Given an array arr[] consisting of N integers, the task is to find the length of the longest increasing absolute even subsequence.
An increasing absolute even subsequence is an increasing subsequence of array elements having absolute difference between adjacent pairs as even.
Examples:
Input: arr[] = {10, 22, 9, 33, 21, 50, 41, 60}
Output: 4
Explanation: The longest increasing absolute even subsequence is {10, 22, 50, 60}. Therefore, the required length is 4.Input: arr[] = {11, -22, 43, -54, 66, 5}
Output: 3
Explanation:
The longest increasing absolute even subsequence is 3 i.e. {-22, -54, 66}. Therefore, the required length is 4.
Naive Approach: The simplest approach is to generate all possible subsequence of the given array and for each subsequence, check if the subsequence is increasing and the absolute difference between adjacent elements is even or not. Print the length of the longest such subsequence.
Time Complexity: O(2N)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is similar to finding the longest increasing subsequence. But the only condition to be changed is to check if the absolute difference between two adjacent elements of the subsequence is even or not. Follow the steps below to solve the problem:
- Initialize an auxiliary array dp[] where all are initially 1.
-
Traverse the given array arr[] using the variable i over the range [0, N), and for each index do the following:
- Iterate using variable j over the range [0, i) and check for the following three conditions:
- If absolute value of arr[i] > arr[j].
- If arr[i] and arr[j] both are even or not.
- If dp[i] < dp[j] + 1.
- If the above three conditions are satisfied for any index j, then update dp[i] = dp[j] + 1.
- Iterate using variable j over the range [0, i) and check for the following three conditions:
- Print the maximum element of the array dp[] as the required result.
Below is the implementation of the above approach:
// C++14 program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the longest // increasing absolute even subsequence void EvenLIS( int arr[], int n)
{ // Stores length of
// required subsequence
int lis[n];
for ( int i = 0; i < n; i++)
lis[i] = 1;
// Traverse the array
for ( int i = 1; i < n; i++)
{
// Traverse prefix of current
// array element
for ( int j = 0; j < i; j++)
{
// Check if the subsequence is
// LIS and have even absolute
// difference of adjacent pairs
if ( abs (arr[i]) > abs (arr[j]) &&
abs (arr[i]) % 2 == 0 &&
abs (arr[j]) % 2 == 0 &&
lis[i] < lis[j] + 1)
// Update lis[]
lis[i] = lis[j] + 1;
}
}
// Stores maximum length
int maxlen = 0;
// Find the length of longest
// absolute even subsequence
for ( int i = 0; i < n; i++)
maxlen = max(maxlen, lis[i]);
// Return the maximum length of
// absolute even subsequence
cout << maxlen << endl;
} // Driver code int main()
{ // Given array arr[] and brr[]
int arr[] = { 11, -22, 43, -54, 66, 5 };
int N = sizeof (arr) / sizeof (arr[0]);
// Function call
EvenLIS(arr, N);
} // This code is contributed by code_hunt |
// Java program for the above approach import java.util.*;
import java.io.*;
class GFG{
// Function to find the longest // increasing absolute even subsequence static void EvenLIS( int arr[])
{ // Length of arr
int n = arr.length;
// Stores length of
// required subsequence
int lis[] = new int [n];
Arrays.fill(lis, 1 );
// Traverse the array
for ( int i = 1 ; i < n; i++)
{
// Traverse prefix of current
// array element
for ( int j = 0 ; j < i; j++)
{
// Check if the subsequence is
// LIS and have even absolute
// difference of adjacent pairs
if (Math.abs(arr[i]) > Math.abs(arr[j]) &&
Math.abs(arr[i]) % 2 == 0 &&
Math.abs(arr[j]) % 2 == 0 &&
lis[i] < lis[j] + 1 )
// Update lis[]
