Longest Common Increasing Subsequence (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[], traverse 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); } |
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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 |
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Python 3
# Python 3 Program to find length of the # Longest Common Increasing Subsequence (LCIS) # Returns the length and the LCIS of two # arrays arr1[0..n-1] and arr2[0..m-1] def LCIS(arr1, n, arr2, m): # table[j] is going to store length of LCIS # ending with arr2[j]. We initialize it as 0, table = [0] * m for j in range(m): table[j] = 0 # Traverse all elements of arr1[] for i in range(n): # Initialize current length of LCIS current = 0 # For each element of arr1[], # traverse all elements of arr2[]. for j in range(m): # 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 result = 0 for i in range(m): if (table[i] > result): result = table[i] return result # Driver Code if __name__ == "__main__": arr1 = [3, 4, 9, 1] arr2 = [5, 3, 8, 9, 10, 2, 1] n = len(arr1) m = len(arr2) print("Length of LCIS is", LCIS(arr1, n, arr2, m)) # This code is contributed by ita_c |
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C#
// A C# Program to find length of the Longest // Common Increasing Subsequence (LCIS) using System; 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() { int []arr1 = {3, 4, 9, 1}; int []arr2 = {5, 3, 8, 9, 10, 2, 1}; int n = arr1.Length; int m = arr2.Length; Console.Write("Length of LCIS is " + LCIS(arr1, n, arr2, m)); } } // This code is contributed by nitin mittal. |
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PHP
<?php // PHP Program to find length of // the Longest Common Increasing // Subsequence (LCIS) // Returns the length and the LCIS // of two arrays arr1[0..n-1] and // arr2[0..m-1] function LCIS($arr1, $n, $arr2, $m) { // table[j] is going to store // length of LCIS ending with // arr2[j]. We initialize it as 0, $table = Array(); //int table[m]; for ($j = 0; $j < $m; $j++) $table[$j] = 0; // Traverse all elements of arr1[] for ($i = 0; $i < $n; $i++) { // Initialize current // length of LCIS $current = 0; // For each element of // arr1[], trvarse all // elements of arr2[]. for ($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 $result = 0; for ($i = 0; $i < $m; $i++) if ($table[$i] > $result) $result = $table[$i]; return $result; } // Driver Code $arr1 = array (3, 4, 9, 1); $arr2 = array (5, 3, 8, 9, 10, 2, 1); $n = sizeof($arr1); $m = sizeof($arr2); echo "Length of LCIS is ", LCIS($arr1, $n, $arr2, $m); // This code is contributed by ajit ?> |
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Output :
Length of LCIS is 2
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