Longest Increasing Odd Even Subsequence
Last Updated :
27 Jul, 2022
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 : 6
{5, 6, 9,4,7,8} is the required longest
increasing odd even subsequence which is the array itself in this case
Input : arr[] = {1, 12, 2, 22, 5, 30, 31, 14, 17, 11}
{1,12,5,30,31,14,17} is the required longest
increasing odd even subsequence
Output : 7
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.
Implementation: A dynamic programming approach has been implemented below for the above mentioned recursive relation.
C++
#include <bits/stdc++.h>
using namespace std;
int longOddEvenIncSeq( int arr[], int n)
{
int lioes[n];
int maxLen = 0;
for ( int i = 0; i < n; i++)
lioes[i] = 1;
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;
for ( int i = 0; i < n; i++)
if (maxLen < lioes[i])
maxLen = lioes[i];
return maxLen;
}
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
import java.util.*;
import java.lang.*;
public class GfG{
public static int longOddEvenIncSeq( int arr[],
int n)
{
int [] lioes = new int [n];
int maxLen = 0 ;
for ( int i = 0 ; i < n; i++)
lioes[i] = 1 ;
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 ;
for ( int i = 0 ; i < n; i++)
if (maxLen < lioes[i])
maxLen = lioes[i];
return maxLen;
}
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));
}
}
|
Python3
def longOddEvenIncSeq( arr , n ):
lioes = list ()
maxLen = 0
for i in range (n):
lioes.append( 1 )
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
for i in range (n):
if maxLen < lioes[i]:
maxLen = lioes[i]
return maxLen
arr = [ 1 , 12 , 2 , 22 , 5 , 30 , 31 , 14 , 17 , 11 ]
n = len (arr)
print ( "Longest Increasing Odd Even " +
"Subsequence: " ,longOddEvenIncSeq(arr, n))
|
C#
using System;
class GFG {
public static int longOddEvenIncSeq( int [] arr,
int n)
{
int [] lioes = new int [n];
int maxLen = 0;
for ( int i = 0; i < n; i++)
lioes[i] = 1;
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;
for ( int i = 0; i < n; i++)
if (maxLen < lioes[i])
maxLen = lioes[i];
return maxLen;
}
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));
}
}
|
PHP
<?php
function longOddEvenIncSeq(& $arr , $n )
{
$lioes = array_fill (0, $n , NULL);
$maxLen = 0;
for ( $i = 0; $i < $n ; $i ++)
$lioes [ $i ] = 1;
for ( $i = 1; $i < $n ; $i ++)
for ( $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;
for ( $i = 0; $i < $n ; $i ++)
if ( $maxLen < $lioes [ $i ])
$maxLen = $lioes [ $i ];
return $maxLen ;
}
$arr = array ( 1, 12, 2, 22, 5, 30,
31, 14, 17, 11) ;
$n = sizeof( $arr );
echo "Longest Increasing Odd Even " .
"Subsequence: " . longOddEvenIncSeq( $arr , $n );
?>
|
Javascript
<script>
function longOddEvenIncSeq(arr, n)
{
let lioes = [];
let maxLen = 0;
for (let i = 0; i < n; i++)
lioes[i] = 1;
for (let i = 1; i < n; i++)
for (let 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;
for (let i = 0; i < n; i++)
if (maxLen < lioes[i])
maxLen = lioes[i];
return maxLen;
}
let arr = [ 1, 12, 2, 22,
5, 30, 31, 14, 17, 11 ];
let n = 10;
document.write( "Longest Increasing Odd"
+ " Even Subsequence: "
+ longOddEvenIncSeq(arr, n));
</script>
|
Output
Longest Increasing Odd Even Subsequence: 5
Time Complexity: O(n2).
Auxiliary Space: O(n).
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