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Find modular node in a linked list in C++
In this problem, we are given a singly linked list LL and a number k. Our task is to Find modular node in a linked list.
Problem Description − we need to find the last node of the linked list for which the index is divisible by k i.e. i % k == 0.
Let’s take an example to understand the problem,
Input
ll = 3 -> 1 -> 9 -> 6 -> 8 -> 2, k = 4
Output
6
Explanation
The element 6 has index 4, which is divisible by 4.
Solution Approach
A simple solution to the problem is by creating a counter to count the elements of the linked list and store the modular node i.e. node for which i % k == 0, and update for all values satisfying the condition till n.
Program to illustrate the working of our solution,
Example
#include <iostream>
using namespace std;
struct Node {
int data;
Node* next;
};
Node* newNode(int data) {
Node* new_node = new Node;
new_node->data = data;
new_node->next = NULL;
return new_node;
}
Node* findModularNodeLL(Node* head, int k) {
if (k <= 0 || head == NULL)
return NULL;
int i = 1;
Node* modNode = NULL;
for (Node* currNode = head; currNode != NULL; currNode =
currNode->next) {
if (i % k == 0)
modNode = currNode;
i++;
}
return modNode;
}
int main() {
Node* head = newNode(3);
head->next = newNode(1);
head->next->next = newNode(9);
head->next->next->next = newNode(6);
head->next->next->next->next = newNode(8);
head->next->next->next->next->next = newNode(2);
int k = 4;
Node* modularNode = findModularNodeLL(head, k);
cout<<"The Modular node of linked list is ";
if (modularNode != NULL)
cout<<modularNode->data;
else
cout<<"Not found!";
return 0;
}Output
The Modular node of linked list is 6
Updated on: 12-Mar-2021
133 Views
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