形成算术级数 (AP) 的最长子序列

给定一个由N 个整数组成的数组arr[] ，任务是找到形成算术级数的最长子序列的长度。

例子：

Input: arr[] = {5, 10, 15, 20, 25, 30}
Output: 6
Explanation:
The whole set is in AP having common difference = 5.
Therefore, the length is 4.

Input: arr[] = { 20, 1, 15, 3, 10, 5, 8 }
Output: 4
Explanation:
The longest subsequence having the same difference is { 20, 15, 10, 5 }.
The above subsequence has same difference for every consecutive pairs i.e., (15 – 20) = (10 – 15) = (5 – 10) = -5.
Therefore, the length is 4.

朴素方法：解决问题的最简单方法是生成给定数组的所有可能子序列，并打印在相邻元素对之间具有相同差异的最长子序列的长度。时间

复杂度： O(N*2 ^N )
辅助空间： O(1)

高效方法：上述方法可以使用动态规划进行优化。以下是步骤：

1. 初始化一个辅助矩阵dp[][] ，其中dp[i][j]表示从i开始的子序列的长度，其公共差为j 。
2. 使用嵌套循环i从N – 1到0和j从i+1到N迭代数组。
   • 假设arr[i]和arr[j]是序列的前两个元素，并让它们的差为d 。
   • 现在，出现两种情况：
     1. dp[j][d] = 0 ：不存在以arr[j]开头且其差为d 的序列。在这种情况下dp[i][d] = 2 ，因为我们只能有arr[i]和arr[j]在序列中。
     2. dp[j][d] > 0 : 存在以arr[j]开头且相邻元素之间的差为d的序列。在这种情况下，关系为dp[i][d] = 1 + dp[j] [d] .
3. 最后，打印形成的所有子序列的最大长度。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function that finds the longest
// arithmetic subsequence having the
// same absolute difference
int lenghtOfLongestAP(int A[], int n)
{
 
    // Stores the length of sequences
    // having same difference
    unordered_map >
        dp;
 
    // Stores the resultant length
    int res = 2;
 
    // Iterate over the array
    for (int i = 0; i < n; ++i) {
 
        for (int j = i + 1; j < n; ++j) {
 
            int d = A[j] - A[i];
 
            // Update length of subsequence
            dp[d][j] = dp[d].count(i)
                           ? dp[d][i] + 1
                           : 2;
 
            res = max(res, dp[d][j]);
        }
    }
 
    // Return res
    return res;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 20, 1, 15, 3, 10, 5, 8 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << lenghtOfLongestAP(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function that finds the longest
// arithmetic subsequence having the
// same absolute difference
static int lenghtOfLongestAP(int A[], int n)
{
     
    // Stores the length of sequences
    // having same difference
    Map> dp = new HashMap>();
   
    // Stores the resultant length
    int res = 2;
 
    // Iterate over the array
    for(int i = 0; i < n; ++i)
    {
        for(int j = i + 1; j < n; ++j)
        {
            int d = A[j] - A[i];
            Map temp;
 
            // Update length of subsequence
            if (dp.containsKey(d))
            {
                temp = dp.get(d);
 
                if (temp.containsKey(i))
                    temp.put(j, temp.get(i) + 1);
                else
                    temp.put(j, 2);
            }
            else
            {
                temp = new HashMap();
                temp.put(j, 2);
            }
            dp.put(d, temp);
            res = Math.max(res, temp.get(j));
        }
    }
     
    // Return res
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 20, 1, 15, 3, 10, 5, 8 };
 
    int N = arr.length;
 
    // Function Call
    System.out.println(lenghtOfLongestAP(arr, N));
}
}
 
// This code is contributed by jithin

Python3

# Python3 program for the above approach
 
# Function that finds the longest
# arithmetic subsequence having the
# same absolute difference
def lenghtOfLongestAP(A, n) :
 
    # Stores the length of sequences
    # having same difference
    dp = {}
 
    # Stores the resultant length
    res = 2
 
    # Iterate over the array
    for i in range(n) :
 
        for j in range(i + 1, n) :
 
            d = A[j] - A[i]
 
            # Update length of subsequence
            if d in dp :
                if i in dp[d] :
                    dp[d][j] = dp[d][i] + 1
                else :
                    dp[d][j] = 2
            else :
                dp[d] = {}
                dp[d][j] = 2
 
            if d in dp :
                if j in dp[d] :
                    res = max(res, dp[d][j])
 
    # Return res
    return res
 
# Given array arr[]
arr = [ 20, 1, 15, 3, 10, 5, 8 ]
 
N = len(arr)
 
# Function Call
print(lenghtOfLongestAP(arr, N))
 
# This code is contributed by divyeshrabadiya07

C#

// C# program for the above approach
using System;
using System.Collections.Generic; 
 
class GFG{
   
// Function that finds the longest
// arithmetic subsequence having the
// same absolute difference
static int lenghtOfLongestAP(int []A, int n)
{
   
    // Stores the length of sequences
    // having same difference
    Dictionary> dp = new Dictionary>();
   
    // Stores the resultant length
    int res = 2;
  
    // Iterate over the array
    for(int i = 0; i < n; ++i)
    {
        for(int j = i + 1; j < n; ++j)
        {
            int d = A[j] - A[i];
  
            // Update length of subsequence
            if (dp.ContainsKey(d))
            {
                 if (dp[d].ContainsKey(i))
                {
                    dp[d][j] = dp[d][i] + 1;   
                }
                else
                {
                    dp[d][j] = 2;
                }
            }
            else
            {
                dp[d] = new Dictionary();
                dp[d][j] = 2;
            }
            res = Math.Max(res, dp[d][j]);
        }
    }
  
    // Return res
    return res;
}
   
// Driver Code
public static void Main(string[] args)
{
   
    // Given array arr[]
    int []arr = { 20, 1, 15, 3, 10, 5, 8 };
  
    int N = arr.Length;
  
    // Function call
    Console.Write(lenghtOfLongestAP(arr, N));
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// JavaScript program for the above approach
 
// Function that finds the longest
// arithmetic subsequence having the
// same absolute difference
function lenghtOfLongestAP(A, n)
{
 
    // Stores the length of sequences
    // having same difference
    var dp = new Map();
 
    // Stores the resultant length
    var res = 2;
 
    // Iterate over the array
    for (var i = 0; i < n; ++i) {
 
        for (var j = i + 1; j < n; ++j) {
 
            var d = A[j] - A[i];
 
             // Update length of subsequence
             if (dp.has(d))
            {
                 if (dp.get(d).has(i))
                {
                    var tmp = dp.get(d);
                    tmp.set(j, dp.get(d).get(i)+1);
                }
                else
                {
                    var tmp = new Map();
                    tmp.set(j, 2);
                    dp.set(d, tmp);
                }
            }
            else
            {
                var tmp = new Map();
                tmp.set(j, 2);
                dp.set(d, tmp);
            }
            res = Math.max(res, dp.get(d).get(j));
        }
    }
 
    // Return res
    return res;
}
 
// Driver Code
 
// Given array arr[]
var arr = [20, 1, 15, 3, 10, 5, 8];
var N = arr.length;
 
// Function Call
document.write( lenghtOfLongestAP(arr, N));
 

输出：

4

时间复杂度： O(N ² )
辅助空间： O(N ² )

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