LeetCode //C - 1137. N-th Tribonacci Number

1137. N-th Tribonacci Number

The Tribonacci sequence Tn is defined as follows:

$T_0=0$, $T_1=1$, $T_2=1$, and $T_{n+3}=T_n+T_{n+1}+T_{n+2}$ for n >= 0.

Given n, return the value of $T_n$.
 

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 <= $2^{31}-1$.

From: LeetCode
Link: 1137. N-th Tribonacci Number

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Solution:

Ideas：

This function first checks for the base cases, i.e., when n is 0, 1, or 2, and returns the corresponding Tribonacci number directly. For other cases, it iteratively calculates the Tribonacci numbers up to n and returns the last computed value. This solution is efficient and avoids the overhead of recursion, making it well-suited for the constraints given in the problem statement.

Caode:

int tribonacci(int n) {
    // Base cases
    if (n == 0) return 0;
    if (n == 1 || n == 2) return 1;

    // Initial values for T_0, T_1, and T_2
    int T0 = 0, T1 = 1, T2 = 1;

    // To store the next Tribonacci number
    int nextT;

    // Compute Tribonacci from T_3 to T_n
    for (int i = 3; i <= n; i++) {
        nextT = T0 + T1 + T2; // Compute the next Tribonacci number
        T0 = T1; // Update T0 to T1's value
        T1 = T2; // Update T1 to T2's value
        T2 = nextT; // Update T2 to the newly computed value
    }

    // T2 now contains T_n
    return T2;
}