Construction of Longest Increasing Subsequence(LIS) and printing LIS sequence
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++
#include <iostream>
#include <vector>
using namespace std;
void printLIS(vector< int >& arr)
{
for ( int x : arr)
cout << x << " " ;
cout << endl;
}
void constructPrintLIS( int arr[], int n)
{
vector<vector< int > > L(n);
L[0].push_back(arr[0]);
for ( int i = 1; i < n; i++)
{
for ( int j = 0; j < i; j++)
{
if ((arr[i] > arr[j]) &&
(L[i].size() < L[j].size() + 1))
L[i] = L[j];
}
L[i].push_back(arr[i]);
}
vector< int > max = L[0];
for (vector< int > x : L)
if (x.size() > max.size())
max = x;
printLIS(max);
}
int main()
{
int arr[] = { 3, 2, 6, 4, 5, 1 };
int n = sizeof (arr) / sizeof (arr[0]);
constructPrintLIS(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG{
static void printLIS(Vector<Integer> arr)
{
for ( int x : arr)
System.out.print(x + " " );
System.out.println();
}
static void constructPrintLIS( int arr[],
int n)
{
Vector<Integer> L[] = new Vector[n];
for ( int i = 0 ; i < L.length; i++)
L[i] = new Vector<Integer>();
L[ 0 ].add(arr[ 0 ]);
for ( int i = 1 ; i < n; i++)
{
for ( int j = 0 ; j < i; j++)
{
if ((arr[i] > arr[j]) &&
(L[i].size() < L[j].size() + 1 ))
L[i] = (Vector<Integer>) L[j].clone();
}
L[i].add(arr[i]);
}
Vector<Integer> max = L[ 0 ];
for (Vector<Integer> x : L)
if (x.size() > max.size())
max = x;
printLIS(max);
}
public static void main(String[] args)
{
int arr[] = { 3 , 2 , 4 , 5 , 1 };
int n = arr.length;
constructPrintLIS(arr, n);
}
}
|
Python3
def printLIS(arr: list ):
for x in arr:
print (x, end = " " )
print ()
def constructPrintLIS(arr: list , n: int ):
l = [[] for i in range (n)]
l[ 0 ].append(arr[ 0 ])
for i in range ( 1 , n):
for j in range (i):
if arr[i] > arr[j] and ( len (l[i]) < len (l[j]) + 1 ):
l[i] = l[j].copy()
l[i].append(arr[i])
maxx = l[ 0 ]
for x in l:
if len (x) > len (maxx):
maxx = x
printLIS(maxx)
if __name__ = = "__main__" :
arr = [ 3 , 2 , 6 , 4 , 5 , 1 ]
n = len (arr)
constructPrintLIS(arr, n)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void printLIS(List< int > arr)
{
foreach ( int x in arr)
{
Console.Write(x + " " );
}
Console.WriteLine();
}
static void constructPrintLIS( int [] arr, int n)
{
List<List< int >> L = new List<List< int >>();
for ( int i = 0; i < n; i++)
{
L.Add( new List< int >());
}
L[0].Add(arr[0]);
for ( int i = 1; i < n; i++)
{
for ( int j = 0; j < i; j++)
{
if ((arr[i] > arr[j]) && (L[i].Count < L[j].Count + 1))
L[i] = L[j];
}
L[i].Add(arr[i]);
}
List< int > max = L[0];
foreach (List< int > x in L)
{
if (x.Count > max.Count)
{
max = x;
}
}
printLIS(max);
}
static void Main()
{
int [] arr = { 3, 2, 4, 5, 1 };
int n = arr.Length;
constructPrintLIS(arr, n);
}
}
|
Javascript
<script>
function printLIS( x)
{
document.write(x+ " " );
}
function constructPrintLIS( arr, n)
{
var L = new Array(n);
for ( var i = 0; i < n;i++){
L[i] = [];
}
L[0].push(arr[0]);
for ( var i = 1; i < n; i++)
{
for ( var j = 0; j < i; j++)
{
if ((arr[i] > arr[j]) && (L[i].length< L[j].length + 1))
L[i] = [...L[j]];
}
L[i].push(arr[i]);
}
var max = 0;
for ( var x=0; x < L.length;x++){
if (L[x].length> L[max].length)
max = x;
}
L[max].forEach(printLIS);
}
var arr = [3, 2, 6, 4, 5, 1 ];
var n = 6;
constructPrintLIS(arr, n);
</script>
|
Note that the time complexity of the above Dynamic Programming (DP) solution is O(n^3) (n^2 for two nested loops and n for copying another vector in a vector eg: L[i] = L[j] contributes O(n) also) and space complexity is O(n^2) as we are using 2d vector to store our LIS 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)
Last Updated :
14 Oct, 2022
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