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Top 35 Fibonacci Sequence Interview Questions in 2026\n\n\u003cdiv\u003e\n\u003cp align=\"center\"\u003e\n\u003ca href=\"https://devinterview.io/questions/data-structures-and-algorithms/\"\u003e\n\u003cimg src=\"https://firebasestorage.googleapis.com/v0/b/dev-stack-app.appspot.com/o/github-blog-img%2Fdata-structures-and-algorithms-github-img.jpg?alt=media\u0026token=fa19cf0c-ed41-4954-ae0d-d4533b071bc6\" alt=\"data-structures-and-algorithms\" width=\"100%\"\u003e\n\u003c/a\u003e\n\u003c/p\u003e\n\n#### You can also find all 35 answers here 👉 [Devinterview.io - Fibonacci Sequence](https://devinterview.io/questions/data-structures-and-algorithms/fibonacci-sequence-interview-questions)\n\n\u003cbr\u003e\n\n## 1. What is the _Fibonacci sequence_?\n\nThe **Fibonacci Sequence** is a series of numbers where each number $F(n)$ is the sum of the two preceding ones, often starting with 0 and 1. That is:\n\n$$\nF(n) = F(n-1) + F(n-2)\n$$\n\nwith initial conditions\n\n$$\nF(0) = 0, \\quad F(1) = 1\n$$\n\n### Golden Ratio\n\nThe ratio of consecutive Fibonacci numbers approximates the Golden Ratio ($\\phi \\approx 1.6180339887$):\n\n$$\n\\lim_{{n \\to \\infty}} \\frac{{F(n+1)}}{{F(n)}} = \\phi\n$$\n\n### Real-World Occurrences\n\nThe sequence frequently manifests in nature, such as in flower petals, seedhead spirals, and seashell growth patterns.\n\n### Visual Representation\n\n![Fibonacci Spiral](https://firebasestorage.googleapis.com/v0/b/dev-stack-app.appspot.com/o/fibonacci-sequence%2Fwhat-is-fibonacci-sequence%20%20.svg?alt=media\u0026token=023579c2-a056-4aa8-8af5-a82082d3a621)\n\n### Code Example: Calculating The Nth Fibonacci Number\n\nHere is the Python code:\n\n```python\n# Using recursion\ndef fibonacci_recursive(n):\n    if n \u003c= 0: return 0\n    elif n == 1: return 1\n    return fibonacci_recursive(n-1) + fibonacci_recursive(n-2)\n    \n# Using dynamic programming for efficiency\ndef fibonacci_dynamic(n):\n    fib = [0, 1]\n    for i in range(2, n+1):\n        fib.append(fib[-1] + fib[-2])\n    return fib[n]\n```\n\u003cbr\u003e\n\n## 2. Write a function to calculate the _nth Fibonacci number_ using a _recursive_ approach.\n\n### Problem Statement\n\nThe task is to write a function that **returns the $n$th Fibonacci number** using a **recursive approach**.\n\n### Solution\n\nThe **naive recursive** solution for the Fibonacci series, while easy to understand, is inefficient due to its exponential time complexity of $O(2^n)$.\n\n**Dynamic Programming** methods like memoization and tabulation result in optimized time complexity.\n\n#### Algorithm Steps\n\n1. Check for the base cases, i.e., if $n$ is 0 or 1.\n2. If not a base case, **recursively** compute $F(n-1)$ and $F(n-2)$.\n3. Return the sum of the two recursive calls.\n\n#### Implementation\n\nHere is the Python code:\n\n```python\ndef fib_recursive(n):\n    if n \u003c= 1:\n        return n\n    return fib_recursive(n-1) + fib_recursive(n-2)\n```\n\n#### Complexity Analysis\n\n- **Time Complexity**: $O(2^n)$. This is due to the two recursive calls made at each level of the recursion tree, resulting in an exponential number of function calls.\n\n- **Space Complexity**: $O(n)$. Despite the inefficient time complexity, the space complexity is $O(n)$ as it represents the depth of the recursion stack.