Construction of Longest Increasing Subsequence using Dynamic Programming

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 previous post, we have discussed about Longest Increasing Subsequence problem. However, the post only covered code related to querying size of LIS, but not the construction of LIS. In this post, we will discuss how to print LIS using 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 an 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 C++ implementation of above idea –

/* Dynamic Programming solution to construct Longest
   Increasing Subsequence */
#include <iostream>
#include <vector>
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<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 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;
}

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)

This article is contributed by Aditya Goel. 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.

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