Construction of Longest Increasing Subsequence(LIS) and printing LIS sequence

Java code:

The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence
of a given sequence such that all elements of the subsequence are sorted in increasing order.
For example, the length of LIS for {10, 22, 9, 33, 21, 50, 41, 60, 80} is 6 and LIS is {10, 22, 33, 50, 60, 80}.

Java

filter_none

edit
close

play_arrow

link
brightness_4
code

package BIT;
  
import java.util.ArrayList;
import java.util.Iterator;
  
public class LongestIncreasingSubsequence {
    public static void main(String[] args) {
//        int array[] = {10, 22, 9, 33, 21, 50, 41, 60, 80};
//        int array[] = {10, 2, 9, 3, 5, 4, 6, 8};
        int array[] = {10, 9, 8, 6, 5, 4};
        ArrayList list = new ArrayList();
        ArrayList longestList = new ArrayList();
        int currentMax;
        int highestCount = 0;
        for(int i = 0; i < array.length;i++)
        {
            currentMax = Integer.MIN_VALUE;
            for(int j = i;j < array.length; j++)
            {
                if(array[j] > currentMax)
                {
                    list.add(array[j]);
                    currentMax = array[j];
                }
            }
              
            //Compare previous highest subsequence 
            if(highestCount < list.size())
            {
                highestCount = list.size();
                longestList = new ArrayList(list);  
            }   
            list.clear();
        } 
        System.out.println();
          
        //Print list
        Iterator itr = longestList.iterator();
        System.out.println("The Longest subsequence");
        while(itr.hasNext())
        {
            System.out.print(itr.next() + " ");
        }
        System.out.println();
        System.out.println("Length of LIS: " + highestCount);
    }
      
}

chevron_right






                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















using System; 


using System.Collections.Generic; 


  


class longestIncreasingSubsequence 


{  


    public static void Main(String[] args) 


    {  


        int []array = {10, 22, 9, 33, 21, 50, 41, 60, 80};  


        // int []array = {10, 2, 9, 3, 5, 4, 6, 8};  


        //int []array = {10, 9, 8, 6, 5, 4};  


        List<int> list = new List<int>();  


        List<int> longestList = new List<int>();  


        int currentMax;  


        int highestCount = 0;  


        for(int i = 0; i < array.Length;i++)  


        {  


            currentMax = int.MinValue;  


            for(int j = i;j < array.Length; j++)  


            {  


                if(array[j] > currentMax)  


                {  


                    list.Add(array[j]);  


                    currentMax = array[j];  


                }  


            }  


              


            // Compare previous highest subsequence  


            if(highestCount < list.Count)  


            {  


                highestCount = list.Count;  


                longestList = new List<int>(list);  


            }  


            list.Clear();  


        }  


        Console.WriteLine();  


          


        // Print list  


        Console.WriteLine("The longest subsequence");  


        foreach(int itr in longestList)  


        {  


            Console.Write(itr + " ");  


        }  


        Console.WriteLine();  


        Console.WriteLine("Length of LIS: " + highestCount);  


    }  


}  


  


// This code is contributed by 29AjayKumar 



















                        chevron_right







                        filter_none










The longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order.


Examples:


Input:  [10, 22, 9, 33, 21, 50, 41, 60, 80]
Output: [10, 22, 33, 50, 60, 80] 
OR [10 22 33 41 60 80] or any other LIS of same length.






In previous post, we have discussed about Longest Increasing Subsequence problem. However, the post only covered code related to querying size of LIS, but not the construction of LIS. In this post, we will discuss how to print LIS using similar DP solution discussed earlier.