\n\u003cbr\u003e\n\n## 3. Provide a non-recursive implementation for generating the _Fibonacci sequence_ to the _nth number_.\n\n### Problem Statement\n\nThe task is to **generate** the Fibonacci sequence to the $n$th number using a **non-recursive approach**.\n\n### Solution\n\nWhile recursion offers a straightforward solution for the Fibonacci sequence, it has performance and stack overflow issues. A **non-recursive** approach, often based on a **loop** or **iterative method**, overcomes these limitations.\n\nHere, I'll present both **binet's formula** and an **iterative method** as the non-recursive solutions.\n\n#### Binet's Formula\n\n$$\nF(n) = \\frac{{\\phi^n - \\psi^n}}{{\\sqrt 5}}\n$$\n\nwhere:\n\n- $\\phi = \\frac{{1 + \\sqrt 5}}{2}$ (the golden ratio)\n- $\\psi = \\frac{{1 - \\sqrt 5}}{2}$\n\n#### Algorithm Steps\n\n1. Compute $\\phi$ and $\\psi$.\n2. Plug values into the formula for $F(n)$.\n\n#### Complexity Analysis\n\n- **Time Complexity**: $O(1)$\n- **Space Complexity**: $O(1)$\n\n**Note**: While Binet's formula offers an elegant non-recursive solution, it's sensitive to **floating-point errors** which can impact accuracy, especially for large $n$.\n\n#### Iterative Method\n\n#### Algorithm Steps\n\n1. Initialize $\\text{prev} = 0$ and $\\text{curr} = 1$. These are the first two Fibonacci numbers.\n2. For $i = 2$ to $n$, update $\\text{prev}$ and $\\text{curr}$ to be $\\text{prev} + \\text{curr}$ and $\\text{prev}$ respectively. These become the next numbers in the sequence.\n\n#### Complexity Analysis\n\n- **Time Complexity**: $O(n)$\n- **Space Complexity**: $O(1)$\n\n#### Implementation\n\nHere's the Python code for both methods:\n\n### Binet's Formula\n\n```python\nimport math\n\ndef fib_binet(n):\n    phi = (1 + math.sqrt(5)) / 2\n    psi = (1 - math.sqrt(5)) / 2\n    return int((phi**n - psi**n) / math.sqrt(5))\n\n# Output\nprint(fib_binet(5))  # Output: 5\n```\n\n### Iterative Method\n\n```python\ndef fib_iterative(n):\n    if n \u003c= 1:\n        return n\n    prev, curr = 0, 1\n    for _ in range(2, n + 1):\n        prev, curr = curr, prev + curr\n    return curr\n\n# Output\nprint(fib_iterative(5))  # Output: 5\n```\n\nBoth functions will return the 5th Fibonacci number.\n\u003cbr\u003e\n\n## 4. What is the _time complexity_ of the recursive Fibonacci solution, and how can this be improved?\n\nThe **naive recursive** implementation of the Fibonacci sequence has a time complexity of $O(2^n)$, which can be optimized using techniques such as **memoization** or employing an **iterative approach**.\n\n### Naive Recursive Approach\n\nThis is the straightforward, but inefficient, method.\n\n#### Algorithm\n\n1. **Base Case**: Return 0 if $n = 0$ and 1 if $n = 1$.\n2.  **Function Call**: Recur on the sum of the $n-1$ and $n-2$ elements.\n\n#### Python Code\n\nHere is the Python code:\n\n```python\ndef fibonacci_recursive(n):\n    if n \u003c= 1:\n        return n\n    return fibonacci_recursive(n-1) + fibonacci_recursive(n-2)\n```\n\n#### Complexity Analysis\n\n- **Time Complexity**: $O(2^n)$ - As each call branches into two more calls (with the exception of the base case), the number of function calls grows exponentially with $n$, resulting in this time complexity.\n- **Space Complexity**: $O(n)$ - The depth of the recursion stack can go up to $n$ due to the $n-1$ and $n-2$ calls.