最大大小为2的最小分区和总和受给定值限制

给定一个数组arr []为正数，请找到满足以下属性的数组中的最小集合数：
1.一个集合中最多可以包含两个元素。这两个元素不必是连续的。
2. set元素的总和应小于或等于给定的Key。可以假定给定键大于或等于最大数组元素。

例子：

Input: arr[] = [10, 20, 12], key = 25
Output: 2
We break into two parts {10, 12} and {2}

Input : arr[] = [3, 5, 3, 4], key=5
Output : 4
Explanation: 4 sets (3), (5), (3), (4)

这个想法是先对数组排序，然后遵循两个指针的方法。我们从排序数组的两个角开始两个指针。如果它们的总和小于或等于给定的键，则我们对它们进行设置，否则我们仅考虑最后一个元素。

下面是上述方法的实现：

C++

// C++ program to count minimum number of partitions
// of size 2 and sum smaller than or equal to given
// key.
#include 
#include 
using namespace std;
  
int minimumSets(int arr[], int n, int key)
{
    int i, j;
  
    // sort the array
    sort(arr, arr + n);
  
    // if sum of ith smaller and jth larger element is
    // less than key, then pack both numbers in a set 
    // otherwise pack the jth larger number
    // alone in the set
    for (i = 0, j = n - 1; i <= j; ++i)
        if (arr[i] + arr[j] <= key)
            j--;
  
    // After ending of loop i will contain minimum
    // number of sets
    return i;
}
  
int main()
{
    int arr[] = { 3, 5, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int key = 5;
    cout << minimumSets(arr, n, key);
    return 0;
}

Java

// Java program to count minimum number of partitions
// of size 2 and sum smaller than or equal to given
// key.
  
import java.util.Arrays; 
class GFG {
  
  
static int minimumSets(int arr[], int n, int key)
{
    int i, j;
  
    // sort the array
    Arrays.sort(arr);
  
    // if sum of ith smaller and jth larger element is
    // less than key, then pack both numbers in a set 
    // otherwise pack the jth larger number
    // alone in the set
    for (i = 0, j = n - 1; i <= j; ++i)
        if (arr[i] + arr[j] <= key)
            j--;
  
    // After ending of loop i will contain minimum
    // number of sets
    return i;
}
  
  
  
    public static void main (String[] args) {
    int []arr = { 3, 5, 3, 4 };
    int n =arr.length;
    int key = 5;
    System.out.println( minimumSets(arr, n, key));
    }
}
// This code is contributed by chandan_jnu.

Python3

# Python 3 program to count minimum number 
# of partitions of size 2 and sum smaller 
# than or equal to given key.
  
def minimumSets(arr, n, key):
      
    # sort the array
    arr.sort(reverse = False)
  
    # if sum of ith smaller and jth larger 
    # element is less than key, then pack 
    # both numbers in a set otherwise pack 
    # the jth larger number alone in the set
    j = n - 1
    for i in range(0, j + 1, 1):
        if (arr[i] + arr[j] <= key):
            j -= 1
  
    # After ending of loop i will 
    # contain minimum number of sets
    return i + 1
  
# Driver Code
if __name__ == '__main__':
    arr = [3, 5, 3, 4]
    n = len(arr)
    key = 5
    print(minimumSets(arr, n, key))
  
# This code is contributed by
# Shashank_Sharma

C#

// C# program to count minimum 
// number of partitions of size 
// 2 and sum smaller than or 
// equal to given key.
using System;
class GFG 
{
  
static int minimumSets(int []arr, 
                       int n, int key)
{
    int i, j;
  
    // sort the array
    Array.Sort(arr);
  
    // if sum of ith smaller and 
    // jth larger element is less 
    // than key, then pack both 
    // numbers in a set otherwise 
    // pack the jth larger number
    // alone in the set
    for (i = 0, j = n - 1; i <= j; ++i)
        if (arr[i] + arr[j] <= key)
            j--;
  
    // After ending of loop i 
    // will contain minimum
    // number of sets
    return i;
}
  
// Driver Code
public static void Main ()
{
    int []arr = { 3, 5, 3, 4 };
    int n =arr.Length;
    int key = 5;
    Console.WriteLine(minimumSets(arr, n, key));
}
}
  
// This code is contributed
// by chandan_jnu.

输出：

4

时间复杂度： O(nlogn)