Longest Common Increasing Subsequence (LCS + LIS)

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.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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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