Longest Increasing Odd Even Subsequence

Given an array of size n. The problem is to find the length of the subsequence in the given array such that all the elements of the subsequence are sorted in increasing order and also they are alternately odd and even.
Note that the subsequence could start either with the odd number or with the even number.

Examples:

Input : arr[] = {5, 6, 9, 4, 7, 8}
Output : 4
{5, 6, 7, 8} is the required longest
increasing odd even subsequence.

Input : arr[] = {1, 12, 2, 22, 5, 30, 31, 14, 17, 11}
Output : 5



Naive Approach: Consider all subsequences and select the ones with alternate odd even numbers in increasing order. Out of them select the longest one. This has an exponential time complexity.

Efficient Approach:
Let L(i) be the length of the LIOES (Longest Increasing Odd Even Subsequence) ending at index i such that arr[i] is the last element of the LIOES.
Then, L(i) can be recursively written as:
L(i) = 1 + max( L(j) ) where 0 < j < i and (arr[j] < arr[i]) and (arr[i]+arr[j])%2 != 0; or
L(i) = 1, if no such j exists.
To find the LIOES for a given array, we need to return max(L(i)) where 0 < i < n.
A dynamic programming approach has been implemented below for the above mentioned recursive relation.

C++

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// C++ implementation to find the longest
// increasing odd even subsequence
#include <bits/stdc++.h>
  
using namespace std;
  
// function to find the longest
// increasing odd even subsequence
int longOddEvenIncSeq(int arr[], int n)
{
    // lioes[i] stores longest increasing odd
    // even subsequence ending at arr[i]
    int lioes[n];
  
    // to store the length of longest increasing
    // odd even subsequence
    int maxLen = 0;
  
    // Initialize LIOES values for all indexes
    for (int i = 0; i < n; i++)
        lioes[i] = 1;
  
    // Compute optimized LIOES values
    // in bottom up manner
    for (int i = 1; i < n; i++)
        for (int j = 0; j < i; j++)
            if (arr[i] > arr[j] && 
               (arr[i] + arr[j]) % 2 != 0
                && lioes[i] < lioes[j] + 1)
                lioes[i] = lioes[j] + 1;
  
    // Pick maximum of all LIOES values
    for (int i = 0; i < n; i++)
        if (maxLen < lioes[i])
            maxLen = lioes[i];
  
    // required maximum length
    return maxLen;
}
  
// Driver program to test above
int main()
{
    int arr[] = { 1, 12, 2, 22, 5, 30, 
                    31, 14, 17, 11 };
    int n = sizeof(arr) / sizeof(n);
    cout << "Longest Increasing Odd Even "
         << "Subsequence: "
         << longOddEvenIncSeq(arr, n);
    return 0;
}

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Java














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// Java implementation to find the longest 


// increasing odd even subsequence 


import java.util.*; 


import java.lang.*; 


  


public class GfG{ 


      


    // function to find the longest 


    // increasing odd even subsequence 


    public static int longOddEvenIncSeq(int arr[],  


                                           int n) 


    { 


        // lioes[i] stores longest increasing odd 


        // even subsequence ending at arr[i] 


        int[] lioes = new int[n]; 


  


        // to store the length of longest  


        // increasing odd even subsequence 


        int maxLen = 0; 


  


        // Initialize LIOES values for all indexes 


        for (int i = 0; i < n; i++) 


            lioes[i] = 1; 


  


        // Compute optimized LIOES values 


        // in bottom up manner 


        for (int i = 1; i < n; i++) 


            for (int j = 0; j < i; j++) 


                if (arr[i] > arr[j] &&  


                (arr[i] + arr[j]) % 2 != 0


                    && lioes[i] < lioes[j] + 1) 


                    lioes[i] = lioes[j] + 1; 


  


        // Pick maximum of all LIOES values 


        for (int i = 0; i < n; i++) 


            if (maxLen < lioes[i]) 


                maxLen = lioes[i]; 


  


        // required maximum length 


        return maxLen; 


    } 


    // driver function 


    public static void main(String argc[]){ 


        int[] arr = new int[]{ 1, 12, 2, 22,  


                     5, 30, 31, 14, 17, 11 }; 


        int n = 10; 


