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Formatted question description: https://leetcode.ca/all/1137.html

1137. N-th Tribonacci Number (Easy)

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.

Related Topics:
Recursion

Similar Questions:

Solution 1.

  • class Solution {
        public int tribonacci(int n) {
            int a = 0, b = 1, c = 1;
            while (n-- > 0) {
                int d = a + b + c;
                a = b;
                b = c;
                c = d;
            }
            return a;
        }
    }
    
  • class Solution {
    public:
        int tribonacci(int n) {
            long long a = 0, b = 1, c = 1;
            while (n--) {
                long long d = a + b + c;
                a = b;
                b = c;
                c = d;
            }
            return (int) a;
        }
    };
    
  • class Solution:
        def tribonacci(self, n: int) -> int:
            a, b, c = 0, 1, 1
            for _ in range(n):
                a, b, c = b, c, a + b + c
            return a
    
    
  • func tribonacci(n int) int {
    	a, b, c := 0, 1, 1
    	for i := 0; i < n; i++ {
    		a, b, c = b, c, a+b+c
    	}
    	return a
    }
    
  • /**
     * @param {number} n
     * @return {number}
     */
    var tribonacci = function (n) {
        let a = 0;
        let b = 1;
        let c = 1;
        while (n--) {
            let d = a + b + c;
            a = b;
            b = c;
            c = d;
        }
        return a;
    };
    
    
  • class Solution {
        /**
         * @param Integer $n
         * @return Integer
         */
        function tribonacci($n) {
            if ($n == 0) {
                return 0;
            } else if ($n == 1 || $n == 2) {
                return 1;
            }
            $dp = [0, 1, 1];
            for ($i = 3; $i <= $n; $i++) {
                $dp[$i] = $dp[$i - 1] + $dp[$i - 2] + $dp[$i - 3];
            }
            return $dp[$n];
        }
    }
    

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