给定一个由N个整数组成的数组arr [] ，任务是找到最长的增加的绝对偶数子序列的长度。
An increasing absolute even subsequence is an increasing subsequence of array elements having absolute difference between adjacent pairs as even.
例子：
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.
天真的方法：最简单的方法是生成给定数组的所有可能的子序列，并针对每个子序列检查子序列是否在增加，并且相邻元素之间的绝对差是否相等。打印最长的此类子序列的长度。
时间复杂度： O(2 N )
辅助空间： O(1)
高效的方法：为了优化上述方法，该想法类似于找到最长的递增子序列。但是唯一要更改的条件是检查子序列的两个相邻元素之间的绝对差是否相等。请按照以下步骤解决问题：
- 初始化一个辅助数组dp [] ，其中所有初始都为1 。
- 使用变量i在[0，N)范围内遍历给定数组arr [] ，并对每个索引执行以下操作：
- 在[0，i)范围内使用变量j进行迭代，并检查以下三个条件：
- 如果arr [i]的绝对值> arr [j] 。
- 如果arr [i]和arr [j]都为偶数或不为偶数。
- 如果dp [i] < dp [j] + 1 。
- 如果对于任何索引j都满足以上三个条件，则更新dp [i] = dp [j] + 1 。
- 在[0，i)范围内使用变量j进行迭代，并检查以下三个条件：
- 将数组dp []的最大元素打印为所需结果。
下面是上述方法的实现：
C++14
// C++14 program for the above approach
#include
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
// abolute 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
// 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
// abolute 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 bikram2001jhaPython3
# 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
# abolute 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#
// 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
// abolute 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输出：
3
时间复杂度： O(N 2 )
辅助空间： O(N)