Let arr[0..n-1] be the input array. We define vector L such that L[i] is itself is a vector that stores LIS of arr that ends with arr[i]. For example, for array [3, 2, 6, 4, 5, 1], 


L[0]: 3
L[1]: 2
L[2]: 2 6
L[3]: 2 4
L[4]: 2 4 5
L[5]: 1



Therefore for an index i, L[i] can be recursively written as –


L[0] = {arr[O]}
L[i] = {Max(L[j])} + arr[i] 
where j < i and arr[j] < arr[i] and if there is no such j then L[i] = arr[i]



Below is the implementation of above idea –




C++















                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















/* Dynamic Programming solution to construct Longest 


   Increasing Subsequence */


#include <iostream> 


#include <vector> 


using namespace std; 


  


// Utility function to print LIS 


void printLIS(vector<int>& arr) 


{ 


    for (int x : arr) 


        cout << x << " "; 


    cout << endl; 


} 


  


// Function to construct and print Longest Increasing 


// Subsequence 


void constructPrintLIS(int arr[], int n) 


{ 


    // L[i] - The longest increasing sub-sequence  


    // ends with arr[i] 


    vector<vector<int> > L(n); 


  


    // L[0] is equal to arr[0] 


    L[0].push_back(arr[0]); 


  


    // start from index 1 


    for (int i = 1; i < n; i++) 


    { 


        // do for every j less than i 


        for (int j = 0; j < i; j++) 


        { 


            /* L[i] = {Max(L[j])} + arr[i] 


            where j < i and arr[j] < arr[i] */


            if ((arr[i] > arr[j]) && 


                    (L[i].size() < L[j].size() + 1)) 


                L[i] = L[j]; 


        } 


  


        // L[i] ends with arr[i] 


        L[i].push_back(arr[i]); 


    } 


  


    // L[i] now stores increasing sub-sequence of 


    // arr[0..i] that ends with arr[i] 


    vector<int> max = L[0]; 


  


    // LIS will be max of all increasing sub- 


    // sequences of arr 


    for (vector<int> x : L) 


        if (x.size() > max.size()) 


            max = x; 


  


    // max will contain LIS 


    printLIS(max); 


} 


  


// Driver function 


int main() 


{ 


    int arr[] = { 3, 2, 6, 4, 5, 1 }; 


    int n = sizeof(arr) / sizeof(arr[0]); 


  


    // construct and print LIS of arr 


    constructPrintLIS(arr, n); 


  


    return 0; 


} 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Dynamic Programming solution to construct Longest 


# Increasing Subsequence 


  


# Utility function to print LIS 


def printLIS(arr: list): 


    for x in arr: 


        print(x, end=" ") 


    print() 


  


# Function to construct and print Longest Increasing 


# Subsequence 


def constructPrintLIS(arr: list, n: int): 


  


    # L[i] - The longest increasing sub-sequence 


    # ends with arr[i] 


    l = [[] for i in range(n)] 


  


    # L[0] is equal to arr[0] 


    l[0].append(arr[0]) 


  


    # start from index 1 


    for i in range(1, n): 


  


        # do for every j less than i 


        for j in range(i): 


  


            # L[i] = {Max(L[j])} + arr[i] 


            # where j < i and arr[j] < arr[i] 


            if arr[i] > arr[j] and (len(l[i]) < len(l[j]) + 1): 


                l[i] = l[j].copy() 


  


        # L[i] ends with arr[i] 


        l[i].append(arr[i]) 


  


    # L[i] now stores increasing sub-sequence of 


    # arr[0..i] that ends with arr[i] 


    maxx = l[0] 


  


    # LIS will be max of all increasing sub- 


    # sequences of arr 


    for x in l: 


        if len(x) > len(maxx): 


            maxx = x 


  


    # max will contain LIS 


    printLIS(maxx) 


  


# Driver Code 


if __name__ == "__main__": 


  


    arr = [3, 2, 6, 4, 5, 1] 


    n = len(arr) 


  


    # construct and print LIS of arr 


    constructPrintLIS(arr, n) 


  


# This code is contributed by 


# sanjeev2552 



















                        chevron_right







                        filter_none










Output:


2 4 5 



Note that the time complexity of the above Dynamic Programming (DP) solution is O(n^2) and there is a O(n Log n) non-DP solution for the LIS problem. See below post for O(n Log n) solution.


Construction of Longest Monotonically Increasing Subsequence (N log N)


This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.





    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts: