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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