Count Special Quadruplets - Leetcode Solution

Count Special Quadruplets is a problem on LeetCode that asks the user to find the number of quadruplets (a, b, c, d) such that a, b, c, and d are integers between 1 and n and satisfy the following conditions:

1. a + b = c
2. b + c = d
3. 1 <= a < b < c < d <= n

The task is to write a function that takes in an integer n and returns the number of quadruplets that satisfy these conditions.

To solve this problem, we first need to understand the conditions given in the problem statement. The conditions require us to find quadruplets (a, b, c, d) where:

1. The sum of the first two elements, a and b, is equal to the third element, c.
2. The sum of the second and third elements, b and c, is equal to the fourth element, d.
3. The elements a, b, c, and d are integers between 1 and n.
4. The elements a, b, c, and d are distinct and arranged in increasing order.

One way to approach this problem is to use nested loops to generate all possible quadruplets and then check if each one satisfies the given conditions. However, this approach has a time complexity of O(n^4) and is not feasible for larger values of n.

A better approach is to use the fact that elements a, b, c, and d are distinct and increasing order to reduce the number of loops. We can start with the smallest value of a and b, which are 1 and 2 respectively, and then increment c and d until we reach the maximum value of n. If the conditions are satisfied, we can count the quadruplet.

Here is pseudocode to find the solution:

1. Initialize a counter variable, say count, to 0.
2. Loop over all possible values of a from 1 to n-3.
3. Loop over all possible values of b from a+1 to n-2.
4. Calculate c as a+b.
5. Loop over all possible values of c from b+1 to n-1.
6. Calculate d as b+c.
7. If d is less than or equal to n, check if a, b, c, d satisfy the conditions.
8. If the conditions are satisfied, increment the counter.
9. Return count.

Here is the python code for the same:

def count_quadruplets(n: int) -> int: count = 0 for a in range(1, n-2): for b in range(a+1, n-1): c = a + b for d in range(b+1, n): if d == c: count += 1 return count

The time complexity of this algorithm is O(n^3) which is much better than the brute force approach.

1