\n\n### Memoization for Improved Efficiency\n\nUsing **memoization** allows for a noticeable performance boost.\n\n#### Algorithm\n\n1. Initialize a cache, `fib_cache`, with default values of -1. \n2. **Base Case**: If the $n$th value is already calculated (i.e., `fib_cache[n] != -1`), return that value. Otherwise, calculate the $n$th value using recursion.\n3. **Cache the Result**: Once the $n$th value is determined, store it in `fib_cache` before returning.\n\n#### Python Code\n\nHere is the Python code:\n\n```python\ndef fibonacci_memo(n, fib_cache={0: 0, 1: 1}):\n    if n not in fib_cache:\n        fib_cache[n] = fibonacci_memo(n-1, fib_cache) + fibonacci_memo(n-2, fib_cache)\n    return fib_cache[n]\n```\n\n#### Performance Analysis\n\n- **Time Complexity**: $O(n)$ - Each $n$ is computed once and then stored in the cache, so subsequent computations are $O(1)$.\n- **Space Complexity**: $O(n)$ - The space used by the cache.\n\n### Iterative Method for Superior Efficiency\n\nThe **iterative approach** shines in terms of time and space efficiency.\n\n#### Algorithm\n\n1. Initialize `a` and `b` as 0 and 1, respectively.\n2. **Loop**: Update `a` and `b` to the next two Fibonacci numbers, replacing them as necessary to ensure they represent the desired numbers.\n3. **Return**: `a` after the loop exits, since it stores the $n$th Fibonacci number.\n\n#### Python Code\n\nHere is the Python code:\n```python\ndef fibonacci_iterative(n):\n    a, b = 0, 1\n    for _ in range(n):\n        a, b = b, a + b\n    return a\n```\n\n#### Performance Analysis\n\n- **Time Complexity**: $O(n)$ - The function iterates a constant number of times, depending on $n$.\n- **Space Complexity**: $O(1)$ - The variables `a` and `b` are updated in place without utilizing any dynamic data structures or recursion stacks.\n\u003cbr\u003e\n\n## 5. Describe the _memoization_ technique as applied to the _Fibonacci sequence_ calculation.\n\n**Memoization** is a technique that makes dynamic programming faster by storing the results of expensive function calls and reusing them.\n\nTo apply memoization to the **Fibonacci sequence**, a list or dictionary (**array** or **hashmap** in terms of computer science) is used to store the intermediate results, essentially turning the calculation process into a more optimized dynamic programming algorithm.\n\n### Code Example: Memoized Fibonacci Calculation\n\nHere is the Python code:\n\n```python\ndef fibonacci_memo(n, memo={0: 0, 1: 1}):\n    if n not in memo:\n        memo[n] = fibonacci_memo(n-1, memo) + fibonacci_memo(n-2, memo)\n    return memo[n]\n\n# Test\nprint(fibonacci_memo(10))  # Outputs: 55\n```\n\u003cbr\u003e\n\n## 6. Implement an _iterative solution_ to generate the _Fibonacci sequence_, discussing its _time_ and _space complexity_.\n\n### Problem Statement\n\nThe task is to **generate the Fibonacci sequence** using an **iterative** approach, and then analyzing its **time and space complexity**.\n\n### Solution\n\n#### Iterative Algorithm  \n\n1. Initialize `a = 0` and `b = 1`.\n2. Use a loop to update `a` and `b`. On each iteration:\n   - Update `a` to the value of `b`.\n   - Update `b` to the sum of its old value and the old value of `a`.\n3. Repeat the loop 'n-1' times, where 'n' is the desired sequence length.