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

1137. N-th Tribonacci Number (Easy)

The Tribonacci sequence T_n is defined as follows: 

T₀ = 0, T₁ = 1, T₂ = 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.

Related Topics:
Recursion

Similar Questions:

• Climbing Stairs (Easy)
• Fibonacci Number (Easy)

Solution 1.

• Java
• C++
• Python
• Go
• Javascript
• Php

• 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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