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# 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++

 `/* Dynamic Programming solution to construct Longest` `   ``Increasing Subsequence */` `#include ` `#include ` `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 > 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;` `}`

## Java

 `// Java program for ` `// the above approach`   `// Dynamic Programming ` `// solution to construct Longest` `// Increasing Subsequence` `import` `java.util.*;` `class` `GFG{`   `// Utility function to print LIS` `static` `void` `printLIS(Vector arr)` `{` `  ``for` `(``int` `x : arr)` `    ``System.out.print(x + ``" "``);` `  ``System.out.println();` `}`   `// 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]` `  ``Vector L[] = ``new` `Vector[n];` `  ``for` `(``int` `i = ``0``; i < L.length; i++)` `    ``L[i] = ``new` `Vector();` `  `  `  ``// 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) 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 max = L[``0``];` `  `  `  ``// LIS will be max of all increasing sub-` `  ``// sequences of arr` `  ``for` `(Vector x : L)` `    ``if` `(x.size() > max.size())` `      ``max = x;`   `  ``// max will contain LIS` `  ``printLIS(max);` `}`   `// Driver function` `public` `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`

## Python3

 `# 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`

## C#

 `// Dynamic Programming solution to construct Longest` `// Increasing Subsequence` `using` `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> L = ``new` `List>();` `        ``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`

## Javascript

 ``

Output

`2 4 5 `

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)