Problem:

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.

Problem Analysis:

High Level Strategy

1. This solution utilizes memoization to avoid redundant calculations when computing the Tribonacci sequence. It starts with a dictionary memo initialized with the base cases of the Tribonacci sequence (0, 1, 1), which are stored as key-value pairs. When computing the Tribonacci number for a given n, it first checks if n is already in the memo dictionary. If it is, it returns the value associated with n. Otherwise, it recursively computes the Tribonacci number for n-1, n-2, and n-3, adds them up, stores the result in the memo dictionary, and returns the computed value. This approach saves time by avoiding redundant calculations, especially for larger values of n.

Time Complexity

1. The time complexity of this solution is O(n), where n is the input parameter. This is because each value of n is computed only once and then stored in the memo dictionary for future use. The space complexity is also O(n) due to the memo dictionary, which can store at most n key-value pairs. Overall, this solution provides an efficient way to compute Tribonacci numbers by trading off space for time.

Solutions:

class Solution:
    def __init__(self):
        self.memo = {0: 0, 1: 1, 2: 1}

    def tribonacci(self, n: int) -> int:
        if n in self.memo:
            return self.memo[n]
            
        self.memo[n] = self.tribonacci(n-1) + self.tribonacci(n-2) + self.tribonacci(n-3)
        return self.memo[n]

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