Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Delete continuous nodes with sum K from a given linked list

  • Difficulty Level : Hard
  • Last Updated : 30 Nov, 2021

Given a singly linked list and an integer K, the task is to remove all the continuous set of nodes whose sum is K from the given linked list. Print the updated linked list after the removal. If no such deletion can occur, print the original 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: Linked List: 1 -> 2 -> -3 -> 3 -> 1, K = 3 



Output: -3 -> 1 

Explanation: 

The nodes with continuous sum 3 are: 

1) 1 -> 2 

2) 3 

Therefore, after removing these chain of nodes Linked List becomes: -3-> 1

Input: Linked List: 1 -> 1 -> -3 -> -3 -> -2, K = 5 
Output: 1 -> 1 -> -3 -> -3 -> -2 
Explanation: 
No continuous nodes exits with sum K 

Approach:  

  1. Append Node with value zero at the starting of the linked list.
  2. Traverse the given linked list.
  3. During traversal store the sum of the node value till that node with the reference of the current node in an unordered_map.
  4. If there is Node with value (sum – K) present in the unordered_map then delete all the nodes from the node corresponding to value (sum – K) stored in map to the current node and update the sum as (sum – K).
  5. If there is no Node with value (sum – K) present in the unordered_map, then stored the current sum with node in the map.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// A Linked List Node
struct ListNode {
    int val;
    ListNode* next;
 
    // Constructor
    ListNode(int x)
        : val(x)
        , next(NULL)
    {
    }
};
 
// Function to create Node
ListNode* getNode(int data)
{
    ListNode* temp;
    temp = (ListNode*)malloc(sizeof(ListNode));
    temp->val = data;
    temp->next = NULL;
    return temp;
}
 
// Function to print the Linked List
void printList(ListNode* head)
{
    while (head->next) {
        cout << head->val << " -> ";
        head = head->next;
    }
    printf("%d", head->val);
}
 
// Function that removes continuous nodes
// whose sum is K
ListNode* removeZeroSum(ListNode* head, int K)
{
    // Root node initialise to 0
    ListNode* root = new ListNode(0);
 
    // Append at the front of the given
    // Linked List
    root->next = head;
 
    // Map to store the sum and reference
    // of the Node
    unordered_map<int, ListNode*> umap;
 
    umap[0] = root;
 
    // To store the sum while traversing
    int sum = 0;
 
    // Traversing the Linked List
    while (head != NULL) {
 
        // Find sum
        sum += head->val;
 
        // If found value with (sum - K)
        if (umap.find(sum - K) != umap.end()) {
 
            ListNode* prev = umap[sum - K];
            ListNode* start = prev;
 
 
             
 
            // Update sum
            sum = sum - K;
            int aux = sum;
            // Traverse till current head
            while (prev != head) {
                prev = prev->next;
                aux += prev->val;
                if (prev != head) {
                    umap.erase(aux);
                }
            }
 
            // Update the start value to
            // the next value of current head
            start->next = head->next;
        }
 
        // If (sum - K) value not found
        else if (umap.find(sum) == umap.end()) {
            umap[sum] = head;
        }
 
        head = head->next;
    }
 
    // Return the value of updated
    // head node
    return root->next;
}
 
// Driver Code
int main()
{
    // head Node
    ListNode* head;
 
    // Create Linked List
    head = getNode(1);
    head->next = getNode(2);
    head->next->next = getNode(-3);
    head->next->next->next = getNode(3);
    head->next->next->next->next = getNode(1);
 
    // Given sum K
    int K = 5;
 
    // Function call to get head node
    // of the updated Linked List
    head = removeZeroSum(head, K);
 
    // Print the updated Linked List
    if (head != NULL)
        printList(head);
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
// A Linked List Node
class ListNode {
    int val;
    ListNode next;
 
    // Constructor
    ListNode(int val)
    {
        this.val = val;
        this.next = null;
    }
}
 
class GFG {
 
    // Function to create Node
    static ListNode getNode(int data)
    {
        ListNode temp = new ListNode(data);
        return temp;
    }
 
    // Function to print the Linked List
    static void printList(ListNode head)
    {
        while (head.next != null) {
            System.out.print(head.val + " -> ");
            head = head.next;
        }
        System.out.print(head.val);
    }
 
    // Function that removes continuous nodes
    // whose sum is K
    static ListNode removeZeroSum(ListNode head, int K)
    {
        // Root node initialise to 0
        ListNode root = new ListNode(0);
 
        // Append at the front of the given
        // Linked List
        root.next = head;
 
        // Map to store the sum and reference
        // of the Node
        Map<Integer, ListNode> umap
            = new HashMap<Integer, ListNode>();
 
        umap.put(0, root);
 
        // To store the sum while traversing
        int sum = 0;
 
        // Traversing the Linked List
        while (head != null) {
 
            // Find sum
            sum += head.val;
 
            // If found value with (sum - K)
            if (umap.containsKey(sum - K)) {
 
                ListNode prev = umap.get(sum - K);
                ListNode start = prev;
 
                // Delete all the node
                // traverse till current node
                int aux = sum;
 
                // Update sum
                sum = sum - K;
 
                // Traverse till current head
                while (prev != head) {
                    prev = prev.next;
                    aux += prev.val;
                    if (prev != head) {
                        umap.remove(aux);
                    }
                }
 
                // Update the start value to
                // the next value of current head
                start.next = head.next;
            }
 
            // If (sum - K) value not found
            else if (!umap.containsKey(sum)) {
                umap.put(sum, head);
            }
 
            head = head.next;
        }
 
        // Return the value of updated
        // head node
        return root.next;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        // head Node
        ListNode head;
 
        // Create Linked List
        head = getNode(1);
        head.next = getNode(2);
        head.next.next = getNode(-3);
        head.next.next.next = getNode(3);
        head.next.next.next.next = getNode(1);
 
        // Given sum K
        int K = 5;
 
        // Function call to get head node
        // of the updated Linked List
        head = removeZeroSum(head, K);
 
        // Print the updated Linked List
        if (head != null)
            printList(head);
    }
}
 
// This code is contributed by jitin.

