Replace each node with its Surpasser Count in Linked List
Given LinkedList, replace each node’s value with its surpasser count. That is the count of elements which are greater towards its right.
Examples:
Input : 10->12->5->40->21->70->1->49->37->NULL
Output : 6->5->5->2->3->0->2->0->0->NULL
Explanation :
Element in the first node is 10 and the number of elements to the right of the node that are greater than 10 is 6. Therefore replace the node with 6.
Element in the first node is 12 and the number of elements to the right of the node that are greater than 12 is 5. Therefore replace the node with 5.
Similarly, replace for all the elements in the list.Input : 5->4->6->3->2->NULL
Output : 1->1->0->0->0->NULL
Simple Approach
- Take two pointers p and x. The pointer p is used to traverse the list and x is used to traverse the right half of the list for every node.
- Initialize a variable count to count the nodes greater than the current nodes.
- Traverse through all the nodes in the list using the pointer p.
- Initialize the count to 0.
- Initialize the pointer x to point the current node p.
- Count the number of nodes that are greater than the current node.
- Replace the current node with the count.
- Repeat step 4 until the list is traversed completely.
Below is the implementation of the above approach:
C++
// C++ program to replace the nodes // with their surpasser count #include <bits/stdc++.h> using namespace std; // A linked list node struct Node { int data; struct Node* next; }; // Utility function to create a new Node Node* newNode(int data) { Node* temp = new Node; temp->data = data; temp->next = NULL; return temp; } // Function to display the linked list void printList(Node* node) { while (node != NULL) { cout << node->data << " "; node = node->next; } } // Function to check Surpasser Count void replaceNodes(Node* head) { // Pointer used to traverse through // all the nodes in the list Node* p = head; // Pointer used to traverse through the right // elements to count the greater elements Node* x = NULL; // Variable to count the number of // elements greater than the // current element on right int count = 0; int i; // Traverse through all the elements // in the list while (p != NULL) { count = 0; i = 0; // Initialize x to current node x = p; // Check or count the number of nodes // that are greater than the current // node on right while (x != NULL) { if (x->data > p->data) count++; x = x->next; } // Replace the node data with the // count of elements greater than // the current element p->data = count; p = p->next; } } // Driver code int main() { // Creating the linked list Node* head = newNode(10); head->next = newNode(12); head->next->next = newNode(5); head->next->next->next = newNode(40); head->next->next->next->next = newNode(21); head->next->next->next->next->next = newNode(70); head->next->next->next->next->next->next = newNode(1); head->next->next->next->next->next->next->next = newNode(49); head->next->next->next->next->next->next->next->next = newNode(37); replaceNodes(head); printList(head); return 0; } |
Java
// Java program to replace the nodes // with their surpasser count class GFG { // A linked list node static class Node { int data; Node next; }; // Utility function to create a new Node static Node newNode(int data) { Node temp = new Node(); temp.data = data; temp.next = null; return temp; } // Function to display the linked list static void printList(Node node) { while (node != null) { System.out.print(node.data+" "); node = node.next; } } // Function to check Surpasser Count static void replaceNodes(Node head) { // Pointer used to traverse through // all the nodes in the list Node p = head; // Pointer used to traverse through the right // elements to count the greater elements Node x = null; // Variable to count the number of // elements greater than the // current element on right int count = 0; int i; // Traverse through all the elements // in the list while (p != null) { count = 0; i = 0; // Initialize x to current node x = p; // Check or count the number of nodes // that are greater than the current // node on right while (x != null) { if (x.data > p.data) count++; x = x.next; } // Replace the node data with the // count of elements greater than // the current element p.data = count; p = p.next; } } // Driver code public static void main(String[] args) { // Creating the linked list Node head = newNode(10); head.next = newNode(12); head.next.next = newNode(5); head.next.next.next = newNode(40); head.next.next.next.next = newNode(21); head.next.next.next.next.next = newNode(70); head.next.next.next.next.next.next = newNode(1); head.next.next.next.next.next.next.next = newNode(49); head.next.next.next.next.next.next.next.next = newNode(37); replaceNodes(head); printList(head); } } // This code has been contributed by 29AjayKumar |
Python3
# Python3 program to replace the nodes # with their surpasser count # A linked list node class Node: def __init__(self, data): self.data = data self.next = None # Function to display the linked list def printList(node): while node != None: print(node.data, end = " ") node = node.next # Function to check Surpasser Count def replaceNodes(head): # Pointer used to traverse through # all the nodes in the list p = head # Pointer used to traverse through # the right elements to count the # greater elements x = None # Variable to count the number of # elements greater than the # current element on right count = 0 # Traverse through all the elements # in the list while p != None: count = 0 # Initialize x to current node x = p # Check or count the number of nodes # that are greater than the current # node on right while x != None: if x.data > p.data: count += 1 x = x.next # Replace the node data with the # count of elements greater than # the current element p.data = count p = p.next # Driver code if __name__ == "__main__": # Creating the linked list head = Node(10) head.next = Node(12) head.next.next = Node(5) head.next.next.next = Node(40) head.next.next.next.next = Node(21) head.next.next.next.next.next = Node(70) head.next.next.next.next.next.next = Node(1) head.next.next.next.next.next.next.next = Node(49) head.next.next.next.next.next.next.next.next = Node(37) replaceNodes(head) printList(head) # This code is contributed by Rituraj Jain |
C#
// C# program to replace the nodes // with their surpasser count using System; class GFG { // A linked list node public class Node { public int data; public Node next; }; // Utility function to create a new Node static Node newNode(int data) { Node temp = new Node(); temp.data = data; temp.next = null; return temp; } // Function to display the linked list static void printList(Node node) { while (node != null) { Console.Write(node.data + " "); node = node.next; } } // Function to check Surpasser Count static void replaceNodes(Node head) { // Pointer used to traverse through // all the nodes in the list Node p = head; // Pointer used to traverse through the right // elements to count the greater elements Node x = null; // Variable to count the number of // elements greater than the // current element on right int count = 0; int i; // Traverse through all the elements // in the list while (p != null) { count = 0; i = 0; // Initialize x to current node x = p; // Check or count the number of nodes // that are greater than the current // node on right while (x != null) { if (x.data > p.data) count++; x = x.next; } // Replace the node data with the // count of elements greater than // the current element p.data = count; p = p.next; } } // Driver code public static void Main() { // Creating the linked list Node head = newNode(10); head.next = newNode(12); head.next.next = newNode(5); head.next.next.next = newNode(40); head.next.next.next.next = newNode(21); head.next.next.next.next.next = newNode(70); head.next.next.next.next.next.next = newNode(1); head.next.next.next.next.next.next.next = newNode(49); head.next.next.next.next.next.next.next.next = newNode(37); replaceNodes(head); printList(head); } } /* This code contributed by PrinciRaj1992 */ |
Output:
6 5 5 2 3 0 2 0 0
Time Complexity: O(N2) where N is the number of nodes in the linked list.
Auxiliary Space: O(1)
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