Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Maximum sum of K consecutive nodes in the given Linked List

  • Difficulty Level : Medium
  • Last Updated : 02 Jun, 2021

Given a linked list, the task is to find the maximum sum obtained by adding any k consecutive nodes of linked list.

Examples:  

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL, K = 5 
Output: 20 
Maximum sum is obtained by adding last 5 nodes



Input: 2 -> 5 -> 3 -> 6 -> 4 -> 1 -> 7 -> NULL, K = 4 
Output: 18 

Approach: The idea is to use a sliding window of size k, keep track of sum of current window and update maximum sum if required. To implement sliding window two pointers can be used to represent starting and ending point. At each step first the value of node pointed by start is subtracted from current sum and the value of node pointed by end is added to current sum. This sum is compared to maximum sum and result is updated if required. The start and end pointers are incremented by one at each step.

Below is the implementation of above approach:  

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Structure of a node
struct Node {
    int data;
    Node* next;
};
 
// Function to create new node
Node* newNode(int data)
{
    Node* node = new Node;
    node->next = NULL;
    node->data = data;
    return node;
}
 
// Function to return the maximum sum of
// k consecutive nodes
int findMaxSum(Node* head, int k)
{
    // To store current window sum
    int sum = 0;
 
    // To store maximum sum
    int maxSum = 0;
 
    // Pointer to the start of window
    Node* start = head;
 
    // Pointer to the end of window
    Node* end = head;
 
    int i;
 
    // Find the sum of first k nodes
    for (i = 0; i < k; i++) {
        sum += end->data;
        end = end->next;
    }
 
    maxSum = sum;
 
    // Move window by one step and
    // update sum. Node pointed by
    // start is excluded from current
    // window so subtract it. Node
    // pointed by end is added to
    // current window so add its value.
    while (end != NULL) {
 
        // Subtract the starting element
        // from previous window
        sum -= start->data;
        start = start->next;
 
        // Add the element next to the end
        // of previous window
        sum += end->data;
        end = end->next;
 
        // Update the maximum sum so far
        maxSum = max(maxSum, sum);
    }
 
    return maxSum;
}
 
// Driver code
int main()
{
    Node* head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(5);
    head->next->next->next->next->next = newNode(6);
 
    int k = 5;
 
    cout << findMaxSum(head, k);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
// Structure of a node
static class Node
{
    int data;
    Node next;
};
 
// Function to create new node
static Node newNode(int data)
{
    Node node = new Node();
    node.next = null;
    node.data = data;
    return node;
}
 
// Function to return the maximum sum of
// k consecutive nodes
static int findMaxSum(Node head, int k)
{
    // To store current window sum
    int sum = 0;
 
    // To store maximum sum
    int maxSum = 0;
 
    // Pointer to the start of window
    Node start = head;
 
    // Pointer to the end of window
    Node end = head;
 
    int i;
 
    // Find the sum of first k nodes
    for (i = 0; i < k; i++)
    {
        sum += end.data;
        end = end.next;
    }
 
    maxSum = sum;
 
    // Move window by one step and
    // update sum. Node pointed by
    // start is excluded from current
    // window so subtract it. Node
    // pointed by end is added to
    // current window so add its value.
    while (end != null)
    {
 
        // Subtract the starting element
        // from previous window
        sum -= start.data;
        start = start.next;
 
        // Add the element next to the end
        // of previous window
        sum += end.data;
        end = end.next;
 
        // Update the maximum sum so far
        maxSum = Math.max(maxSum, sum);
    }
    return maxSum;
}
 
// Driver code
public static void main(String[] args)
{
    Node head = newNode(1);
    head.next = newNode(2);
    head.next.next = newNode(3);
    head.next.next.next = newNode(4);
    head.next.next.next.next = newNode(5);
    head.next.next.next.next.next = newNode(6);
 
    int k = 5;
    System.out.print(findMaxSum(head, k));
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 implementation of the approach
 
