Longest Increasing Subsequence using Longest Common Subsequence Algorithm

Given an array arr[] of N integers, the task is to find and print the Longest Increasing Subsequence.

Examples:

Input: arr[] = {12, 34, 1, 5, 40, 80}
Output: 4
{12, 34, 40, 80} and {1, 5, 40, 80} are the
longest increasing subsequences.

Input: arr[] = {10, 22, 9, 33, 21, 50, 41, 60, 80}
Output: 6

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Prerequisite: LCS, LIS

Approach: The longest increasing subsequence of any sequence is the subsequence of the sorted sequence of itself. It can be solved using a Dynamic Programming approach. The approach is the same as the classical LCS problem but instead of the second sequence, given sequence is taken again in its sorted form.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to return the size of the 
// longest increasing subsequence 
int LISusingLCS(vector<int>& seq) 
{ 
    int n = seq.size(); 
  
    // Create an 2D array of integer 
    // for tabulation 
    vector<vector<int> > L(n + 1, vector<int>(n + 1)); 
  
    // Take the second sequence as the sorted 
    // sequence of the given sequence 
    vector<int> sortedseq(seq); 
  
    sort(sortedseq.begin(), sortedseq.end()); 
  
    // Classical Dynamic Programming algorithm 
    // for Longest Common Subsequence 
    for (int i = 0; i <= n; i++) { 
        for (int j = 0; j <= n; j++) { 
            if (i == 0 || j == 0) 
                L[i][j] = 0; 
  
            else if (seq[i - 1] == sortedseq[j - 1]) 
                L[i][j] = L[i - 1][j - 1] + 1; 
  
            else
                L[i][j] = max(L[i - 1][j], L[i][j - 1]); 
        } 
    } 
  
    // Return the ans 
    return L[n][n]; 
} 
  
// Driver code 
int main() 
{ 
  
    vector<int> sequence{ 12, 34, 1, 5, 40, 80 }; 
  
    cout << LISusingLCS(sequence) << endl; 
  
    return 0; 
} 

Java

//Java implementation of above approach 
import java.util.*; 
  
class GFG 
{ 
      
// Function to return the size of the 
// longest increasing subsequence 
static int LISusingLCS(Vector<Integer> seq) 
{ 
    int n = seq.size(); 
  
    // Create an 2D array of integer 
    // for tabulation 
    int L[][] = new int [n + 1][n + 1]; 
  
    // Take the second sequence as the sorted 
    // sequence of the given sequence 
    Vector<Integer> sortedseq = new Vector<Integer>(seq); 
  
    Collections.sort(sortedseq); 
  
    // Classical Dynamic Programming algorithm 
    // for Longest Common Subsequence 
    for (int i = 0; i <= n; i++)  
    { 
        for (int j = 0; j <= n; j++)  
        { 
            if (i == 0 || j == 0) 
                L[i][j] = 0; 
  
            else if (seq.get(i - 1) == sortedseq.get(j - 1)) 
                L[i][j] = L[i - 1][j - 1] + 1; 
  
            else
                L[i][j] = Math.max(L[i - 1][j],  
                                   L[i][j - 1]); 
        } 
    } 
  
    // Return the ans 
    return L[n][n]; 
} 
  
// Driver code 
public static void main(String args[]) 
{ 
    Vector<Integer> sequence = new Vector<Integer>(); 
    sequence.add(12); 
    sequence.add(34); 
    sequence.add(1); 
    sequence.add(5); 
    sequence.add(40); 
    sequence.add(80); 
  
    System.out.println(LISusingLCS(sequence)); 
} 
} 
  
// This code is contributed by Arnab Kundu 

Python3

# Python3 implementation of the approach 
  
# Function to return the size of the 
# longest increasing subsequence 
def LISusingLCS(seq): 
    n = len(seq) 
  
    # Create an 2D array of integer 
    # for tabulation 
    L = [[0 for i in range(n + 1)]  
            for i in range(n + 1)] 
      
    # Take the second sequence as the sorted 
    # sequence of the given sequence 
    sortedseq = sorted(seq) 
  
    # Classical Dynamic Programming algorithm 
    # for Longest Common Subsequence 
    for i in range(n + 1): 
        for j in range(n + 1): 
            if (i == 0 or j == 0): 
                L[i][j] = 0
  
            elif (seq[i - 1] == sortedseq[j - 1]): 
                L[i][j] = L[i - 1][j - 1] + 1
  
            else: 
                L[i][j] = max(L[i - 1][j],  
                              L[i][j - 1]) 
  
    # Return the ans 
    return L[n][n] 
  
# Driver code 
sequence = [12, 34, 1, 5, 40, 80] 
  
print(LISusingLCS(sequence)) 
  
# This code is contributed by mohit kumar 

C#

// C# implementation of above approach 
using System; 
using System.Collections.Generic; 
  
class GFG 
{ 
      
// Function to return the size of the 
// longest increasing subsequence 
static int LISusingLCS(List<int> seq) 
{ 
    int n = seq.Count; 
  
    // Create an 2D array of integer 
    // for tabulation 
    int [,]L = new int [n + 1, n + 1]; 
  
    // Take the second sequence as the sorted 
    // sequence of the given sequence 
    List<int> sortedseq = new List<int>(seq); 
  
    sortedseq.Sort(); 
  
    // Classical Dynamic Programming algorithm 
    // for longest Common Subsequence 
    for (int i = 0; i <= n; i++)  
    { 
        for (int j = 0; j <= n; j++)  
        { 
            if (i == 0 || j == 0) 
                L[i, j] = 0; 
  
            else if (seq[i - 1] == sortedseq[j - 1]) 
                L[i, j] = L[i - 1, j - 1] + 1; 
  
            else
                L[i,j] = Math.Max(L[i - 1, j],  
                                L[i, j - 1]); 
        } 
    } 
  
    // Return the ans 
    return L[n, n]; 
} 
  
// Driver code 
public static void Main(String []args) 
{ 
    List<int> sequence = new List<int>(); 
    sequence.Add(12); 
    sequence.Add(34); 
    sequence.Add(1); 
    sequence.Add(5); 
    sequence.Add(40); 
    sequence.Add(80); 
  
    Console.WriteLine(LISusingLCS(sequence)); 
} 
} 
  
// This code is contributed by 29AjayKumar 

Output:

Time Complexity: O(n²) where n is the length of the sequence.

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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