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++

// 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; }

## Java

// 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 */

## Python3

# 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".

## C#

// 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

Output:

Longest Increasing Odd Even Subsequence: 5

Time Complexity: O(n^{2}).

Auxiliary Space : O(n).

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