# Node of Linked List
class Node:
     
    def __init__(self, x):
         
        self.data = x
        self.next = None
 
# Function to return the maximum sum of
# k consecutive nodes
def findMaxSum(head, k):
     
    # To store current window sum
    sum = 0
 
    # To store maximum sum
    maxSum = 0
 
    # Pointer to the start of window
    start = head
 
    # Pointer to the end of window
    end = head
     
    # Find the sum of first k nodes
    for i in range(k):
        sum += end.data
        end = end.next
 
    maxSum = sum
 
    # Move window by one step and
    # update sum. Node pointed by
    # start is excluded from current
    # window so subtract it. Node
    # pointed by end is added to
    # current window so add its value.
    while (end != None):
 
        # Subtract the starting element
        # from previous window
        sum -= start.data
        start = start.next
 
        # Add the element next to the end
        # of previous window
        sum += end.data
        end = end.next
 
        # Update the maximum sum so far
        maxSum = max(maxSum, sum)
 
    return maxSum
 
# Driver code
if __name__ == '__main__':
     
    head = Node(1)
    head.next = Node(2)
    head.next.next = Node(3)
    head.next.next.next = Node(4)
    head.next.next.next.next = Node(5)
    head.next.next.next.next.next = Node(6)
 
    k = 5
 
    print(findMaxSum(head, k))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Structure of a node
public class Node
{
    public int data;
    public Node next;
};
 
// Function to create new node
static Node newNode(int data)
{
    Node node = new Node();
    node.next = null;
    node.data = data;
    return node;
}
 
// Function to return the maximum sum of
// k consecutive nodes
static int findMaxSum(Node head, int k)
{
    // To store current window sum
    int sum = 0;
 
    // To store maximum sum
    int maxSum = 0;
 
    // Pointer to the start of window
    Node start = head;
 
    // Pointer to the end of window
    Node end = head;
 
    int i;
 
    // Find the sum of first k nodes
    for (i = 0; i < k; i++)
    {
        sum += end.data;
        end = end.next;
    }
 
    maxSum = sum;
 
    // Move window by one step and
    // update sum. Node pointed by
    // start is excluded from current
    // window so subtract it. Node
    // pointed by end is added to
    // current window so add its value.
    while (end != null)
    {
 
        // Subtract the starting element
        // from previous window
        sum -= start.data;
        start = start.next;
 
        // Add the element next to the end
        // of previous window
        sum += end.data;
        end = end.next;
 
        // Update the maximum sum so far
        maxSum = Math.Max(maxSum, sum);
    }
    return maxSum;
}
 
// Driver code
public static void Main(String[] args)
{
    Node head = newNode(1);
    head.next = newNode(2);
    head.next.next = newNode(3);
    head.next.next.next = newNode(4);
    head.next.next.next.next = newNode(5);
    head.next.next.next.next.next = newNode(6);
 
    int k = 5;
    Console.Write(findMaxSum(head, k));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// JavaScript implementation of the approach   
// Structure of a node
class Node {
    constructor(val) {
        this.data = val;
        this.next = null;
    }
}
 
    // Function to create new node
    function newNode(data) {
        var node = new Node();
        node.next = null;
        node.data = data;
        return node;
    }
 
    // Function to return the maximum sum of
    // k consecutive nodes
    function findMaxSum(head , k) {
        // To store current window sum
        var sum = 0;
 
        // To store maximum sum
        var maxSum = 0;
 
        // Pointer to the start of window
        var start = head;
 
        // Pointer to the end of window
        var end = head;
 
        var i;
 
        // Find the sum of first k nodes
        for (i = 0; i < k; i++) {
            sum += end.data;
            end = end.next;
        }
 
        maxSum = sum;
 
        // Move window by one step and
        // update sum. Node pointed by
        // start is excluded from current
        // window so subtract it. Node
        // pointed by end is added to
        // current window so add its value.
        while (end != null) {
 
            // Subtract the starting element
            // from previous window
            sum -= start.data;
            start = start.next;
 
            // Add the element next to the end
            // of previous window
            sum += end.data;
            end = end.next;
 
            // Update the maximum sum so far
            maxSum = Math.max(maxSum, sum);
        }
        return maxSum;
    }
 
    // Driver code
     
        var head = newNode(1);
        head.next = newNode(2);
        head.next.next = newNode(3);
        head.next.next.next = newNode(4);
        head.next.next.next.next = newNode(5);
        head.next.next.next.next.next = newNode(6);
 
        var k = 5;
        document.write(findMaxSum(head, k));
 
// This code contributed by aashish1995
 
</script>
Output: 
20

 

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




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!