1137. N-th Tribonacci Number

Easy | DP | 24 ms (94.39%), 14.2 MB (43.64%)

Source: LeetCode - N-th Tribonacci Number GitHub: Solution / Performance

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.

Constraints:

0 <= n <= 37
The answer is guaranteed to fit within a 32-bit integer, ie. answer <= 2^31 - 1.

Input: n = 4
Output: 4
Explanation:
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4

Input: n = 25
Output: 1389537

Same as 0509. Fibonacci Number

class Solution:
    def tribonacci(self, n: int) -> int:
        # (base case)
        if n == 0: return 0
        if n == 1: return 1
        if n == 2: return 1
        
        # ==================================================
        #  Dynamic Programming                             =
        # ==================================================
        # time  : O(n)
        # space : O(1)
        
        first, second, third = 0, 1, 1
        
        for i in range(n - 2):
            ans = first + second + third
            first = second
            second = third
            third = ans
            
        return third

class Solution {
    /**
     * @time  : O(n)
     * @space : O(1)
     */

    public int tribonacci(int n) {
        /* base case */
        if(n == 0) return 0;
        if(n == 1 || n == 2) return 1;
        
        int first = 0, second = 1, third = 1;
        
        for(int i=0 ; i<n-2 ; i++) {
            int ans = first + second + third;
            first = second;
            second = third;
            third = ans;
        }
        
        return third; 
    }
}

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