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 nodestruct Node { int data; struct Node* next;};// Utility function to create a new NodeNode* newNode(int data){ Node* temp = new Node; temp->data = data; temp->next = NULL; return temp;}// Function to display the linked listvoid printList(Node* node){ while (node != NULL) { cout << node->data << " "; node = node->next; }}// Function to check Surpasser Countvoid 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 codeint 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 countclass GFG {// A linked list nodestatic class Node { int data; Node next;}; // Utility function to create a new Nodestatic Node newNode(int data){ Node temp = new Node(); temp.data = data; temp.next = null; return temp;} // Function to display the linked liststatic void printList(Node node){ while (node != null) { System.out.print(node.data+" "); node = node.next; }} // Function to check Surpasser Countstatic 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 codepublic 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 countusing System;class GFG {// A linked list nodepublic class Node{ public int data; public Node next;};// Utility function to create a new Nodestatic Node newNode(int data){ Node temp = new Node(); temp.data = data; temp.next = null; return temp;}// Function to display the linked liststatic void printList(Node node){ while (node != null) { Console.Write(node.data + " "); node = node.next; }}// Function to check Surpasser Countstatic 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 codepublic 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 */ |
Javascript
<script>// JavaScript program to replace the nodes// with their surpasser count// A linked list nodeclass Node { constructor() { this.data = 0; this.next = null; } } // Utility function to create a new Nodefunction newNode( data){ var temp = new Node(); temp.data = data; temp.next = null; return temp;}// Function to display the linked listfunction printList( node){ while (node != null) { document.write(node.data+" "); node = node.next; }}// Function to check Surpasser Countfunction replaceNodes( head){ // Pointer used to traverse through // all the nodes in the list var p = head; // Pointer used to traverse through the right // elements to count the greater elements var x = null; // Variable to count the number of // elements greater than the // current element on right let count = 0; let 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// Creating the linked listvar 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);</script> |
Output
6 5 5 2 3 0 2 0 0
Complexity Analysis:
- Time Complexity: O(N2) where N is the number of nodes in the linked list.
- Auxiliary Space: O(1)
Another Approach with PBDS and Hash map
We can list down our requirements and approach to observe and find a solution. Our approach:
- Traverse the linked list from backwards (so we do not concern about elements to the left of current)
- Store the current element in a sorted data structure (any point of time, elements are sorted in our data structure)
- Find total number of elements greater than the current element in our sorted data structure and replace the current element with it
To traverse the linked list backwards, we will use recursion.
We can use PBDS. With PBDS we can find elements strictly smaller than a particular key. With strictly smaller, we can add current element and subtract it from all the elements. So basically:
Greater elements than current = Total Elements in PBDS – (Query in PBDS to find elements strictly smaller) – (Current element + element equal to current)
PBDS does not allow duplicate elements. But for the counting part, we need duplicate elements. So we will insert a pair in PBDS (first = element, second = index) to make each entry unique. After that, to find total elements equal to the current element, we will use a hash map. In hash map we will store total occurrences of an element (a basic integer to integer map).
NOTE: We can also construct our own data structure that fulfills the current requirements as we stated. Insert for duplicates in sorted order and query for elements just greater than the key. But we will use C++ STL DS to do it in this post.
Implementation:
C++
#include <iostream>#include <unordered_map>// library for PBS#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>// PBDS of pair<int, int>#define oset tree<pair<int, int>, null_type,less<pair<int, int>>, rb_tree_tag,tree_order_statistics_node_update>using namespace std; using namespace __gnu_pbds;// Linked list storage class Node { public: int value; Node* next; Node(int value) { this->value = value; next = NULL; }};/*head => head is the head of the linked listos => unordered_set or PBDSmp => Hash map to store count of each elementcount => count of element from backwards*/void solve(Node* head, oset& os, unordered_map<int, int>& mp, int& count) { if(head == NULL) return; solve(head->next, os, mp, count); // we first call our recursion and then compute to traverse the linked list backwards on recursion return // increment count backwards count++; // inserting elements into PBDS with indexes starting from backwards os.insert({head->value, count}); // mapping elements to count mp[head->value]++; // formula: total elements in PBDS - total number of elements equal to current - elements smaller than current int numberOfElements = count - mp[head->value] - os.order_of_key({head->value, -1}); // replace the linked list current element to the number of elements greater than the current head->value = numberOfElements;}void printList(Node* head) { while(head) { cout << head->value << (head->next ? "->" : ""); head = head->next; } cout << "\n";}int main() { Node* head = new Node(43); head->next = new Node(65); head->next->next = new Node(12); head->next->next->next = new Node(46); head->next->next->next->next = new Node(68); head->next->next->next->next->next = new Node(85); head->next->next->next->next->next->next = new Node(59); head->next->next->next->next->next->next->next = new Node(85); head->next->next->next->next->next->next->next->next = new Node(37); oset os; unordered_map<int, int> mp; int count = 0; printList(head); solve(head, os, mp, count); printList(head); return 0; } |
Output
43->65->12->46->68->85->59->85->37 6->3->6->4->2->0->1->0->0
Complexity Analysis:
- Time complexity: O(n*log(n)) as insert and order_of_key of PBDS uses log(n) and hash map takes O(1).
- Space complexity: O(n) as we store the elements in PBDS and hash to count.
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