        System.out.println("Longest Increasing Odd"


                          + " Even Subsequence: "


                       + longOddEvenIncSeq(arr, n)); 


    } 


} 


  


/* This code is contributed by Sagar Shukla */



















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Python3














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# Python3 implementation to find the longest 


# increasing odd even subsequence 


  


# function to find the longest 


# increasing odd even subsequence 


def longOddEvenIncSeq( arr , n ): 


  


    # lioes[i] stores longest increasing odd 


    # even subsequence ending at arr[i] 


    lioes = list() 


      


    # to store the length of longest increasing 


    # odd even subsequence 


    maxLen = 0


      


    # Initialize LIOES values for all indexes 


    for i in range(n): 


        lioes.append(1) 


          


    # Compute optimized LIOES values 


    # in bottom up manner 


    i=1


    for i in range(n): 


        for j in range(i): 


            if (arr[i] > arr[j] and


                (arr[i] + arr[j]) % 2 != 0 and


                lioes[i] < lioes[j] + 1): 


                    lioes[i] = lioes[j] + 1


      


    # Pick maximum of all LIOES values 


    for i in range(n): 


        if maxLen < lioes[i]: 


            maxLen = lioes[i] 


              


    # required maximum length 


    return maxLen 


      


# Driver to test above 


arr = [ 1, 12, 2, 22, 5, 30, 31, 14, 17, 11 ] 


n = len(arr) 


print("Longest Increasing Odd Even " +


      "Subsequence: ",longOddEvenIncSeq(arr, n)) 


                  


# This code is contributed by "Sharad_Bhardwaj". 



















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C#














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// C# implementation to find the longest 


// increasing odd even subsequence 


using System; 


  


class GFG { 


  


    // function to find the longest 


    // increasing odd even subsequence 


    public static int longOddEvenIncSeq(int[] arr, 


                                            int n) 


    { 


        // lioes[i] stores longest increasing odd 


        // even subsequence ending at arr[i] 


        int[] lioes = new int[n]; 


  


        // to store the length of longest 


        // increasing odd even subsequence 


        int maxLen = 0; 


  


        // Initialize LIOES values for all indexes 


        for (int i = 0; i < n; i++) 


            lioes[i] = 1; 


  


        // Compute optimized LIOES values 


        // in bottom up manner 


        for (int i = 1; i < n; i++) 


            for (int j = 0; j < i; j++) 


                if (arr[i] > arr[j] && 


                   (arr[i] + arr[j]) % 2 != 0 && 


                    lioes[i] < lioes[j] + 1) 


                      


                    lioes[i] = lioes[j] + 1; 


  


        // Pick maximum of all LIOES values 


        for (int i = 0; i < n; i++) 


            if (maxLen < lioes[i]) 


                maxLen = lioes[i]; 


  


        // required maximum length 


        return maxLen; 


    } 


      


    // driver function 


    public static void Main() 


    { 


        int[] arr = new int[]{ 1, 12, 2, 22, 


                               5, 30, 31, 14, 17, 11 }; 


        int n = 10; 


        Console.Write("Longest Increasing Odd"


                    + " Even Subsequence: "


                    + longOddEvenIncSeq(arr, n)); 


    } 


} 


  


// This code is contributed by Sam007 



















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PHP





 $arr[$j] &&

			($arr[$i] + $arr[$j]) % 2 != 0 &&

			    $lioes[$i] < $lioes[$j] + 1)
				$lioes[$i] = $lioes[$j] + 1;

	// Pick maximum of all LIOES values
	for ($i = 0; $i < $n; $i++)
		if ($maxLen < $lioes[$i])
			$maxLen = $lioes[$i];

	// required maximum length
	return $maxLen;
}

// Driver Code
$arr = array( 1, 12, 2, 22, 5, 30, 
					31, 14, 17, 11) ;
$n = sizeof($arr);
echo "Longest Increasing Odd Even ".
		"Subsequence: " . longOddEvenIncSeq($arr, $n);

// This code is contributed 
// by ChitraNayal
?>




Output:


Longest Increasing Odd Even Subsequence: 5



Time Complexity: O(n2).

Auxiliary Space: O(n).



          
          
          
          

            

    

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