# 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 ` `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 > 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 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 sortedseq = ``new` `Vector(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 sequence = ``new` `Vector(); ` `    ``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:

```4
```

Time Complexity: O(n2) where n is the length of the sequence. My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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 Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.