Given two integers **L** and **R**, the task is to find the number of unique triplets whose values lie in the range **[L, R]**, such that the sum of any two numbers is equal to the third number.

**Examples:**

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

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

**Naive Approach:** The simplest approach to solve the problem is to generate all possible triplets over the range **[L, R]** and count those triplets having sum of any two numbers from the pair equal to the third number. After checking for all the triplets, print the total count obtained. **Time Complexity:** O((R – L)^{3})**Auxiliary Space:** O(1)

**Efficient Approach:** The above approach can be optimized based on the following observations:

- If the difference between the given range is less than
**L**, then there doesn’t exist any such triplet having the sum of two numbers equal to the third number. - If the difference between the given range is
**at least L**, then**(R – L)**lies in the range**L**and**R**, i.e.**{L, (R – L), R}**. It can be observed that the sum of**L**and any other number in the range**[L, R – L]**is**at most R**. - Therefore, the total number of possible valid triplets, where the first element is
**L**is given by**(R – L – L + 1)**. - Similarly, when the first element is
**(L + 1)**, then the number of triplets is**(R – L – L)**, and so on.

From the above observations, the total number of triplets forms an AP with the terms **(R – L – L + 1), (R – L – L), (R – L – L – 1), ………** consisting of (R – L – L + 1) number of terms. Therefore, the total number of triplets is given by:

where, N is (R – L – L + 1)

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `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

`<script>` ` ` `// 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));` `</script>` |

**Output:**

6

**Time Complexity:** O(1)**Auxiliary Space:** O(1)

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