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 instead of summing the last two numbers, we sum the last three numbers. This problem is significant in various applications such as dynamic programming and algorithm optimization.
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 recursively calculating T(n) by summing T(n-1), T(n-2), and T(n-3). However, this approach has exponential time complexity due to overlapping subproblems.
We can optimize the solution by using an iterative approach with three variables to keep track of the last three Tribonacci numbers. This way, we can compute the result in O(n) time and O(1) space.
Here is a step-by-step breakdown of the optimized iterative algorithm:
public class Tribonacci {
public int tribonacci(int n) {
// Base cases
if (n == 0) return 0;
if (n == 1 || n == 2) return 1;
// Initialize the first three Tribonacci numbers
int t0 = 0, t1 = 1, t2 = 1;
// Variable to store the current Tribonacci number
int current = 0;
// Iterate from 3 to n
for (int i = 3; i <= n; i++) {
// Calculate the current Tribonacci number
current = t0 + t1 + t2;
// Update the last three Tribonacci numbers
t0 = t1;
t1 = t2;
t2 = current;
}
// Return the nth Tribonacci number
return current;
}
public static void main(String[] args) {
Tribonacci tribonacci = new Tribonacci();
System.out.println(tribonacci.tribonacci(3)); // Output: 2
System.out.println(tribonacci.tribonacci(4)); // Output: 4
System.out.println(tribonacci.tribonacci(5)); // Output: 7
}
}
The time complexity of the optimized iterative solution is O(n) because we iterate from 3 to n. The space complexity is O(1) because we only use a constant amount of extra space to store the last three Tribonacci numbers.
Potential edge cases include:
These edge cases are handled by the base cases in the code.
To test the solution comprehensively, we should include a variety of test cases:
We can use JUnit or any other testing framework to automate the testing process.
When approaching such problems, it is essential to:
In this blog post, we discussed the Tribonacci number problem, explored different approaches to solve it, and provided an optimized iterative solution in Java. Understanding and solving such problems is crucial for improving algorithmic thinking and coding skills.
For further reading and practice, consider the following resources: