Given a sequence of numbers in the form of a linked list lis. Find the length of the Longest Increasing Subsequence(LIS) of the given Linked List.
Examples:
Input: list = 3->10->2->1->20
Output: 3
Explanation: The longest increasing subsequence is 3->10-> 20Input: list = 3-> 2
Output: 1
Explanation: The longest increasing subsequence are 3 and 2Input: list = 50->3->10->7->40->80
Output: Length of LIS = 4
Explanation: The longest increasing subsequence is {3->7->40->80} or {3->10->40->80}
Approach: The basic intuition of the solution is to start iterating from the first node to the end of linked list .In the process of moving calculate length of LIS ending at every node and store it in a count variable. Finally, calculate maximum count value among all nodes. Follow the steps mentioned below to solve the problem:
- Traverse the linked list from the starting node.
- LIS length of a linked list with one node is 1 .So we initialize every node count variable to 1.
- For every ith node traverse the first (i-1) nodes and do the following:
- if value of current node is greater than the value of previous node, extend sequence length.
- As maximum length ending with current node is required. select node from first (i-1) nodes which satisfy the previous condition and have maximum count value .
- Once all the nodes are traversed following the above procedure .find maximum count value among all nodes.
- The maximum count value is the required length of the LIS.
Below is the implementation of the above approach:
C++
// C++ program to find LIS on LinkedList#include <bits/stdc++.h>using namespace std;// Structure of a nodeclass Node {public: int data; struct Node* next; // "count" variable is to keep track // of LIS_LENGTH ending with // that particular element int count;};// Function to find the length of the LISint LIS(struct Node* head){ // If linked list is empty length is 0 if (head == NULL) return 0; // If linked list has only one node // LIS length is 1 if (head->next == NULL) return 1; Node* curr_p = head->next; // This loop calculates what is // LIS_LENGTH ending with each and // every node and stores in // curr->count variable while (curr_p != NULL) { int maxi = 0; Node* prev_p = head; // This while loop traverse all nodes // before curr_p and finds which node // to extend so that maximum LIS // length ending with curr_P can be while (prev_p != curr_p) { // Only extend if present data // greater than previous value if (curr_p->data > prev_p->data) { if (prev_p->count > maxi) { maxi = prev_p->count; } } prev_p = prev_p->next; } curr_p->count = 1 + maxi; curr_p = curr_p->next; } int LIS_length = 0; curr_p = head; // Finding Maximum LIS_LENGTH while (curr_p != NULL) { if (LIS_length < curr_p->count) { LIS_length = curr_p->count; } curr_p = curr_p->next; } return LIS_length;}// Function to push a node in linked listvoid push(struct Node** head_ref, int new_data){ // Allocate node Node* new_node = new Node(); // Put in the data new_node->data = new_data; // Link the old list with the new node new_node->next = (*head_ref); // Assign count value to 1 new_node->count = 1; // Move the head to point the new node (*head_ref) = new_node;}// Driver codeint main(){ // Start with the empty list struct Node* head = NULL; // Create a linked list // Created linked list will be // 3->10->2->1->20 push(&head, 20); push(&head, 1); push(&head, 2); push(&head, 10); push(&head, 3); // Call LIS function which calculates // LIS of Linked List int ans = LIS(head); cout << ans; return 0;} |
Java
// Java program to find LIS on LinkedListimport java.util.*;class GFG{// Structure of a nodestatic class Node { int data; Node next; // "count" variable is to keep track // of LIS_LENGTH ending with // that particular element int count;};// Function to find the length of the LISstatic int LIS(Node head){ // If linked list is empty length is 0 if (head == null) return 0; // If linked list has only one node // LIS length is 1 if (head.next == null) return 1; Node curr_p = head.next; // This loop calculates what is // LIS_LENGTH ending with each and // every node and stores in // curr.count variable while (curr_p != null) { int maxi = 0; Node prev_p = head; // This while loop traverse all nodes // before curr_p and finds which node // to extend so that maximum LIS // length ending with curr_P can be while (prev_p != curr_p) { // Only extend if present data // greater than previous value if (curr_p.data > prev_p.data) { if (prev_p.count > maxi) { maxi = prev_p.count; } } prev_p = prev_p.next; } curr_p.count = 1 + maxi; curr_p = curr_p.next; } int LIS_length = 0; curr_p = head; // Finding Maximum LIS_LENGTH while (curr_p != null) { if (LIS_length < curr_p.count) { LIS_length = curr_p.count; } curr_p = curr_p.next; } return LIS_length;}// Function to push a node in linked liststatic Node push(Node head_ref, int new_data){ // Allocate node Node new_node = new Node(); // Put in the data new_node.data = new_data; // Link the old list with the new node new_node.next = head_ref; // Assign count value to 1 new_node.count = 1; // Move the head to point the new node head_ref = new_node; return head_ref;}// Driver codepublic static void main(String[] args){ // Start with the empty list Node head = null; // Create a linked list // Created linked list will be // 3.10.2.1.20 head = push(head, 20); head = push(head, 1); head = push(head, 2); head = push(head, 10); head = push(head, 3); // Call LIS function which calculates // LIS of Linked List int ans = LIS(head); System.out.print(ans);}}// This code is contributed by 29AjayKumar |