\n\n#### Visual Representation\n\nHere's how the first few Fibonacci numbers are computed:\n\n- Iteration 1: $a = 0, \\, b = 1, \\, b = 0 + 1 = 1$\n- Iteration 2: $a = 1, \\, b = 1, \\, b = 1 + 1 = 2$\n- Iteration 3: $a = 1, \\, b = 2, \\, b = 2 + 1 = 3$\n- Iteration 4: $a = 2, \\, b = 3, \\, b = 3 + 2 = 5$\n- And so on...\n\n#### Complexity Analysis\n\n- **Time Complexity**: $O(n)$ — This is more efficient than the recursive approach which is $O(2^n)$.\n- **Space Complexity**: $O(1)$ — The space used is constant, regardless of the input 'n'.\n\n#### Implementation\n\nHere is the Python code:\n\n```python\ndef fibonacci_iterative(n):\n    if n \u003c= 0:\n        return \"Invalid input. n must be a positive integer.\"\n\n    fib_sequence = [0] if n \u003e= 1 else []\n\n    a, b = 0, 1\n    for _ in range(2, n+1):\n        fib_sequence.append(b)\n        a, b = b, a + b\n\n    return fib_sequence\n\n# Example usage\nprint(fibonacci_iterative(10)) # Output: [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]\n```\n\u003cbr\u003e\n\n## 7. Explain the concept of _dynamic programming_ as it relates to the _Fibonacci sequence_.\n\n**Dynamic Programming** is a powerful algorithmic technique that is widely used to optimize **recursive problems**, like computing the Fibonacci sequence, by avoiding redundant computations.\n\n### Efficient Fibonacci Calculation\n\nThe naive method of Fibonacci computation is highly inefficient, often taking exponential time. Dynamic Programming offers better time complexity, often linear or even constant, without sacrificing accuracy.\n\n#### Memoization and Caching\n\nThe most common way of optimizing Fibonacci computations is through **memoization**, where function calls are stored with their results for future reference. In Python, you can employ decorators or dictionaries to achieve this.\n\nHere is the Python code:\n\n```python\ndef memoize(fib):\n    cache = {}\n    def wrapper(n):\n        if n not in cache:\n            cache[n] = fib(n)\n        return cache[n]\n    return wrapper\n\n@memoize\ndef fib(n):\n    if n \u003c 2:\n        return n\n    return fib(n-1) + fib(n-2)\n\n# Test\nprint(fib(10))  # 55\n```\n\u003cbr\u003e\n\n## 8. Solve the _nth Fibonacci number_ problem using _matrix exponentiation_ and analyze its efficiency.\n\n### Problem Statement\n\nThe task is to compute the $n$th Fibonacci number using **matrix exponentiation** and analyze its efficiency.\n\n### Solution\n\nMatrix exponentiation offers an optimal $O(\\log n)$ solution for the Fibonacci sequence, in contrast to the traditional recursive method that has a time complexity of $O(2^n)$.\n\n#### Algorithm Steps\n\n1. Represent the **Fibonacci transformation** as $F = \\begin{bmatrix} 1 \u0026 1 \\\\ 1 \u0026 0 \\end{bmatrix}$.\n2. Utilize **exponentiation by squaring** to efficiently compute $F^n$ for large $n$.\n3. Extract the $n$-th Fibonacci number as the top right element of the resultant matrix.\n\n#### Complexity Analysis\n\n- **Time Complexity**: $O(\\log n)$ due to the efficiency of exponentiation by squaring.\n- **Space Complexity**: $O(\\log n)$\n\n#### Implementation\n\nHere is the Python code:\n\n```python\nimport numpy as np\n\ndef matrix_power(A, n):\n    if n == 1:\n        return A\n    if n % 2 == 0:\n        half = matrix_power(A, n // 2)\n        return np.dot(half, half)\n    else:\n        half = matrix_power(A, (n - 1) // 2)\n        return np.dot(np.dot(half, half), A)\n\ndef fibonacci(n):\n    if n \u003c= 0:\n        return 0\n    F = np.array([[1, 1], [1, 0]])\n    result = matrix_power(F, n - 1)\n    return result[0, 0]\n\n# Example usage\nprint(fibonacci(6))  # Output: 8\n```\n\u003cbr\u003e\n\n## 9. Can the _nth Fibonacci number_ be found using the _golden ratio_, and if so, how would you implement this method?