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Delete continuous nodes with sum K from a given linked list
  • Difficulty Level : Expert
  • Last Updated : 30 Dec, 2020

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:  

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 0.
  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++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// A Linked List Node
struct ListNode {
    int val;
    ListNode* next;
 
    // Contructor
    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 continuos 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;
 
            // Delete all the node
            // traverse till current node
            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;
 
            // Update sum to zero
            sum = 0;
        }
 
        // 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;
}

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Java

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// Java program for the above approach
import java.io.*;
import java.util.*;
 
// A Linked List Node
class ListNode {
    int val;
    ListNode next;
 
    // Contructor
    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 continuos 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;
 
                // 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;
 
                // Update sum to zero
                sum = 0;
            }
 
            // 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.

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Python3

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# 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 continuos 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;
  
            # 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;
  
            # Update sum to zero
            sum = 0
  
        # 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

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C#

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// 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;
     
    // Contructor
    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 continuos 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;
 
            // 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;
 
            // Update sum to zero
            sum = 0;
        }
 
        // 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

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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.
 

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