1到N之间没有相邻元素的整数子集的计数

给定整数N ，任务是计算从1到N的整数数组形成的子集的数量，该子集不包含相邻元素。如果满足不相邻元素条件，则无法选择子集，但是可以添加更多元素。
例子：

Input: N = 4
Output: 3
Explanation:
Array is {1, 2, 3, 4}
So to satisfy the condition, the subsets formed are :
{1, 3}, {2, 4}, {1, 4}

Input: N = 5
Output: 4

方法：
通过使用动态编程可以解决此问题。对于最后一个元素，我们有两个选择，要么包含它，要么排除它。令DP [i]为以索引i结尾的期望子集的数量。
所以下一个子问题变成DP [i-3]
因此，DP关系变为：

DP[i] = DP[i-2] + DP[i-3]  

但是，我们需要观察基本情况：

• 当N = 0时，我们不能形成任何具有0个数字的子集。
• 当N = 1时，我们可以形成1个子集{1}
• 当N = 2时，我们可以形成2个子集{1} ， {2}
• 当N = 3时，我们可以形成2个子集{1，3} ， {2}

下面是上述方法的实现：

C++

// C++ Code to count subsets not containing
// adjacent elements from 1 to N
 
#include 
using namespace std;
 
// Function to count subsets
int countSubsets(int N)
{
     
    if(N <= 2)
        return N;
         
    if(N == 3)
        return 2;
     
    int DP[N + 1] = {0};
     
    DP[0] = 0, DP[1] = 1, DP[2] = 2, DP[3] = 2;
     
    for (int i = 4; i <= N; i++) {
 
        DP[i] = DP[i - 2] + DP[i - 3];
    }
     
    return DP[N];
}
 
// Driver Code
int main()
{
    int N = 20;
     
    cout << countSubsets(N);
     
    return 0;
}

Java

// Java code to count subsets not containing
// adjacent elements from 1 to N
class GFG{
 
// Function to count subsets
static int countSubsets(int N)
{
    if(N <= 2)
       return N;
         
    if(N == 3)
       return 2;
     
    int []DP = new int[N + 1];
     
    DP[0] = 0;
    DP[1] = 1;
    DP[2] = 2;
    DP[3] = 2;
     
    for(int i = 4; i <= N; i++)
    {
       DP[i] = DP[i - 2] + DP[i - 3];
    }
    return DP[N];
}
 
// Driver code
public static void main(String[] args)
{
    int N = 20;
     
    System.out.print(countSubsets(N));
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 Code to count subsets
# not containing adjacent elements
# from 1 to N
 
# Function to count subsets
def countSubsets(N):
 
    if(N <= 2):
        return N
 
    if(N == 3):
        return 2
 
    DP = [0] * (N + 1)
 
    DP[0] = 0
    DP[1] = 1
    DP[2] = 2
    DP[3] = 2
 
    for i in range(4, N + 1):
 
        DP[i] = DP[i - 2] + DP[i - 3]
 
    return DP[N]
 
# Driver Code
if __name__ == '__main__':
    N = 20
 
    print(countSubsets(N))
     
# This code is contributed by Mohit Kumar

C#

// C# code to count subsets not containing
// adjacent elements from 1 to N
using System;
class GFG{
 
// Function to count subsets
static int countSubsets(int N)
{
    if(N <= 2)
        return N;
         
    if(N == 3)
        return 2;
     
    int []DP = new int[N + 1];
     
    DP[0] = 0;
    DP[1] = 1;
    DP[2] = 2;
    DP[3] = 2;
     
    for(int i = 4; i <= N; i++)
    {
        DP[i] = DP[i - 2] + DP[i - 3];
    }
    return DP[N];
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 20;
     
    Console.Write(countSubsets(N));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

// javascript code to count subsets not containing
// adjacent elements from 1 to N   
// Function to count subsets
    function countSubsets(N) {
        if (N <= 2)
            return N;
 
        if (N == 3)
            return 2;
 
        var DP = Array(N + 1).fill(0);
 
        DP[0] = 0;
        DP[1] = 1;
        DP[2] = 2;
        DP[3] = 2;
 
        for (i = 4; i <= N; i++) {
            DP[i] = DP[i - 2] + DP[i - 3];
        }
        return DP[N];
    }
 
    // Driver code
     
        var N = 20;
 
        document.write(countSubsets(N));
 
// This code contributed by gauravrajput1

输出：

265

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