\n\nThe $n$-th term of the **Fibonacci sequence** can be approximated using the **Golden Ratio** $(\\phi \\approx 1.61803)$ through Binet's formula:\n\n$$ F(n) \\approx \\frac{{\\phi^n}}{{\\sqrt{5}}} $$\n\n### Code Example: Fibonacci with Golden Ratio\n\nHere is the Python code:\n\n```python\nimport math\n\ndef approximate_fibonacci(n):\n    golden_ratio = (1 + math.sqrt(5)) / 2\n    return round(golden_ratio ** n / math.sqrt(5))\n\n# Test\nprint(approximate_fibonacci(10))  # Output: 55\n```\n\u003cbr\u003e\n\n## 10. Present an approach to precomputing _Fibonacci numbers_ to answer multiple _nth Fibonacci queries_ efficiently.\n\n### Problem Statement\n\nThe objective is to compute $\\text{Fib}(n)$, the $n$th Fibonacci number, efficiently for **multiple queries**.\n\n### Solution\n\nA **precomputation** strategy is suitable for scenarios where `Fibonacci` numbers are repeatedly requested over a **fixed range**.\n\n#### Precomputation Table\n\n1. Generate and store Fibonacci numbers from 0 to the highest $n$ using an **array** or **dictionary**. This operation has a time complexity of $O(n)$.\n2. Subsequent queries are answered directly from the precomputed table with a time complexity of $O(1)$.\n\n#### Complexity Analysis\n\n- Precomputation: $O(\\text{max\\_n})$\n- Query time: $O(1)$\n\n#### Realization\n\nLet's consider Python as our programming language.\n\n#### Code\n\nHere is a Python function that precomputes Fibonacci numbers up to a certain limit and then returns the $n$th Fibonacci number based on the precomputed table.\n\n```python\ndef precompute_fibonacci(max_n):\n    fib = [0, 1]\n    a, b = 0, 1\n\n    while b \u003c max_n:\n        a, b = b, a + b\n        fib.append(b)\n\n    return fib\n\ndef fibonacci(n, fib_table):\n    return fib_table[n] if n \u003c len(fib_table) else -1\n\n# Usage\nmax_n = 100\nfib_table = precompute_fibonacci(max_n)\nprint(fibonacci(10, fib_table))  # 55\n```\n\u003cbr\u003e\n\n## 11. How might the _Fibonacci sequence_ be altered to start with two arbitrary initial values? Provide an algorithm for such a sequence.\n\nModifying the **Fibonacci sequence** to start with arbitrary initial values still leads to a unique sequence.\n\nThe iterative approach can handle custom starting values and **compute any term** in the sequence through the specified algorithm.\n\n### Algorithm: Custom Fibonacci Sequence\n\n1. **Input**: Start values `a` and `b` (with a not equal to b) and target term `n`.\n2. Check if `n` is 1 or 2. If it is, return the corresponding start value.\n3. Otherwise, execute a loop `n-2` times and update the start values.\n4. Compute the `n`-th term once the loop concludes.\n\n### Code Example: Custom Fibonacci Sequence\n\nHere is the Python code:\n\n```python\ndef custom_fibonacci(a, b, n):\n    if n == 1:\n        return a\n    elif n == 2:\n        return b\n    \n    for _ in range(n-2):\n        a, b = b, a+b\n    return b\n\n# Example with start values 3 and 4 for the 6th term\nresult = custom_fibonacci(3, 4, 6)  # Output: 10\n```\n\u003cbr\u003e\n\n## 12. Explain an algorithm to compute the _sum of the first n Fibonacci numbers_ without generating the entire sequence.\n\nThe **Fibonacci sequence** is a classic mathematical series defined by the following:\n\n$$\nF(n) = \n\\begin{cases} \n0 \u0026 \\text{if } n = 0 \\\\\n1 \u0026 \\text{if } n = 1 \\\\\nF(n-1) + F(n-2) \u0026 \\text{if } n \u003e 1 \n\\end{cases}\n$$\n\n### Direct Formulas for Sum and Generalization\n\n1. **Sum of First n Numbers**: The sum of the first $n$ Fibonacci numbers is equal to $F(n+2) - 1$.\n  \n  This can be computed using a simple, **iterative function**:\n\n  $$ \\text{Sum}(n) = F(n+2) - 1 $$\n\n2. **n-th Fibonacci Number**: It can be calculated using **two seed values** $F(0)$ and $F(1)$. \n\n  The general form is:\n\n  $$ F(n) = F(0)A + F(1)B $$\n\n  **Matrix Representation**:\n  $$\n  \\begin{bmatrix}\n  F(n) \\\\\n  F(n-1) \\\\\n  \\end{bmatrix}\n  = \n  \\begin{bmatrix}\n  1 \u0026 1 \\\\\n  1 \u0026 0 \\\\\n  \\end{bmatrix}^{(n-1)}\n  \\begin{bmatrix}\n  F(1) \\\\\n  F(0) \\\\\n  \\end{bmatrix}\n  $$\n\n  Where:\n  - $A$ and $B$ vary based on the initial seed values.\n  - The matrix is multiplied by itself $(n-1)$ times.\n\n### Code Example: Sum of First $n$ Fibonacci Numbers\n\nHere is the Python code:\n\n  ```python\n  def fibonacci_sum(n):\n      fib_curr, fib_next, fib_sum = 0, 1, 0\n\n      for _ in range(n):\n          fib_sum += fib_curr\n          fib_curr, fib_next = fib_next, fib_curr + fib_next\n\n      return fib_sum\n  ```\n\nThe time complexity is $O(n)$, which is significantly better than $O(n\\log n)$ when computing the $n$-th Fibonacci number using matrix exponentiation.\n\u003cbr\u003e\n\n## 13. Define the _Lucas sequence_ and detail how a program can generate it.\n\nThe **Lucas Sequence**, a variant of the Fibonacci sequence, starts with 2 and 1, rather than 0 and 1, and follows the same recursive structure as the classic sequence:\n\n$$\n\\text{Lucas}(n) = \\begin{cases}\n2, \u0026 \\text{if } n = 0, \\\\\n1, \u0026 \\text{if } n = 1, \\\\\n\\text{Lucas}(n-1) + \\text{Lucas}(n-2), \u0026 \\text{otherwise}.\n\\end{cases}\n$$\n\n### Advantages of the Lucas Sequence\n\nCompared to the Fibonacci sequence, the Lucas sequence offers:\n\n- **Simpler Recurrence Relation**: The Lucas sequence uses only addition for its recursive relation, which can be computationally more efficient than Fibonacci's addition and subtraction.\n\n- **Alternate Closed-Form Expression**: While the closed-form formula for the $n$th term of the Fibonacci sequence involves radicals, the Lucas sequence provides an alternate expression that can be easier to work with.\n\u003cbr\u003e\n\n## 14. How can the _Fibonacci sequence_ be used to solve the _tiling problem_, where you need to cover a _2xn rectangle_ with _2x1 tiles_?\n\nThe **Fibonacci sequence** is closely related to the problem of covering a 2xN rectangle with 2x1 tiles, often referred to as the \"tiling problem\". It in fact offers a direct solution to this problem.\n\n### Relationship Between Tiling and Fibonacci Sequence\n\nCovering a 2xN rectangle $R_{1}$ with 2x1 tiles can be understood in terms of the number of ways to cover the last column, whether with a single vertical tile or two horizontal tiles:\n\n$$\nW_{1}(2xN) = W(2x(N-1)) + W(2x(N-2))\n$$\n\nThis is a recursive relationship similar to the one used to define the Fibonacci numbers, making it clear that there is a connection between the two.