The Tribonacci numbers, commonly denoted T(n) form a sequence, called the Tribonacci sequence, such that each number is the sum of the two preceding ones, starting from 0, 1 and 1. That is,
T(0) = 0, T(1) = 1, T(2) = 1 T(n) = T(n - 1) + T(n - 2) + T(n - 3), for n > 2.
Given n, calculate and return T(n).
Example 1:
Input: n = 3 Output: 2 Explanation: T(3) = T(2) + T(1) + T(0) = 1 + 1 + 0 = 2.
Example 2:
Input: n = 4 Output: 4 Explanation: T(4) = T(3) + T(2) + T(1) = 2 + 1 + 1 = 4.
Example 3:
Input: n = 5 Output: 7 Explanation: T(5) = T(4) + T(3) + T(2) = 4 + 2 + 1 = 7.
Note:
Your algorithm should run in O(n) time and use O(1) space.
The core challenge of this problem is to compute the nth Tribonacci number efficiently. The Tribonacci sequence is similar to the Fibonacci sequence but sums the last three numbers instead of the last two. This sequence has applications in various mathematical and computational problems.
To solve this problem, we can start with a naive recursive solution, but it will be highly inefficient due to repeated calculations. Instead, we can use an iterative approach to achieve O(n) time complexity and O(1) space complexity.
The naive approach involves a simple recursive function:
function tribonacci(n) {
if (n === 0) return 0;
if (n === 1 || n === 2) return 1;
return tribonacci(n - 1) + tribonacci(n - 2) + tribonacci(n - 3);
}
However, this approach has exponential time complexity, O(3^n), due to repeated calculations.
We can optimize the solution by using an iterative approach with three variables to keep track of the last three Tribonacci numbers. This approach ensures O(n) time complexity and O(1) space complexity.
1. Initialize three variables to store the first three Tribonacci numbers: T0 = 0, T1 = 1, T2 = 1.
2. Iterate from 3 to n, updating the variables to store the current Tribonacci number.
3. Return the nth Tribonacci number.
function tribonacci(n) {
// Base cases
if (n === 0) return 0;
if (n === 1 || n === 2) return 1;
// Initialize the first three Tribonacci numbers
let T0 = 0, T1 = 1, T2 = 1;
// Iterate from 3 to n
for (let i = 3; i <= n; i++) {
// Calculate the next Tribonacci number
let Tn = T0 + T1 + T2;
// Update the variables
T0 = T1;
T1 = T2;
T2 = Tn;
}
// Return the nth Tribonacci number
return T2;
}
The time complexity of the iterative solution is O(n) because we iterate from 3 to n. The space complexity is O(1) because we use a constant amount of space to store the last three Tribonacci numbers.
1. n = 0: The function should return 0.
2. n = 1 or n = 2: The function should return 1.
3. Large values of n: The function should handle large values efficiently due to its O(n) time complexity.
To test the solution comprehensively, we can use a variety of test cases:
console.log(tribonacci(0)); // 0
console.log(tribonacci(1)); // 1
console.log(tribonacci(2)); // 1
console.log(tribonacci(3)); // 2
console.log(tribonacci(4)); // 4
console.log(tribonacci(5)); // 7
console.log(tribonacci(25)); // 1389537
1. Break down the problem into smaller parts and understand the base cases.
2. Consider both recursive and iterative approaches and analyze their time and space complexities.
3. Practice solving similar problems to improve problem-solving skills.
Understanding and solving the Tribonacci number problem helps in developing efficient algorithms for sequence-based problems. The iterative approach ensures optimal time and space complexity, making it suitable for large inputs.
1. Generalizations of Fibonacci numbers
2. LeetCode: N-th Tribonacci Number
3. GeeksforGeeks: Tribonacci Numbers