How do you solve N-th Tribonacci Number on LeetCode?

N-th Tribonacci Number (LeetCode #1137) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Dynamic programming can significantly reduce time complexity by storing intermediate results.

What is the brute force solution for N-th Tribonacci Number?

The brute force approach for N-th Tribonacci Number is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves recursively calculating the Tribonacci numbers. We directly implement the recursive formula, but this leads to repeated calculations, making it inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve N-th Tribonacci Number?

N-th Tribonacci Number uses the following concepts: Math, Dynamic Programming, Memoization. The recommended patterns to study are: Dynamic Programming, Memoization, Array.

What are the interview tips for N-th Tribonacci Number?

Always consider edge cases, especially for base cases in recursion. Explain your thought process clearly when transitioning from brute force to optimal solutions. Practice implementing both recursive and iterative solutions to build a strong foundation.

What are common mistakes when solving N-th Tribonacci Number?

Not recognizing the need for base cases in recursion. Failing to optimize recursive solutions with memoization or dynamic programming.

#1137

N-th Tribonacci Number

Easy

MathDynamic ProgrammingMemoizationDynamic ProgrammingMemoizationArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to store previously calculated Tribonacci numbers, avoiding redundant calculations and significantly improving efficiency.

⚙️

Algorithm

4 steps

1Step 1: Create an array F of length n+1 to store Tribonacci numbers.
2Step 2: Initialize F[0] = 0, F[1] = 1, F[2] = 1.
3Step 3: Use a loop from 3 to n, setting F[i] = F[i-1] + F[i-2] + F[i-3].
4Step 4: Return F[n].

solution.py6 lines

1def tribonacci(n):
2    F = [0] * (n + 1)
3    F[0], F[1], F[2] = 0, 1, 1
4    for i in range(3, n + 1):
5        F[i] = F[i - 1] + F[i - 2] + F[i - 3]
6    return F[n]

ℹ

Complexity note: The time complexity is O(n) because we compute each Tribonacci number once. The space complexity is O(n) due to the array storing the results of each computation.

1Dynamic programming can significantly reduce time complexity by storing intermediate results.
2Understanding the recursive relationships helps in formulating the optimal solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.