\n\n### Code Example: Tiling a 2xN Rectangle\n\nHere is the Python code:\n\n```python\ndef tiling_ways(n):\n    a, b = 1, 1\n    for _ in range(n):\n        a, b = b, a + b\n    return a\n\nn = 5\nways_to_tile = tiling_ways(n)\nprint(f'Number of ways to tile a 2x{n} rectangle: {ways_to_tile}')\n```\n\nIn this example, `a, b = b, a + b` is a compact way to **update** `a` and `b`. It relies on the fact that the right-hand side of the assignment is _evaluated first_ before being assigned to the left-hand side.\n\nThis approach has a **time complexity** of $O(N)$ and a **space complexity** of $O(1)$ since it only requires two variables.\n\u003cbr\u003e\n\n## 15. How to calculate large _Fibonacci numbers_ without encountering _integer overflow_ issues?\n\n**Fibonacci numbers** grow at a rapid rate, which can lead to **integer overflow**. Numerous strategies exist to circumvent this issue.\n\n### Various Techniques to Handle Overflow \n\n1. **Using a Data Type with Larger Capacity**:\n    The `long long int` data type in C/C++ provides up to 19 digits of precision, thus accommodating Fibonacci numbers up to $F(92)$.\n  \n2. **Using Built-in Arbitrary Precision Libraries**:\n    Certain programming languages such as Python and Ruby come with **arbitrary-precision** arithmetic support, making them suited for such computations.\n\n3. **Implementing Custom Arithmetic**:\n    Libraries like GMP (GNU Multiple Precision Arithmetic Library) and `bigInt` in Java enable the handling of arbitrary-precision operations. Additionally, rolling out a custom arithmetic procedure through **arrays, linked lists, or strings** is viable.\n\n### Code Example: Using `long long int` for Larger Capacity\n\nHere is the C++ code:\n\n  ```cpp\n    #include \u003ciostream\u003e\n    #include \u003cvector\u003e\n    using namespace std;\n    \n    long long int fibonacci(int n) {\n        if (n \u003c= 1) return n;\n    \n        long long int a = 0, b = 1, c;\n        for (int i = 2; i \u003c= n; ++i) {\n            c = a + b;\n            a = b;\n            b = c;\n        }\n        return b;\n    }\n  \n    int main() {\n        int n = 100;\n        cout \u003c\u003c \"Fibonacci number at position \" \u003c\u003c n \u003c\u003c \" is: \" \n             \u003c\u003c fibonacci(n) \u003c\u003c endl;\n        return 0;\n    }\n  ```\n\n\n#### Explore all 35 answers here 👉 [Devinterview.io - Fibonacci Sequence](https://devinterview.io/questions/data-structures-and-algorithms/fibonacci-sequence-interview-questions)\n\n\u003cbr\u003e\n\n\u003ca href=\"https://devinterview.io/questions/data-structures-and-algorithms/\"\u003e\n\u003cimg src=\"https://firebasestorage.googleapis.com/v0/b/dev-stack-app.appspot.com/o/github-blog-img%2Fdata-structures-and-algorithms-github-img.jpg?alt=media\u0026token=fa19cf0c-ed41-4954-ae0d-d4533b071bc6\" alt=\"data-structures-and-algorithms\" width=\"100%\"\u003e\n\u003c/a\u003e\n\u003c/p\u003e\n\n","project_url":"https://awesome.ecosyste.ms/api/v1/projects/github.com%2Fdevinterview-io%2Ffibonacci-sequence-interview-questions","html_url":"https://awesome.ecosyste.ms/projects/github.com%2Fdevinterview-io%2Ffibonacci-sequence-interview-questions","lists_url":"https://awesome.ecosyste.ms/api/v1/projects/github.com%2Fdevinterview-io%2Ffibonacci-sequence-interview-questions/lists"}