lis[i] = lis[j] + 1 ;
}
}
// Stores maximum length
int maxlen = 0 ;
// Find the length of longest
// absolute even subsequence
for ( int i = 0 ; i < n; i++)
maxlen = Math.max(maxlen, lis[i]);
// Return the maximum length of
// absolute even subsequence
System.out.println(maxlen);
} // Driver code public static void main(String args[])
{ // Given array arr[] and brr[]
int arr[] = { 11 , - 22 , 43 , - 54 , 66 , 5 };
int N = arr.length;
// Function call
EvenLIS(arr);
} } // This code is contributed by bikram2001jha |
# Python3 program for the above approach # Function to find the longest # increasing absolute even subsequence def EvenLIS(arr):
# Length of arr
n = len (arr)
# Stores length of
# required subsequence
lis = [ 1 ] * n
# Traverse the array
for i in range ( 1 , n):
# Traverse prefix of current
# array element
for j in range ( 0 , i):
# Check if the subsequence is
# LIS and have even absolute
# difference of adjacent pairs
if abs (arr[i]) > abs (arr[j]) \
and abs (arr[i] % 2 ) = = 0 \
and abs (arr[j] % 2 ) = = 0 \
and lis[i] < lis[j] + 1 :
# Update lis[]
lis[i] = lis[j] + 1
# Stores maximum length
maxlen = 0
# Find the length of longest
# absolute even subsequence
for i in range (n):
maxlen = max (maxlen, lis[i])
# Return the maximum length of
# absolute even subsequence
print (maxlen)
# Driver Code # Given arr[] arr = [ 11 , - 22 , 43 , - 54 , 66 , 5 ]
# Function Call EvenLIS(arr) |
// C# program for the above approach using System;
class GFG{
// Function to find the longest // increasing absolute even subsequence static void EvenLIS( int []arr)
{ // Length of arr
int n = arr.Length;
// Stores length of
// required subsequence
int []lis = new int [n];
for ( int i = 0; i < n; i++)
lis[i] = 1;
// Traverse the array
for ( int i = 1; i < n; i++)
{
// Traverse prefix of current
// array element
for ( int j = 0; j < i; j++)
{
// Check if the subsequence is
// LIS and have even absolute
// difference of adjacent pairs
if (Math.Abs(arr[i]) > Math.Abs(arr[j]) &&
Math.Abs(arr[i]) % 2 == 0 &&
Math.Abs(arr[j]) % 2 == 0 &&
lis[i] < lis[j] + 1)
// Update lis[]
lis[i] = lis[j] + 1;
}
}
// Stores maximum length
int maxlen = 0;
// Find the length of longest
// absolute even subsequence
for ( int i = 0; i < n; i++)
maxlen = Math.Max(maxlen, lis[i]);
// Return the maximum length of
// absolute even subsequence
Console.WriteLine(maxlen);
} // Driver code public static void Main(String []args)
{ // Given array []arr and brr[]
int []arr = { 11, -22, 43, -54, 66, 5 };
int N = arr.Length;
// Function call
EvenLIS(arr);
} } // This code is contributed by Amit Katiyar |
<script> // Javascript program to implement // the above approach // Function to find the longest // increasing absolute even subsequence function EvenLIS(arr)
{ // Length of arr
let n = arr.length;
// Stores length of
// required subsequence
let lis = new Array(n).fill(1);
// Traverse the array
for (let i = 1; i < n; i++)
{
// Traverse prefix of current
// array element
for (let j = 0; j < i; j++)
{
// Check if the subsequence is
// LIS and have even absolute
// difference of adjacent pairs
if (Math.abs(arr[i]) > Math.abs(arr[j]) &&
Math.abs(arr[i]) % 2 == 0 &&
Math.abs(arr[j]) % 2 == 0 &&
lis[i] < lis[j] + 1)
// Update lis[]
lis[i] = lis[j] + 1;
}
}
// Stores maximum length
let maxlen = 0;
// Find the length of longest
// absolute even subsequence
for (let i = 0; i < n; i++)
maxlen = Math.max(maxlen, lis[i]);
// Return the maximum length of
// absolute even subsequence
document.write(maxlen);
} // Driver Code // Given array arr[] and brr[]
let arrr = [ 11, -22, 43, -54, 66, 5 ];
let N = arrr.length;
// Function call
EvenLIS(arrr);
</script> |
3
Time Complexity: O(N2)
Auxiliary Space: O(N)