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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