Longest Common Increasing Subsequence (LCS + LIS)

4

Prerequisites : LCS, LIS

Given two arrays, find length of the longest common increasing subsequence [LCIS] and print one of such sequences (multiple sequences may exist)

Suppose we consider two arrays –
arr1[] = {3, 4, 9, 1} and
arr2[] = {5, 3, 8, 9, 10, 2, 1}

Our answer would be {3, 9} as this is the longest common subsequence which is increasing also.

The idea is to use dynamic programming here as well. We store the longest common increasing sub-sequence ending at each index of arr2[]. We create an auxiliary array table[] such that table[j] stores length of LCIS ending with arr2[j]. At the end, we return maximum value from this table. For filling values in this table, we traverse all elements of arr1[] and for every element arr1[i], we traverse all elements of arr2[]. If we find a match, we update table[j] with length of current LCIS. To maintain current LCIS, we keep checking valid table[j] values.

Below is the program to find length of LCIS.

C++

// A C++ Program to find length of the Longest Common
// Increasing Subsequence (LCIS)
#include<bits/stdc++.h>
using namespace std;

// Returns the length and the LCIS of two
// arrays arr1[0..n-1] and arr2[0..m-1]
int LCIS(int arr1[], int n, int arr2[], int m)
{
    // table[j] is going to store length of LCIS
    // ending with arr2[j]. We initialize it as 0,
    int table[m];
    for (int j=0; j<m; j++)
        table[j] = 0;

    // Traverse all elements of arr1[]
    for (int i=0; i<n; i++)
    {
        // Initialize current length of LCIS
        int current = 0;

        // For each element of arr1[], trvarse all
        // elements of arr2[].
        for (int j=0; j<m; j++)
        {
            // If both the array have same elements.
            // Note that we don't break the loop here.
            if (arr1[i] == arr2[j])
                if (current + 1 > table[j])
                    table[j] = current + 1;

            /* Now seek for previous smaller common
               element for current element of arr1 */
            if (arr1[i] > arr2[j])
                if (table[j] > current)
                    current = table[j];
        }
    }

    // The maximum value in table[] is out result
    int result = 0;
    for (int i=0; i<m; i++)
        if (table[i] > result)
           result = table[i];

    return result;
}

/* Driver program to test above function */
int main()
{
    int arr1[] = {3, 4, 9, 1};
    int arr2[] = {5, 3, 8, 9, 10, 2, 1};

    int n = sizeof(arr1)/sizeof(arr1[0]);
    int m = sizeof(arr2)/sizeof(arr2[0]);

    cout << "Length of LCIS is "
         << LCIS(arr1, n, arr2, m);
    return (0);
}

Java

// A Java Program to find length of the Longest
// Common Increasing Subsequence (LCIS)
import java.io.*;

class GFG {

    // Returns the length and the LCIS of two
    // arrays arr1[0..n-1] and arr2[0..m-1]
    static int LCIS(int arr1[], int n, int arr2[],
                                         int m)
    {
        // table[j] is going to store length of 
        // LCIS ending with arr2[j]. We initialize
        // it as 0,
        int table[] = new int[m];
        for (int j = 0; j < m; j++)
            table[j] = 0;

        // Traverse all elements of arr1[]
        for (int i = 0; i < n; i++)
        {
            // Initialize current length of LCIS
            int current = 0;

            // For each element of arr1[], trvarse 
            // all elements of arr2[].
            for (int j = 0; j < m; j++)
            {
                // If both the array have same 
                // elements. Note that we don't
                // break the loop here.
                if (arr1[i] == arr2[j])
                    if (current + 1 > table[j])
                        table[j] = current + 1;

                /* Now seek for previous smaller
                common element for current 
                element of arr1 */
                if (arr1[i] > arr2[j])
                    if (table[j] > current)
                        current = table[j];
            }
        }

        // The maximum value in table[] is out
        // result
        int result = 0;
        for (int i=0; i<m; i++)
            if (table[i] > result)
            result = table[i];

        return result;
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int arr1[] = {3, 4, 9, 1};
        int arr2[] = {5, 3, 8, 9, 10, 2, 1};

        int n = arr1.length;
        int m = arr2.length;

    System.out.println("Length of LCIS is " +
                       LCIS(arr1, n, arr2, m));
    }
}
// This code is contributed by Prerna Saini


Output :
Length of LCIS is 2

 
How to print a LCIS?
To print the longest common increasing subsequence we keep track of the parent of each element in the longest common increasing subsequence.