Python3
# Python code for the above approach# Structure of a nodeclass Node: def __init__(self, d): self.data = d self.next = None self.count = 1 # "count" variable is to keep track # of LIS_LENGTH ending with # that particular element# Function to find the length of the LISdef LIS(head): # If linked list is empty length is 0 if (head == None): return 0 # If linked list has only one node # LIS length is 1 if (head.next == None): return 1 curr_p = head.next # This loop calculates what is # LIS_LENGTH ending with each and # every node and stores in # curr.count variable while (curr_p != None): maxi = 0 prev_p = head # This while loop traverse all nodes # before curr_p and finds which node # to extend so that maximum LIS # length ending with curr_P can be while (prev_p != curr_p): # Only extend if present data # greater than previous value if (curr_p.data > prev_p.data): if (prev_p.count > maxi): maxi = prev_p.count prev_p = prev_p.next curr_p.count = 1 + maxi curr_p = curr_p.next LIS_length = 0 curr_p = head # Finding Maximum LIS_LENGTH while (curr_p != None): if (LIS_length < curr_p.count): LIS_length = curr_p.count curr_p = curr_p.next return LIS_length# Driver code# Start with the empty list# Create a linked list# Created linked list will be# 3->10->2->1->20head = Node(3)head.next = Node(10)head.next.next = Node(2)head.next.next.next = Node(1)head.next.next.next.next = Node(20)# Call LIS function which calculates# LIS of Linked Listans = LIS(head)print(ans)# This code is contributed by Saurabh Jaiswal |
C#
// C# program to find LIS on Listusing System;using System.Collections.Generic;public class GFG { // Structure of a node public class Node { public int data; public Node next; // "count" variable is to keep track // of LIS_LENGTH ending with // that particular element public int count; }; // Function to find the length of the LIS static int LIS(Node head) { // If linked list is empty length is 0 if (head == null) return 0; // If linked list has only one node // LIS length is 1 if (head.next == null) return 1; Node curr_p = head.next; // This loop calculates what is // LIS_LENGTH ending with each and // every node and stores in // curr.count variable while (curr_p != null) { int maxi = 0; Node prev_p = head; // This while loop traverse all nodes // before curr_p and finds which node // to extend so that maximum LIS // length ending with curr_P can be while (prev_p != curr_p) { // Only extend if present data // greater than previous value if (curr_p.data > prev_p.data) { if (prev_p.count > maxi) { maxi = prev_p.count; } } prev_p = prev_p.next; } curr_p.count = 1 + maxi; curr_p = curr_p.next; } int LIS_length = 0; curr_p = head; // Finding Maximum LIS_LENGTH while (curr_p != null) { if (LIS_length < curr_p.count) { LIS_length = curr_p.count; } curr_p = curr_p.next; } return LIS_length; } // Function to push a node in linked list static Node push(Node head_ref, int new_data) { // Allocate node Node new_node = new Node(); // Put in the data new_node.data = new_data; // Link the old list with the new node new_node.next = head_ref; // Assign count value to 1 new_node.count = 1; // Move the head to point the new node head_ref = new_node; return head_ref; } // Driver code public static void Main(String[] args) { // Start with the empty list Node head = null; // Create a linked list // Created linked list will be // 3.10.2.1.20 head = push(head, 20); head = push(head, 1); head = push(head, 2); head = push(head, 10); head = push(head, 3); // Call LIS function which calculates // LIS of Linked List int ans = LIS(head); Console.Write(ans); }}// This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript code for the above approach // Structure of a node class Node { constructor(d) { this.data = d; this.next = null; this.count = 1; } // "count" variable is to keep track // of LIS_LENGTH ending with // that particular element }; // Function to find the length of the LIS function LIS(head) { // If linked list is empty length is 0 if (head == null) return 0; // If linked list has only one node // LIS length is 1 if (head.next == null) return 1; let curr_p = head.next; // This loop calculates what is // LIS_LENGTH ending with each and // every node and stores in // curr.count variable while (curr_p != null) { let maxi = 0; let prev_p = head; // This while loop traverse all nodes // before curr_p and finds which node // to extend so that maximum LIS // length ending with curr_P can be while (prev_p != curr_p) { // Only extend if present data // greater than previous value if (curr_p.data > prev_p.data) { if (prev_p.count > maxi) { maxi = prev_p.count; } } prev_p = prev_p.next; } curr_p.count = 1 + maxi; curr_p = curr_p.next; } let LIS_length = 0; curr_p = head; // Finding Maximum LIS_LENGTH while (curr_p != null) { if (LIS_length < curr_p.count) { LIS_length = curr_p.count; } curr_p = curr_p.next; } return LIS_length; } // Driver code // Start with the empty list // Create a linked list // Created linked list will be // 3->10->2->1->20 let head = new Node(3); head.next = new Node(10); head.next.next = new Node(2); head.next.next.next = new Node(1); head.next.next.next.next = new Node(20); // Call LIS function which calculates // LIS of Linked List let ans = LIS(head); document.write(ans); // This code is contributed by Potta Lokesh </script> |
Output
3
Time Complexity: O(N * N) where N is the length of the linked list
Auxiliary Space: O(N)
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