Python3




# Python3 program for the above approach
 
# A Linked List Node
class ListNode:
    def __init__(self, val):
        self.val = val
        self.next = None
 
# Function to create Node
def getNode(data):
    temp = ListNode(data)
    temp.next = None
    return temp
 
# Function to print the Linked List
def printList(head):
    while (head.next):
        print(head.val, end=' -> ')
        head = head.next
    print(head.val, end='')
 
# Function that removes continuous nodes
# whose sum is K
def removeZeroSum(head, K):
 
    # Root node initialise to 0
    root = ListNode(0)
 
    # Append at the front of the given
    # Linked List
    root.next = head
 
    # Map to store the sum and reference
    # of the Node
    umap = dict()
 
    umap[0] = root
 
    # To store the sum while traversing
    sum = 0
 
    # Traversing the Linked List
    while (head != None):
 
        # Find sum
        sum += head.val
 
        # If found value with (sum - K)
        if ((sum - K) in umap):
 
            prev = umap[sum - K]
            start = prev
 
            # Delete all the node
            # traverse till current node
            aux = sum
 
            # Update sum
            sum = sum - K
 
            # Traverse till current head
            while (prev != head):
                prev = prev.next
                aux += prev.val
                if (prev != head):
                    umap.remove(aux)
 
            # Update the start value to
            # the next value of current head
            start.next = head.next
 
        # If (sum - K) value not found
        else:
            umap[sum] = head
 
        head = head.next
 
    # Return the value of updated
    # head node
    return root.next
 
 
# Driver Code
if __name__ == '__main__':
 
    # Create Linked List
    head = getNode(1)
    head.next = getNode(2)
    head.next.next = getNode(-3)
    head.next.next.next = getNode(3)
    head.next.next.next.next = getNode(1)
 
    # Given sum K
    K = 5
 
    # Function call to get head node
    # of the updated Linked List
    head = removeZeroSum(head, K)
 
    # Print the updated Linked List
    if(head != None):
        printList(head)
 
    # This code is contributed by pratham76

C#




// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
// A Linked List Node
class ListNode {
    public int val;
    public ListNode next;
 
    // Constructor
    public ListNode(int val)
    {
        this.val = val;
        this.next = null;
    }
}
 
class GFG {
 
    // Function to create Node
    static ListNode getNode(int data)
    {
        ListNode temp = new ListNode(data);
        return temp;
    }
 
    // Function to print the Linked List
    static void printList(ListNode head)
    {
        while (head.next != null) {
            Console.Write(head.val + " -> ");
            head = head.next;
        }
        Console.Write(head.val);
    }
 
    // Function that removes continuous nodes
    // whose sum is K
    static ListNode removeZeroSum(ListNode head, int K)
    {
 
        // Root node initialise to 0
        ListNode root = new ListNode(0);
 
        // Append at the front of the given
        // Linked List
        root.next = head;
 
        // Map to store the sum and reference
        // of the Node
        Dictionary<int, ListNode> umap
            = new Dictionary<int, ListNode>();
 
        umap.Add(0, root);
 
        // To store the sum while traversing
        int sum = 0;
 
        // Traversing the Linked List
        while (head != null) {
 
            // Find sum
            sum += head.val;
 
            // If found value with (sum - K)
            if (umap.ContainsKey(sum - K)) {
                ListNode prev = umap[sum - K];
                ListNode start = prev;
 
                // Delete all the node
                // traverse till current node
                int aux = sum;
 
                // Update sum
                sum = sum - K;
 
                // Traverse till current head
                while (prev != head) {
                    prev = prev.next;
                    aux += prev.val;
 
                    if (prev != head) {
                        umap.Remove(aux);
                    }
                }
 
                // Update the start value to
                // the next value of current head
                start.next = head.next;
            }
 
            // If (sum - K) value not found
            else if (!umap.ContainsKey(sum)) {
                umap.Add(sum, head);
            }
 
            head = head.next;
        }
 
        // Return the value of updated
        // head node
        return root.next;
    }
 
    // Driver code
    public static void Main(string[] args)
    {
 
        // head Node
        ListNode head;
 
        // Create Linked List
        head = getNode(1);
        head.next = getNode(2);
        head.next.next = getNode(-3);
        head.next.next.next = getNode(3);
        head.next.next.next.next = getNode(1);
 
        // Given sum K
        int K = 5;
 
        // Function call to get head node
        // of the updated Linked List
        head = removeZeroSum(head, K);
 
        // Print the updated Linked List
        if (head != null)
            printList(head);
    }
}
 
// This code is contributed by rutvik_56
Output
1 -> 2 -> -3 -> 3 -> 1

Time Complexity: O(N), where N is the number of Node in the Linked List. 
Auxiliary Space Complexity: O(N), where N is the number of Node in the Linked List.




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!