# Replace each node with its Surpasser Count in Linked List

• Difficulty Level : Medium
• Last Updated : 24 Nov, 2021

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

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Simple Approach

1. 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.
2. Initialize a variable count to count the nodes greater than the current nodes.
3. 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.
4. 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 ``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 */`

## Javascript

 ``
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)

Another Approach with PBDS and Hash map

We can list down our requirements and approach to observe and find a solution. Our approach:

1. Traverse the linked list from backwards (so we do not concern about elements to the left of current)

2. Store the current element in a sorted data structure (any point of time, elements are sorted in our data structure)

3. 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 substract 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.

## C++

 `#include ``#include ` `// library for PBS``#include ``#include ` `// PBDS of pair``#define oset tree, null_type,less>, 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 list``os => unordered_set or PBDS``mp => Hash map to store count of each element``count => 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``    ``// increament 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```

Space complexity of our algorithm is O(n) as we store the elements in PBDS and hash to count. Time complexity is O(n*log(n)) as insert and order_of_key of PBDS  uses log(n) and hash map takes O(1).

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