What is the time complexity of N-th Tribonacci Number?

The optimal solution has a time complexity of O(n).

What data structures are used to solve N-th Tribonacci Number?

This problem uses Math, Dynamic Programming, Memoization. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving N-th Tribonacci Number?

**Recursive Definition Implies Dynamic Programming:** The problem's definition `T_n = T_{n-1} + T_{n-2} + T_{n-3}` immediately reveals an optimal substructure and overlapping subproblems. This pattern is a canonical signal for dynamic programming, where storing previously computed values prevents redundant calculations and improves efficiency dramatically over naive recursion. **Precise Handling of Base Cases:** Getting the initial conditions `T_0 = 0, T_1 = 1, T_2 = 1` absolutely correct is foundational. Any error or omission in these base cases will propagate throughout the entire sequence, leading to an incorrect `T_n`. The solution explicitly handles `n=0`, `n=1`, and `n=2` before entering the iterative loop, ensuring the sequence starts correctly.

Which companies ask LeetCode 1137. N-th Tribonacci Number in interviews?

This problem is frequently asked at Accenture, Coursera, tcs.

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N-th Tribonacci Number - LeetCode 1137 Solution

N-th Tribonacci Number - Complete Solution Guide

N-th Tribonacci Number is LeetCode problem 1137, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

The Tribonacci sequence T n is defined as follows: T 0 = 0, T 1 = 1, T 2 = 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 .

Detailed Explanation

The "N-th Tribonacci Number" problem challenges us to generate a specific term from the Tribonacci sequence, a close cousin to the more famous Fibonacci sequence. Instead of each number being the sum of the two preceding ones, a Tribonacci number is the sum of the *three* preceding numbers. The problem provides the fundamental building blocks: `T_0 = 0`, `T_1 = 1`, and `T_2 = 1`. From `n >= 0`, the sequence then proceeds with `T_{n+3} = T_n + T_{n+1} + T_{n+2}`. Our task is to calculate and return `T_n` for a given input `n`.

Solution Approach

The provided solution leverages an iterative, bottom-up dynamic programming strategy to efficiently compute the N-th Tribonacci number. Instead of blindly following the recursive definition, which would cause redundant computations, this approach systematically builds up the sequence from its base cases. It handles the initial `n=0`, `n=1`, `n=2` cases directly. For `n >= 3`, it initializes an array (or list in Python) `trib` with the first three Tribonacci numbers: `[0, 1, 1]`, corresponding to `T_0`, `T_1`, and `T_2` respectively. The core insight here is that to calculate `T_i`, we only need the three immediately preceding values: `T_{i-1}, T_{i-2}, T_{i-3}`. Since we're building the sequence *up* from `T_0`, these preceding values will always have already been computed and stored in our `trib` array. The solution then simply iterates from `i = 3` up to `n`. In each step, it appends the sum of `trib[i-1] + trib[i-2] + trib[i-3]` to the `trib` array, effectively calculating `T_i`. This guarantees that each Tribonacci number is computed exactly once. Once the loop finishes, `trib[n]` holds the desired N-th Tribonacci number, which is then returned. This elegant method transforms a potentially exponential problem into a linear one by remembering past results.

Step-by-Step Algorithm

Handle base cases: If n is 0, return 0. If n is 1 or 2, return 1.
Initialize: Set the first three Tribonacci numbers (T0, T1, T2). This can be done using an array or individual variables.
Iterate: Loop from 3 to n (inclusive). In each iteration, calculate the next Tribonacci number by summing the three previous numbers.
Update: Update the variables (or array) to store the newly calculated Tribonacci number and shift older values.
Return: After the loop completes, return the n-th Tribonacci number.

Key Insights

**Recursive Definition Implies Dynamic Programming:** The problem's definition `T_n = T_{n-1} + T_{n-2} + T_{n-3}` immediately reveals an optimal substructure and overlapping subproblems. This pattern is a canonical signal for dynamic programming, where storing previously computed values prevents redundant calculations and improves efficiency dramatically over naive recursion.
**Precise Handling of Base Cases:** Getting the initial conditions `T_0 = 0, T_1 = 1, T_2 = 1` absolutely correct is foundational. Any error or omission in these base cases will propagate throughout the entire sequence, leading to an incorrect `T_n`. The solution explicitly handles `n=0`, `n=1`, and `n=2` before entering the iterative loop, ensuring the sequence starts correctly.
**Iterative Bottom-Up Build-up for O(n) Efficiency:** The solution's iterative approach, starting from the known base cases and incrementally calculating `T_i` up to `T_n`, ensures each Tribonacci number is computed in constant time (O(1)) and exactly once. This leads to an overall time complexity of O(n) and space complexity of O(n) (to store the sequence), which is highly efficient and perfectly suited for the given constraints of `n <= 37`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Math, Dynamic Programming, Memoization.

