1)
            return findLevel(root.right, a, level + 1);
        return left;
    }
 
    public static int findDistance(Node root, int a, int b)
    {
        Node lca = LCA(root, a, b);
 
        int d1 = findLevel(lca, a, 0);
        int d2 = findLevel(lca, b, 0);
 
        return d1 + d2;
    }
     
    // Driver program to test above functions
    public static void main(String[] args) {
         
        // Let us create binary tree given in
        // the above example
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.left = new Node(6);
        root.right.right = new Node(7);
        root.right.left.right = new Node(8);
        System.out.println("Dist(4, 5) = "
                             + findDistance(root, 4, 5));
                              
        System.out.println("Dist(4, 6) = "
                             + findDistance(root, 4, 6));
                              
        System.out.println("Dist(3, 4) = "
                             + findDistance(root, 3, 4));
                              
        System.out.println("Dist(2, 4) = "
                             + findDistance(root, 2, 4));
                              
        System.out.println("Dist(8, 5) = "
                             + findDistance(root, 8, 5));
 
    }
}
 
// This code is contributed by Srinivasan Jayaraman.

Python3

"""
A python program to find distance between n1 
and n2 in binary tree
"""
# binary tree node
class Node:
    # Constructor to create new node
    def __init__(self, data):
        self.data = data
        self.left = self.right = None
 
 
# This function returns pointer to LCA of 
# two given values n1 and n2.
def LCA(root, n1, n2):
     
    # Base case
    if root is None:
        return None
 
    # If either n1 or n2 matches with root's
    # key, report the presence by returning
    # root 
    if root.data == n1 or root.data == n2:
        return root
    if root.data == None or root.data == None:
        return None
 
    # Look for keys in left and right subtrees
    left = LCA(root.left, n1, n2)
    right = LCA(root.right, n1, n2)
 
    if left is not None and right is not None:
        return root
 
    # Otherwise check if left subtree or 
    # right subtree is LCA
    if left:
        return left
    else:
        return right
 
 
# function to find distance of any node
# from root
def findLevel(root, data, d, level):
     
    # Base case when tree is empty
    if root is None:
        return
 
    # Node is found then append level
    # value to list and return
    if root.data == data:
        d.append(level)
        return
 
    findLevel(root.left, data, d, level + 1)
    findLevel(root.right, data, d, level + 1)
 
# function to find distance between two
# nodes in a binary tree
def findDistance(root, n1, n2):
     
    lca = LCA(root, n1, n2)
     
    # to store distance of n1 from lca
    d1 = [] 
     
    # to store distance of n2 from lca
    d2 = [] 
 
    # if lca exist
    if lca:
         
        # distance of n1 from lca
        findLevel(lca, n1, d1, 0) 
         
        # distance of n2 from lca
        findLevel(lca, n2, d2, 0) 
        return d1[0] + d2[0]
    else:
        return -1
 
 
# Driver program to test above function
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
root.right.left.right = Node(8)
 
print("Dist(4,5) = ", findDistance(root, 4, 5))
print("Dist(4,6) = ", findDistance(root, 4, 6))
print("Dist(3,4) = ", findDistance(root, 3, 4))
print("Dist(2,4) = ", findDistance(root, 2, 4))
print("Dist(8,5) = ", findDistance(root, 8, 5))
 
# This article is contributed by Shweta Singh.

C#

using System;
 
/* C# Program to find distance between n1 and n2 
   using one traversal */
public class GFG
{
 
    public class Node
    {
        public int value;
        public Node left;
        public Node right;
 
        public Node(int value)
        {
            this.value = value;
        }
    }
 
    public static Node LCA(Node root, int n1, int n2)
    {
        if (root == null)
        {
            return root;
        }
        if (root.value == n1 || root.value == n2)
        {
            return root;
        }
 
        Node left = LCA(root.left, n1, n2);
        Node right = LCA(root.right, n1, n2);
 
        if (left != null && right != null)
        {
            return root;
        }
         if (left == null && right == null)
        {
            return null;
        }
       
        if (left != null)
        {
            return LCA(root.left, n1, n2);
        }
        else
        {
            return LCA(root.right, n1, n2);
        }
    }
 
    // Returns level of key k if it is present in 
    // tree, otherwise returns -1 
    public static int findLevel(Node root, int a, int level)
    {
        if (root == null)
        {
            return -1;
        }
        if (root.value == a)
        {
            return level;
        }
        int left = findLevel(root.left, a, level + 1);
        if (left == -1)
        {
            return findLevel(root.right, a, level + 1);
        }
        return left;
    }
 
    public static int findDistance(Node root, int a, int b)
    {
        Node lca = LCA(root, a, b);
 
        int d1 = findLevel(lca, a, 0);
        int d2 = findLevel(lca, b, 0);
 
        return d1 + d2;
    }
 
    // Driver program to test above functions 
    public static void Main(string[] args)
    {
 
        // Let us create binary tree given in 
        // the above example 
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.left = new Node(6);
        root.right.right = new Node(7);
        root.right.left.right = new Node(8);
        Console.WriteLine("Dist(4, 5) = " + findDistance(root, 4, 5));
 
        Console.WriteLine("Dist(4, 6) = " + findDistance(root, 4, 6));
 
        Console.WriteLine("Dist(3, 4) = " + findDistance(root, 3, 4));
 
        Console.WriteLine("Dist(2, 4) = " + findDistance(root, 2, 4));
 
        Console.WriteLine("Dist(8, 5) = " + findDistance(root, 8, 5));
 
    }
}
 
  //  This code is contributed by Shrikant13

Output :

Dist(4, 5) = 2
Dist(4, 6) = 4
Dist(3, 4) = 3
Dist(2, 4) = 1
Dist(8, 5) = 5

Thanks to NILMADHAB MONDAL for suggesting this solution.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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