# Solving Leetcode Interviews in Seconds with AI: N-th Tribonacci Number


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1137" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> The Tribonacci sequence Tn is defined as follows:  T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0. Given n, return the value of Tn.   Example 1:  Input: n = 4 Output: 4 Explanation: T_3 = 0 + 1 + 1 = 2 T_4 = 1 + 1 + 2 = 4  Example 2:  Input: n = 25 Output: 1389537    Constraints:  0 <= n <= 37 The answer is guaranteed to fit within a 32-bit integer, ie. answer <= 2^31 - 1.  

	# Explanation
	Here's an efficient solution to calculate the nth Tribonacci number:

*   **Iterative Approach:** We'll use an iterative approach to calculate the Tribonacci numbers. We will start from the base cases (T0, T1, T2) and iteratively compute subsequent values up to Tn.
*   **Constant Space Optimization:** To optimize space, we will maintain only the last three Tribonacci numbers calculated. These will be updated in each iteration to produce the next number in the sequence.

*   **Runtime Complexity:** O(n), **Storage Complexity:** O(1)

	
	# Code
	```python
	def tribonacci(n: int) -> int:
    """
    Calculates the nth Tribonacci number.

    Args:
        n: The index of the Tribonacci number to calculate.

    Returns:
        The nth Tribonacci number.
    """
    if n == 0:
        return 0
    if n == 1 or n == 2:
        return 1

    t0 = 0
    t1 = 1
    t2 = 1

    for _ in range(3, n + 1):
        tn = t0 + t1 + t2
        t0 = t1
        t1 = t2
        t2 = tn

    return t2
	```