总和值不超过K的最长递增子序列

给定一个大小为N的整数数组arr []和一个整数K。任务是找到最长子序列的总和小于或等于K的长度。

例子：

Input: arr[] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15} K = 40
Output: 5
Explanation:
If we select subsequence {0, 1, 3, 7, 15} then total sum will be 26, which is less than 40. Hence, the longest increasing possible subsequence length is 5.Input: arr[] = {5, 8, 3, 7, 9, 1} K = 4
Output: 1

方法：

1. 可以使用递归来解决上述问题。
2. 在数组的任何位置，有两种选择：
   • 如果总和小于K ，则在该位置选择元素，然后浏览其余项。
   • 将元素保留在该位置，然后探索其余元素。

递归关系将给出为：

Recurrence relation:
T(N) = max(sove(arr, N, arr[i], i+1, K-arr[i])+1, solve(arr, N, prevele, i+1, K));
Base conditions:
if(i >= N || K <= 0)
return 0

这是上述方法的实现：

C++

// C++ program to find the Longest
// Increasing Subsequence having sum
// value atmost K
#include 
using namespace std;
  
int solve(int arr[], int N,
          int prevele, int i, int K)
{
    // check for base cases
    if (i >= N || K <= 0)
        return 0;
  
    // check if it is possible to take
    // current elements
    if (arr[i] <= prevele
        || (K - arr[i] < 0)) {
  
        return solve(arr, N, prevele,
                     i + 1, K);
    }
  
    // if current element is ignored
    else {
        int ans = max(
            solve(arr, N, arr[i],
                  i + 1, K - arr[i])
                + 1,
            solve(arr, N, prevele,
                  i + 1, K));
        return ans;
    }
}
  
// Driver Code
int main()
{
    int N = 16;
    int arr[N]
        = { 0, 8, 4, 12,
            2, 10, 6, 14,
            1, 9, 5, 13,
            3, 11, 7, 15 };
    int K = 40;
  
    cout << solve(arr, N,
                  INT_MIN, 0, K)
         << endl;
}

Java

// Java program to find the Longest 
// Increasing Subsequence having sum 
// value atmost K 
import java.io.*; 
  
class GFG{ 
      
static int solve(int arr[], int N,
                 int prevele, int i, int K)
{
      
    // Check for base cases
    if (i >= N || K <= 0)
        return 0;
  
    // Check if it is possible to take
    // current elements
    if (arr[i] <= prevele ||
       (K - arr[i] < 0))
    {
        return solve(arr, N, prevele,
                     i + 1, K);
    }
  
    // If current element is ignored
    else
    {
        int ans = Math.max(solve(arr, N, arr[i],
                              i + 1, K - arr[i]) + 1,
                           solve(arr, N, prevele,
                                 i + 1, K));
                                   
        return ans;
    }
} 
  
// Driver code 
public static void main (String[] args) 
{ 
    int N = 16;
    int arr[] = new int[]{ 0, 8, 4, 12,
                           2, 10, 6, 14,
                           1, 9, 5, 13,
                           3, 11, 7, 15 };
    int K = 40;
  
    System.out.print(solve(arr, N,
          Integer.MIN_VALUE, 0, K)); 
} 
} 
  
// This code is contributed by Pratima Pandey

Python3

# Python3 program to find the Longest
# Increasing Subsequence having sum
# value atmost K
import sys
  
def solve(arr, N, prevele, i, K):
      
    # Check for base cases
    if (i >= N or K <= 0):
        return 0;
  
    # Check if it is possible to take
    # current elements
    if (arr[i] <= prevele or 
       (K - arr[i] < 0)):
        return solve(arr, N, prevele,
                     i + 1, K);
  
    # If current element is ignored
    else:
        ans = max(solve(arr, N, arr[i],
                     i + 1, K - arr[i]) + 1,
                  solve(arr, N, prevele, 
                        i + 1, K));
  
        return ans;
  
# Driver code
if __name__ == '__main__':
      
    N = 16;
    arr = [ 0, 8, 4, 12,
            2, 10, 6, 14,
            1, 9, 5, 13,
            3, 11, 7, 15 ];
    K = 40;
  
    print(solve(arr, N, -sys.maxsize, 0, K));
  
# This code is contributed by 29AjayKumar

C#

// C# program to find the Longest 
// Increasing Subsequence having sum 
// value atmost K 
using System; 
  
class GFG{ 
      
static int solve(int[] arr, int N,
                 int prevele, int i, int K)
{
      
    // Check for base cases
    if (i >= N || K <= 0)
        return 0;
  
    // Check if it is possible to take
    // current elements
    if (arr[i] <= prevele ||
       (K - arr[i] < 0))
    {
        return solve(arr, N, prevele,
                     i + 1, K);
    }
  
    // If current element is ignored
    else
    {
        int ans = Math.Max(solve(arr, N, arr[i],
                                 i + 1, K - arr[i]) + 1,
                           solve(arr, N, prevele,
                                 i + 1, K));
                                  
        return ans;
    }
} 
  
// Driver code 
public static void Main () 
{ 
    int N = 16;
    int[] arr = new int[]{ 0, 8, 4, 12,
                           2, 10, 6, 14,
                           1, 9, 5, 13,
                           3, 11, 7, 15 };
    int K = 40;
  
    Console.Write(solve(arr, N,
        Int32.MinValue, 0, K)); 
} 
}
  
// This code is contributed by sanjoy_62

输出：

5

时间复杂度： O(2 ^N )
辅助空间： O(1)

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。