最小和最大元素之和小于K的子集数

给定一个整数数组arr []和一个整数K ，任务是找到非空子集S的数量，使得min(S)+ max(S) 。

例子：

Input: arr[] = {2, 4, 5, 7} K = 8
Output: 4
Explanation:
The possible subsets are {2}, {2, 4}, {2, 4, 5} and {2, 5}

Input:: arr[] = {2, 4, 2, 5, 7} K = 10
Output: 26

方法

1. 首先对输入数组进行排序。
2. 现在，使用“两指针技术”来计数子集的数量。
3. 让我们左右两个指针，并设置left = 0和right = N-1。

4. if (arr[left] + arr[right] < K )
   Increment the left pointer by 1 and add 2 ^{j – i} into answer, because the left and right values make up a potential end values of a subset. All the values from [i, j – 1] also make up end of subsets which will have the sum < K. So, we need to calculate all the possible subsets for left = i and right ∊ [i, j]. So, after suming up values 2 ^{j – i + 1} + 2 ^{j – i – 2} + … + 2 ⁰ of the GP, we get 2 ^{j – i} .
   if( arr[left] + arr[right] >= K )
   Decrement the right pointer by 1.

5. 重复以下过程，直到left <= right为止。

下面是上述方法的实现：

C++

// C++ program to print count
// of subsets S such that
// min(S) + max(S) < K
  
#include 
using namespace std;
  
// Function that return the
// count of subset such that
// min(S) + max(S) < K
int get_subset_count(int arr[], int K,
                     int N)
{
    // Sorting the array
    sort(arr, arr + N);
  
    int left, right;
    left = 0;
    right = N - 1;
  
    // ans stores total number of subsets
    int ans = 0;
  
    while (left <= right) {
        if (arr[left] + arr[right] < K) {
  
            // add all posible subsets
            // between i and j
            ans += 1 << (right - left);
            left++;
        }
        else {
            // Decrease the sum
            right--;
        }
    }
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 4, 5, 7 };
    int K = 8;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << get_subset_count(arr, K, N);
    return 0;
}

Java

// Java program to print count
// of subsets S such that
// Math.min(S) + Math.max(S) < K
import java.util.*;
  
class GFG{
  
// Function that return the
// count of subset such that
// Math.min(S) + Math.max(S) < K
static int get_subset_count(int arr[], int K,
                                       int N)
{
      
    // Sorting the array
    Arrays.sort(arr);
  
    int left, right;
    left = 0;
    right = N - 1;
  
    // ans stores total number
    // of subsets
    int ans = 0;
  
    while (left <= right)
    {
        if (arr[left] + arr[right] < K)
        {
  
            // Add all posible subsets
            // between i and j
            ans += 1 << (right - left);
            left++;
        }
        else 
        {
              
            // Decrease the sum
            right--;
        }
    }
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 4, 5, 7 };
    int K = 8;
    int N = arr.length;
      
    System.out.print(get_subset_count(arr, K, N));
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 program to print 
# count of subsets S such 
# that min(S) + max(S) < K 
  
# Function that return the
# count of subset such that
# min(S) + max(S) < K 
def get_subset_count(arr, K, N):
  
    # Sorting the array 
    arr.sort() 
  
    left = 0; 
    right = N - 1; 
  
    # ans stores total number of subsets 
    ans = 0; 
  
    while (left <= right):
        if (arr[left] + arr[right] < K):
              
            # Add all posible subsets 
            # between i and j 
            ans += 1 << (right - left); 
            left += 1; 
        else:
              
            # Decrease the sum 
            right -= 1; 
      
    return ans; 
  
# Driver code 
arr = [ 2, 4, 5, 7 ]; 
K = 8; 
  
print(get_subset_count(arr, K, 4))
  
# This code is contributed by grand_master

C#

// C# program to print count
// of subsets S such that
// Math.Min(S) + Math.Max(S) < K
using System;
  
class GFG{
  
// Function that return the
// count of subset such that
// Math.Min(S) + Math.Max(S) < K
static int get_subset_count(int []arr, int K,
                                       int N)
{
      
    // Sorting the array
    Array.Sort(arr);
  
    int left, right;
    left = 0;
    right = N - 1;
  
    // ans stores total number
    // of subsets
    int ans = 0;
  
    while (left <= right)
    {
        if (arr[left] + arr[right] < K)
        {
              
            // Add all posible subsets
            // between i and j
            ans += 1 << (right - left);
            left++;
        }
        else
        {
              
            // Decrease the sum
            right--;
        }
    }
    return ans;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 4, 5, 7 };
    int K = 8;
    int N = arr.Length;
      
    Console.Write(get_subset_count(arr, K, N));
}
}
  
// This code is contributed by gauravrajput1

输出：

4

时间复杂度： O(N * log N)
辅助空间： O(1)