C++

// A C++ Program to find length of the Longest Common
// Increasing Subsequence (LCIS)
#include<bits/stdc++.h>
using namespace std;

// Returns the length and the LCIS of two
// arrays arr1[0..n-1] and arr2[0..m-1] and
// prints LCIS
int LCIS(int arr1[], int n, int arr2[], int m)
{
    // table[j] is going to store length of LCIS
    // ending with arr2[j]. We initialize it as 0,
    int table[m], parent[m];
    for (int j=0; j<m; j++)
        table[j] = 0;

    // Traverse all elements of arr1[]
    for (int i=0; i<n; i++)
    {
        // Initialize current length of LCIS
        int current = 0, last = -1;

        // For each element of arr1[], trvarse all
        // elements of arr2[].
        for (int j=0; j<m; j++)
        {
            // If both the array have same elements.
            // Note that we don't break the loop here.
            if (arr1[i] == arr2[j])
            {
                if (current + 1 > table[j])
                {
                    table[j] = current + 1;
                    parent[j] = last;
                }
            }

            /* Now seek for previous smaller common
               element for current element of arr1 */
            if (arr1[i] > arr2[j])
            {
                if (table[j] > current)
                {
                    current = table[j];
                    last = j;
                }
            }
        }
    }

    // The maximum value in table[] is out result
    int result = 0, index = -1;
    for (int i=0; i<m; i++)
    {
        if (table[i] > result)
        {
           result = table[i];
           index = i;
        }
    }

    // LCIS is going to store elements of LCIS
    int lcis[result];
    for (int i=0; index != -1; i++)
    {
        lcis[i] = arr2[index];
        index = parent[index];
    }

    cout << "The LCIS is : ";
    for (int i=result-1; i>=0; i--)
        printf ("%d ", lcis[i]);

    return result;
}

/* Driver program to test above function */
int main()
{
    int arr1[] = {3, 4, 9, 1};
    int arr2[] = {5, 3, 8, 9, 10, 2, 1};

    int n = sizeof(arr1)/sizeof(arr1[0]);
    int m = sizeof(arr2)/sizeof(arr2[0]);

    cout << "\nLength of LCIS is "
         << LCIS(arr1, n, arr2, m);
    return (0);
}

Java

// Java Program to find length of the Longest
// Common Increasing Subsequence (LCIS)
import java.io.*;

class GFG {

    // Returns the length and the LCIS of
    // two arrays arr1[0..n-1] and arr2[0..m-1]
    // and prints LCIS
    static int LCIS(int arr1[], int n, int arr2[],
                                         int m)
    {   
        // table[j] is going to store length of 
        // LCIS ending with arr2[j]. We 
        // initialize it as 0.
        int table[] = new int[m];
        int parent[] = new int[m];
        for (int j = 0; j < m; j++)
            table[j] = 0;

        // Traverse all elements of arr1[]
        for (int i = 0; i < n; i++)
        {
            // Initialize current length of LCIS
            int current = 0, last = -1;

            // For each element of arr1[],
            // trvarse all elements of arr2[].
            for (int j = 0; j < m; j++)
            {
                // If both the array have same
                // elements. Note that we don't 
                // break the loop here.
                if (arr1[i] == arr2[j])
                {   
                    if (current + 1 > table[j])
                    {
                        table[j] = current + 1;
                        parent[j] = last;
                    }
                }

                /* Now seek for previous smaller
                common element for current element
                of arr1 */
                if (arr1[i] > arr2[j])
                {
                    if (table[j] > current)
                    {
                        current = table[j];
                        last = j;
                    }
                }
            }
        }

        // The maximum value in table[] is out
        // result
        int result = 0, index = -1;
        for (int i = 0; i < m; i++)
        {
            if (table[i] > result)
            {
            result = table[i];
            index = i;
            }
        }

        // LCIS is going to store elements 
        // of LCIS
        int lcis[] = new int[result];
        for (int i = 0; index != -1; i++)
        {
            lcis[i] = arr2[index];
            index = parent[index];
        }

        System.out.print("The LCIS is : ");
        for (int i = result - 1; i >= 0; i--)
            System.out.print(lcis[i] + " ");
    
        return result;
    }

    /* Driver program to test above function */ 
    public static void main(String[] args)
    {
        int arr1[] = {3, 4, 9, 1};
        int arr2[] = {5, 3, 8, 9, 10, 2, 1};

        int n = arr1.length;
        int m = arr2.length;

        System.out.println("\nLength of LCIS is "+
                          LCIS(arr1, n, arr2, m));
    }
}
// This code is contributed by Prerna Saini


Output :
The LCIS is : 3 9 
Length of LCIS is 2

Time Complexity : O(m*n)
Auxiliary Space : O(m)

This article is contributed Rachit Belwariar. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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