计算给定范围内三元组的两个数字之和等于第三个数字的三胞胎

给定两个整数L和R ，任务是查找值在[L，R]范围内的唯一三元组的数量，以使任何两个数字的总和等于第三个数字。

例子：

Input: L = 1, R = 3
Output: 3
Explanation: Three such triplets satisfying the necessary conditions are (1, 1, 2), (1, 2, 3) and (2, 1, 3).

Input: L = 2, R = 6
Output: 6

天真的方法：解决问题的最简单方法是生成[L，R]范围内的所有可能的三胞胎，并计算那些从该对中任意两个数之和等于第三数的三胞胎。检查完所有三元组后，打印获得的总数。
时间复杂度： O((R – L) ³ )
辅助空间： O(1)

高效的方法：可以基于以下观察来优化上述方法：

• 如果给定范围之间的差小于L ，则不存在任何这样的三元组，其两个数之和等于第三个数。
• 如果给定范围之间的差至少为L ，则(R – L)位于L和R的范围内，即{L，(R – L)，R} 。可以看出， L和[L，R – L]范围内的任何其他数字的总和最多为R。
• 因此，第一个元素为L的可能有效三元组的总数由(R – L – L + 1)给出。
• 同样，当第一个元素为(L + 1)时，三元组的数量为(R – L – L) ，依此类推。

根据上述观察，三元组的总数构成了一个AP，其术语为(R – L – L + 1)，(R – L – L)，(R – L – L – 1)，………，由( R – L – L +1)项数。因此，三元组的总数由下式给出：

$\frac{N*(N + 1)}{2}$
where, N is (R – L – L + 1)

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the number of triplets
// from the range [L, R] having sum of two
// numbers from the triplet equal to the third number
int totalCombination(int L, int R)
{
    // Stores the total number of triplets
    int count = 0;
 
    // Find the difference of the range
    int K = R - L;
 
    // Case 1: If triplets can't
    // be formed, then return 0
    if (K < L)
        return 0;
 
    // Otherwise
    int ans = K - L;
 
    // Update the total number of triplets
    count = ((ans + 1) * (ans + 2)) / 2;
 
    // Return the count
    return count;
}
 
// Driver Code
int main()
{
    int L = 2, R = 6;
    cout << totalCombination(L, R);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
  // Function to find the number of triplets
  // from the range [L, R] having sum of two
  // numbers from the triplet equal to the third number
  static int totalCombination(int L, int R)
  {
 
    // Stores the total number of triplets
    int count = 0;
 
    // Find the difference of the range
    int K = R - L;
 
    // Case 1: If triplets can't
    // be formed, then return 0
    if (K < L)
      return 0;
 
    // Otherwise
    int ans = K - L;
 
    // Update the total number of triplets
    count = ((ans + 1) * (ans + 2)) / 2;
 
    // Return the count
    return count;
  }
 
  // Driven Code
  public static void main(String[] args)
  {
    int L = 2, R = 6;
    System.out.print(totalCombination(L, R));
  }
}
 
// This code is contributed by susmitakundugoaldanga.

Python3

# Python3 program for the above approach
 
 
# Function to find the number of triplets
# from the range [L, R] having sum of two
# numbers from the triplet equal to the third number
def totalCombination(L, R):
   
    # Stores the total number of triplets
    count = 0
 
    # Find the difference of the range
    K = R - L
 
    # Case 1: If triplets can't
    # be formed, then return 0
    if (K < L):
        return 0
 
    # Otherwise
    ans = K - L
 
    # Update the total number of triplets
    count = ((ans + 1) * (ans + 2)) // 2
 
    # Return the count
    return count
 
# Driver Code
if __name__ == '__main__':
    L, R = 2, 6
    print (totalCombination(L, R))
 
# This code is contributed by mohit kumar 29.

C#

// C# program to implement
// the above approach
using System;
class GFG
{
   
  // Function to find the number of triplets
  // from the range [L, R] having sum of two
  // numbers from the triplet equal to the third number
  static int totalCombination(int L, int R)
  {
     
    // Stores the total number of triplets
    int count = 0;
 
    // Find the difference of the range
    int K = R - L;
 
    // Case 1: If triplets can't
    // be formed, then return 0
    if (K < L)
      return 0;
 
    // Otherwise
    int ans = K - L;
 
    // Update the total number of triplets
    count = ((ans + 1) * (ans + 2)) / 2;
 
    // Return the count
    return count;
  }
 
 
  // Driver Code
  public static void Main()
  {
    int L = 2, R = 6;
    Console.WriteLine(totalCombination(L, R));
  }
}
 
// This code is contributed by sauravghosh0416.

Javascript

    // Javascript program for the above approach
     
    // Function to find the number of triplets
    // from the range [L, R] having sum of two
    // numbers from the triplet equal to the third number
    function totalCombination(L, R)
    {
        // Stores the total number of triplets
        let count = 0;
 
        // Find the difference of the range
        let K = R - L;
 
        // Case 1: If triplets can't
        // be formed, then return 0
        if (K < L)
            return 0;
 
        // Otherwise
        let ans = K - L;
 
        // Update the total number of triplets
        count = ((ans + 1) * (ans + 2)) / 2;
 
        // Return the count
        return count;
    }
     
    let L = 2, R = 6;
    document.write(totalCombination(L, R));
 

输出：

6

时间复杂度： O(1)
辅助空间： O(1)