Companies

Asked at: Accenture, Coursera, tcs.

N-th Tribonacci Number | LeetCode 1137

Easy

Math Dynamic Programming Memoization

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int tribonacci(int n) {
        if (n == 0) return 0;
        if (n == 1 || n == 2) return 1;
        int[] trib = new int[n + 1];
        trib[0] = 0;
        trib[1] = 1;
        trib[2] = 1;
        for (int i = 3; i <= n; i++) {
            trib[i] = trib[i - 1] + trib[i - 2] + trib[i - 3];
        }
        return trib[n];
    }
}

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Problem Info

Description

Problem 1137: N-th Tribonacci Number - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Memoization

Companies

AccentureCourseratcs

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:1137

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Handle base cases: If n is 0, return 0. If n is 1 or 2, return 1.
2
Initialize: Set the first three Tribonacci numbers (T0, T1, T2). This can be done using an array or individual variables.
3
Iterate: Loop from 3 to n (inclusive). In each iteration, calculate the next Tribonacci number by summing the three previous numbers.
4
Update: Update the variables (or array) to store the newly calculated Tribonacci number and shift older values.
5
Return: After the loop completes, return the n-th Tribonacci number.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The algorithm iterates through the sequence once, from 3 to n. This gives a time complexity of O(n). The C++ and C solutions use a constant amount of extra space to store three integers. The Python and Java solutions use an array of size n+1, resulting in O(n) space complexity. However, the space complexity can be improved to O(1) by only storing the three most recent Tribonacci numbers.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive approach with memoization

A recursive solution can be implemented, but it would be less efficient without memoization to avoid redundant calculations. Memoization stores previously computed results to avoid recomputation.

Time: O(n)Space: O(n) for the memoization table and O(n) for the recursive call stack in the worst case

Trade-offs: Recursive solutions with memoization have the same time complexity but higher space complexity than the iterative approach due to recursive call stack.

Matrix exponentiation

This advanced technique can solve the problem in O(log n) time complexity using matrix exponentiation to calculate the nth Tribonacci number.

Time: O(log n)Space: O(1)

Trade-offs: More complex to implement but significantly faster for very large values of n.

✨

Key Insights

1**Recursive Definition Implies Dynamic Programming:** The problem's definition `T_n = T_{n-1} + T_{n-2} + T_{n-3}` immediately reveals an optimal substructure and overlapping subproblems. This pattern is a canonical signal for dynamic programming, where storing previously computed values prevents redundant calculations and improves efficiency dramatically over naive recursion.
2**Precise Handling of Base Cases:** Getting the initial conditions `T_0 = 0, T_1 = 1, T_2 = 1` absolutely correct is foundational. Any error or omission in these base cases will propagate throughout the entire sequence, leading to an incorrect `T_n`. The solution explicitly handles `n=0`, `n=1`, and `n=2` before entering the iterative loop, ensuring the sequence starts correctly.
3**Iterative Bottom-Up Build-up for O(n) Efficiency:** The solution's iterative approach, starting from the known base cases and incrementally calculating `T_i` up to `T_n`, ensures each Tribonacci number is computed in constant time (O(1)) and exactly once. This leads to an overall time complexity of O(n) and space complexity of O(n) (to store the sequence), which is highly efficient and perfectly suited for the given constraints of `n <= 37`.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

n = 0: The solution correctly returns 0.
n = 1 or n = 2: The solution correctly returns 1.
n = 3: The solution correctly returns 2 (0 + 1 + 1).

⚠️

Common Mistakes to Avoid

✗Forgetting the base cases (n=0, 1, 2) leading to incorrect results for small n.

✗Incorrectly calculating the next Tribonacci number by using the wrong indices or not updating the variables properly in the iterative approach.

✗Implementing a recursive solution without memoization, leading to exponential time complexity.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Modeling population growth of certain species where the growth depends on the population of the previous three generations.

🏦Financial modeling where the value at time t depends on the value at times t-1, t-2, and t-3.

🎮Analyzing branching processes in various scientific fields.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 28, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "N-th Tribonacci Number"?

For this easy-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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