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

25 Apr, 2022
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}.
C++
Java
Python3
C#
Javascript
#include <bits/stdc++.h>using namespace std; void LIS(int a[], int n){    vector<int> list;    vector<int> longestlist;         // Highest count    int hc = 0;         // Current max    int cm;    for(int i = 0; i < n; i++)    {        cm = INT_MIN;        for(int j = i; j < n; j++)        {            if (a[j] > cm)            {                list.push_back(a[j]);                cm = a[j];            }        }                 // Compare previous highest subsequence        if (hc < list.size())        {            hc = list.size();            for(int k = 0; k < list.size(); k++)            {                longestlist.push_back(list[k]);            }        }        list.clear();    }     for(int i = 0; i < longestlist.size(); i++)    {        cout << longestlist[i] << ' ';    }    cout << endl;    cout << "length of longestlist is " << hc;} // Driver codeint main(){    int a[] = { 10, 22, 9, 33, 21,                50, 41, 60, 80 };    int n = sizeof(a) / sizeof(a[0]);    LIS(a, n);     return 0;} // This code is contributed by Harshit Srivastava
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);    }     }
import sys def LIS(a, n):     list = []    longestlist = []         # Highest count    hc = 0         # Current max    for i in range(n):        cm = -sys.maxsize -1        for j in range(i,n):            if (a[j] > cm):                list.append(a[j])                cm = a[j]                 # Compare previous highest subsequence        if (hc < len(list)):             hc = len(list)            for k in range(len(list)):                longestlist.append(list[k])         list = []     for i in range(len(longestlist)):        print(longestlist[i] ,end = ' ')    print()    print("length of longestlist is " + str(hc)) # Driver code a = [ 10, 22, 9, 33, 21, 50, 41, 60, 80 ]n = len(a)LIS(a, n) # This code is contributed by shinjanpatra
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
<script>function LIS(a, n){    let list = [];    let longestlist = [];         // Highest count    let hc = 0;         // Current max    let cm;    for(let i = 0; i < n; i++)    {        cm = Number.MIN_VALUE;        for(let j = i; j < n; j++)        {            if (a[j] > cm)            {                list.push(a[j]);                cm = a[j];            }        }                 // Compare previous highest subsequence        if (hc < list.length)        {            hc = list.length;            for(let k = 0; k < list.length; k++)            {                longestlist.push(list[k]);            }        }        list = [];    }     for(let i = 0; i < longestlist.length; i++)    {        document.write(longestlist[i] + ' ');    }    document.write("</br>")    document.write("length of longestlist is " + hc);} // Driver code let a = [ 10, 22, 9, 33, 21, 50, 41, 60, 80 ];let n = a.length;LIS(a, n); // This code is contributed by shinjanpatra</script>
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 the previous post, we have discussed The Longest Increasing Subsequence problem. However, the post only covered code related to the querying size of LIS, but not the construction of LIS. In this post, we will discuss how to print LIS using a 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 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 the above idea – 
C++
Java
Python3
C#
/* Dynamic Programming solution to construct Longest   Increasing Subsequence */#include <iostream>#include <vector>using namespace std; // Utility function to print LISvoid printLIS(vector<int>& arr){    for (int x : arr)        cout << x << " ";    cout << endl;} // Function to construct and print Longest Increasing// Subsequencevoid 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 functionint 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;}
// Java program for// the above approach // Dynamic Programming// solution to conLongest// Increasing Subsequenceimport java.util.*;class GFG{ // Utility function to print LISstatic void printLIS(Vector<Integer> arr){  for (int x : arr)    System.out.print(x + " ");  System.out.println();} // Function to conand print// Longest Increasing Subsequencestatic void constructPrintLIS(int arr[],                              int n){  // L[i] - The longest increasing  // sub-sequence ends with arr[i]  Vector<Integer> L[] = new Vector[n];  for (int i = 0; i < L.length; i++)    L[i] = new Vector<Integer>();     // L[0] is equal to arr[0]  L[0].add(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] = (Vector<Integer>) L[j].clone();  //deep copy    }     // L[i] ends with arr[i]    L[i].add(arr[i]);  }   // L[i] now stores increasing sub-sequence of  // arr[0..i] that ends with arr[i]  Vector<Integer> max = L[0];     // LIS will be max of all increasing sub-  // sequences of arr  for (Vector<Integer> x : L)    if (x.size() > max.size())      max = x;   // max will contain LIS  printLIS(max);} // Driver functionpublic static void main(String[] args){  int arr[] = {3, 2, 4, 5, 1};  int n = arr.length;   // print LIS of arr  constructPrintLIS(arr, n);}} // This code is contributed by gauravrajput1
# Dynamic Programming solution to construct Longest# Increasing Subsequence # Utility function to print LISdef printLIS(arr: list):    for x in arr:        print(x, end=" ")    print() # Function to construct and print Longest Increasing# Subsequencedef 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 Codeif __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
// Dynamic Programming solution to construct Longest// Increasing Subsequenceusing System;using System.Collections.Generic;class GFG{         // Utility function to print LIS    static void printLIS(List<int> arr)    {        foreach(int x in arr)        {            Console.Write(x + " ");        }        Console.WriteLine();    }          // Function to construct and print Longest Increasing    // Subsequence    static void constructPrintLIS(int[] arr, int n)    {               // L[i] - The longest increasing sub-sequence        // ends with arr[i]        List<List<int>> L = new List<List<int>>();        for(int i = 0; i < n; i++)        {            L.Add(new List<int>());        }              // L[0] is equal to arr[0]        L[0].Add(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].Count < L[j].Count + 1))                    L[i] = L[j];            }                  // L[i] ends with arr[i]            L[i].Add(arr[i]);        }              // L[i] now stores increasing sub-sequence of        // arr[0..i] that ends with arr[i]        List<int> max = L[0];              // LIS will be max of all increasing sub-        // sequences of arr        foreach(List<int> x in L)        {            if (x.Count > max.Count)            {                max = x;            }        }              // max will contain LIS        printLIS(max);    }   // Driver code  static void Main()  {    int[] arr = { 3, 2, 4, 5, 1 };    int n = arr.Length;      // construct and print LIS of arr    constructPrintLIS(arr, n);  }} // This code is contributed by divyesh072019
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)
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