Find the balanced node in a Linked List

Last Updated : 17 Aug, 2021

Given a linked list, the task is to find the balanced node in a linked list. A balanced node is a node where the sum of all the nodes on its left is equal to the sum of all the node on its right, if no such node is found then print -1.

Examples:

Input: 1 -> 2 -> 7 -> 10 -> 1 -> 6 -> 3 -> NULL
Output: 10
Sum of nodes on the left of 10 is 1 + 2 + 7 = 10
And, to the right of 10 is 1 + 6 + 3 = 10

Input: 1 -> 5 -> 5 -> 10 -> -3 -> NULL
Output: -1

Approach:

First, find the total sum of the all node values.
Now, traverse the linked list one by one and while traversing keep track of all the previous nodes value sum and find the sum of the remaining node by subtracting current node value and the sum of the previous nodes value from the total sum.
Compare both the sums, if they are equal then current node is the required node else print -1.

Below is the implementation of the above approach:

C++

   // C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Structure of a node of linked list
class Node { 
public: 
    int data; 
    Node* next;
    Node(int data)
    {
        this->data = data;
        this->next = NULL;
    }
};
  
// Push the new node to front of 
// the linked list
Node* push(Node* head, int data)
{
      
    // Return new node as head if 
    // head is empty
    if (head == NULL)
    {
        return new Node(data);
    }
    Node* temp = new Node(data);
    temp->next = head;
    head = temp;
    return head;
}
  
// Function to find the balanced node
int findBalancedNode(Node* head)
{
    int tsum = 0;
    Node* curr_node = head;
      
    // Traverse through all node 
    // to find the total sum
    while (curr_node != NULL)
    {
        tsum += curr_node->data;
        curr_node = curr_node->next;
    }
  
    // Set current_sum and remaining
    // sum to zero 
    int current_sum = 0;
    int remaining_sum = 0;
    curr_node = head;
  
    // Traversing the list to 
    // check balanced node
    while (curr_node != NULL)
    {
        remaining_sum = tsum - (current_sum + 
                               curr_node->data);
  
        // If sum of the nodes on the left and 
        // the current node is equal to the sum
        // of the nodes on the right
        if (current_sum == remaining_sum)
        {
            return curr_node->data;
        }
        current_sum += curr_node->data;
        curr_node = curr_node->next;
    }
    return -1;
}
 
// Driver code
int main()
{
    Node* head = NULL;
    head = push(head, 3);
    head = push(head, 6);
    head = push(head, 1);
    head = push(head, 10);
    head = push(head, 7);
    head = push(head, 2);
    head = push(head, 1); 
    cout << findBalancedNode(head);
    return 0;
}
 
// This code is contributed by divyehrabadiya07
 
  

Java

   // Java implementation of the approach
class GFG{
     
// Structure of a node of linked list
static class Node
{
    int data;
    Node next;
     
    Node(int data)
    {
        this.data = data;
        this.next = null;
    }
}
 
// Push the new node to front of 
// the linked list
static Node push(Node head, int data)
{
     
    // Return new node as head if 
    // head is empty
    if (head == null)
    {
        return new Node(data);
    }
    Node temp = new Node(data);
    temp.next = head;
    head = temp;
    return head;
}
 
// Function to find the balanced node
static int findBalancedNode(Node head)
{
    int tsum = 0;
    Node curr_node = head;
     
    // Traverse through all node 
    // to find the total sum
    while (curr_node != null)
    {
        tsum += curr_node.data;
        curr_node = curr_node.next;
    }
 
    // Set current_sum and remaining
    // sum to zero 
    int current_sum = 0;
    int remaining_sum = 0;
    curr_node = head;
 
    // Traversing the list to 
    // check balanced node
    while (curr_node != null)
    {
        remaining_sum = tsum - (current_sum + 
                               curr_node.data);
 
        // If sum of the nodes on the left and 
        // the current node is equal to the sum
        // of the nodes on the right
        if (current_sum == remaining_sum)
        {
            return curr_node.data;
        }
        current_sum += curr_node.data;
        curr_node = curr_node.next;
    }
    return -1;
}
 
// Driver code
public static void main(String []args)
{
    Node head = null;
    head = push(head, 3);
    head = push(head, 6);
    head = push(head, 1);
    head = push(head, 10);
    head = push(head, 7);
    head = push(head, 2);
    head = push(head, 1);
 
    System.out.println(findBalancedNode(head));
}
}
 
// This code is contributed by rutvik_56
 
  

Python3

   # Python3 implementation of the approach
import sys
import math
 
# Structure of a node of linked list 
class Node:
    def __init__(self, data):
        self.next = None
        self.data = data
 
# Push the new node to front of the linked list
def push(head, data):
 
    # Return new node as head if head is empty
    if not head:
        return Node(data)
    temp = Node(data)
    temp.next = head
    head = temp
    return head
 
# Function to find the balanced node
def findBalancedNode(head):
    tsum = 0
    curr_node = head
     
    # Traverse through all node 
    # to find the total sum
    while curr_node:
        tsum+= curr_node.data
        curr_node = curr_node.next
     
    # Set current_sum and remaining sum to zero 
    current_sum, remaining_sum = 0, 0
    curr_node = head
 
    # Traversing the list to check balanced node
    while(curr_node):
        remaining_sum = tsum-(current_sum + curr_node.data)
 
        # If sum of the nodes on the left and the current node 
        # is equal to the sum of the nodes on the right
        if current_sum == remaining_sum:
            return curr_node.data
        current_sum+= curr_node.data
        curr_node = curr_node.next
     
    return -1
 
# Driver code
if __name__=='__main__':
    head = None
    head = push(head, 3)
    head = push(head, 6)
    head = push(head, 1)
    head = push(head, 10)
    head = push(head, 7)
    head = push(head, 2)
    head = push(head, 1)
 
    print(findBalancedNode(head))
 
  

C#

   // C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
 
  // Structure of a node of linked list
  class Node
  {
    public int data;
    public Node next;
 
    public Node(int data)
    {
      this.data = data;
      this.next = null;
    }
  }
 
  // Push the new node to front of 
  // the linked list
  static Node push(Node head, int data)
  {
 
    // Return new node as head if 
    // head is empty
    if (head == null)
    {
      return new Node(data);
    }
    Node temp = new Node(data);
    temp.next = head;
    head = temp;
    return head;
  }
 
  // Function to find the balanced node
  static int findBalancedNode(Node head)
  {
    int tsum = 0;
    Node curr_node = head;
 
    // Traverse through all node 
    // to find the total sum
    while (curr_node != null)
    {
      tsum += curr_node.data;
      curr_node = curr_node.next;
    }
 
    // Set current_sum and remaining
    // sum to zero 
    int current_sum = 0;
    int remaining_sum = 0;
    curr_node = head;
 
    // Traversing the list to 
    // check balanced node
    while (curr_node != null)
    {
      remaining_sum = tsum - (current_sum + 
                              curr_node.data);
 
      // If sum of the nodes on the left and 
      // the current node is equal to the sum
      // of the nodes on the right
      if (current_sum == remaining_sum)
      {
        return curr_node.data;
      }
      current_sum += curr_node.data;
      curr_node = curr_node.next;
    }
    return -1;
  }
 
  // Driver code
  public static void Main(string []args)
  {
    Node head = null;
    head = push(head, 3);
    head = push(head, 6);
    head = push(head, 1);
    head = push(head, 10);
    head = push(head, 7);
    head = push(head, 2);
    head = push(head, 1);
    Console.Write(findBalancedNode(head));
  }
}
 
// This code is contributed by pratham76
 
  

Javascript

   <script>
 
// Javascript implementation of the approach
 
// Structure of a node of linked list
class Node {
        constructor(data) {
                this.data = data;
                this.next = null;
             }
        }
         
         
// Push the new node to front of
// the linked list
function push( head, data)
{
     
    // Return new node as head if
    // head is empty
    if (head == null)
    {
        return new Node(data);
    }
    var temp = new Node(data);
    temp.next = head;
    head = temp;
    return head;
}
 
// Function to find the balanced node
function findBalancedNode( head)
{
    let tsum = 0;
    let curr_node = head;
     
    // Traverse through all node
    // to find the total sum
    while (curr_node != null)
    {
        tsum += curr_node.data;
        curr_node = curr_node.next;
    }
 
    // Set current_sum and remaining
    // sum to zero
    let current_sum = 0;
    let remaining_sum = 0;
    curr_node = head;
 
    // Traversing the list to
    // check balanced node
    while (curr_node != null)
    {
        remaining_sum = tsum - (current_sum +
                            curr_node.data);
 
        // If sum of the nodes on the left and
        // the current node is equal to the sum
        // of the nodes on the right
        if (current_sum == remaining_sum)
        {
            return curr_node.data;
        }
        current_sum += curr_node.data;
        curr_node = curr_node.next;
    }
    return -1;
}
 
    // Driver Code
 
    var head = null;
    head = push(head, 3);
    head = push(head, 6);
    head = push(head, 1);
    head = push(head, 10);
    head = push(head, 7);
    head = push(head, 2);
    head = push(head, 1);
 
    document.write(findBalancedNode(head));
 
// This code is contributed by Jana_sayantan.
</script>
 
  

Output:

Time Complexity: O(n)
Auxiliary Space: O(1)

XOR Linked List - Find Nth Node from the end

Vikash Kumar 37

Improve

Article Tags :

Practice Tags :

Find the balanced node in a Linked List

C++

Java

Python3

C#

Javascript

Similar Reads