[Easy] 509. Fibonacci Number

The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,

F(0) = 0, F(1) = 1
F(n) = F(n - 1) + F(n - 2), for n > 1.

Given n, calculate F(n).

Examples

Example 1:

Input: n = 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.

Example 2:

Input: n = 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.

Example 3:

Input: n = 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.

Constraints

0 <= n <= 30

Clarification Questions

Before diving into the solution, here are 5 important clarifications and assumptions to discuss during an interview:

Fibonacci definition: What is the Fibonacci sequence? (Assumption: F(0) = 0, F(1) = 1, F(n) = F(n-1) + F(n-2) for n > 1)
Base cases: What are the base cases? (Assumption: F(0) = 0, F(1) = 1 - standard Fibonacci)
Return value: What should we return? (Assumption: Integer - F(n) - nth Fibonacci number)
Input range: What is the range of n? (Assumption: Per constraints, 0 <= n <= 30 - small range)
Time complexity: What time complexity is expected? (Assumption: O(n) - linear time with DP, O(2^n) naive recursion)

Interview Deduction Process (10 minutes)

Step 1: Brute-Force Approach (2 minutes)

Initial Thought: “I need to compute Fibonacci. Let me use recursive definition directly.”

Naive Solution: Recursive function: F(n) = F(n-1) + F(n-2) with base cases F(0)=0, F(1)=1.

Complexity: O(2^n) time, O(n) space

Issues:

Exponential time complexity
Recomputes same values many times
Very inefficient
Doesn’t leverage memoization

Step 2: Semi-Optimized Approach (3 minutes)

Insight: “I can use memoization to cache computed values.”

Improved Solution: Use recursion with memoization. Store computed Fibonacci values in hash map/array to avoid recomputation.

Complexity: O(n) time, O(n) space

Improvements:

Memoization eliminates recomputation
O(n) time is much better
Still uses recursion stack
Can optimize space

Step 3: Optimized Solution (5 minutes)

Final Optimization: “I can use iterative DP to avoid recursion stack.”

Best Solution: Iterative DP (bottom-up). Use two variables to track F(n-1) and F(n-2), compute F(n) iteratively.

Complexity: O(n) time, O(1) space

Key Realizations:

DP is natural approach for Fibonacci
O(n) time is optimal
O(1) space is optimal with iterative approach
Bottom-up avoids recursion overhead

Solution: Dynamic Programming (Bottom-Up)

Time Complexity: O(n) - Single pass through the array
Space Complexity: O(n) - Cache array (can be optimized to O(1))

This solution uses bottom-up dynamic programming with memoization to avoid recalculating Fibonacci numbers.

Solution: DP with Cache Array

class Solution {
public:
    int fib(int n) {
        vector<int> cache(n + 1, 0);
        
        if(n <= 0) return 0;
        if(n == 1) return 1;
        
        cache[0] = 0;
        cache[1] = 1;
        
        for(int i = 2; i <= n; i++) {
            cache[i] = cache[i - 1] + cache[i - 2];
        }
        
        return cache[n];
    }
};

How the Algorithm Works

Step-by-Step Example: `n = 5`

Initial: cache = [0, 0, 0, 0, 0, 0]
         cache[0] = 0
         cache[1] = 1

i = 2: cache[2] = cache[1] + cache[0] = 1 + 0 = 1
        cache = [0, 1, 1, 0, 0, 0]

i = 3: cache[3] = cache[2] + cache[1] = 1 + 1 = 2
        cache = [0, 1, 1, 2, 0, 0]

i = 4: cache[4] = cache[3] + cache[2] = 2 + 1 = 3
        cache = [0, 1, 1, 2, 3, 0]

i = 5: cache[5] = cache[4] + cache[3] = 3 + 2 = 5
        cache = [0, 1, 1, 2, 3, 5]

Result: F(5) = 5

Visual Representation

Fibonacci Sequence:
F(0) = 0
F(1) = 1
F(2) = F(1) + F(0) = 1 + 0 = 1
F(3) = F(2) + F(1) = 1 + 1 = 2
F(4) = F(3) + F(2) = 2 + 1 = 3
F(5) = F(4) + F(3) = 3 + 2 = 5

Sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

Key Insights

Bottom-Up DP: Build solution from base cases upward
Memoization: Store previously computed values to avoid recalculation
Base Cases: F(0) = 0 and F(1) = 1 are the foundation
Recurrence Relation: F(n) = F(n-1) + F(n-2) for n > 1
Overlapping Subproblems: Each Fibonacci number depends on previous two

Algorithm Breakdown

int fib(int n) {
    // Create cache array for memoization
    vector<int> cache(n + 1, 0);
    
    // Handle base cases
    if(n <= 0) return 0;
    if(n == 1) return 1;
    
    // Initialize base values
    cache[0] = 0;
    cache[1] = 1;
    
    // Build solution bottom-up
    for(int i = 2; i <= n; i++) {
        cache[i] = cache[i - 1] + cache[i - 2];
    }
    
    return cache[n];
}

Edge Cases

n = 0: Return 0
n = 1: Return 1
n = 2: Return 1 (first non-base Fibonacci number)
n = 30: Maximum constraint value

Alternative Approaches

Approach 2: Space-Optimized Iterative (O(1) Space)

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

Only keep track of the last two values instead of the entire array:

class Solution {
public:
    int fib(int n) {
        if(n <= 0) return 0;
        if(n == 1) return 1;
        
        int prev2 = 0;  // F(0)
        int prev1 = 1;  // F(1)
        
        for(int i = 2; i <= n; i++) {
            int curr = prev1 + prev2;
            prev2 = prev1;
            prev1 = curr;
        }
        
        return prev1;
    }
};

Pros:

O(1) space complexity
More memory efficient
Same time complexity

Cons:

Can’t access previous Fibonacci numbers after computation

Approach 3: Recursive with Memoization

Time Complexity: O(n)
Space Complexity: O(n) due to recursion stack and memoization

class Solution {
public:
    int fib(int n) {
        vector<int> memo(n + 1, -1);
        return fibHelper(n, memo);
    }
    
private:
    int fibHelper(int n, vector<int>& memo) {
        if(n <= 0) return 0;
        if(n == 1) return 1;
        
        if(memo[n] != -1) return memo[n];
        
        memo[n] = fibHelper(n - 1, memo) + fibHelper(n - 2, memo);
        return memo[n];
    }
};

Pros:

Top-down approach (more intuitive for some)
Natural recursive structure

Cons:

O(n) space for recursion stack
Function call overhead

Approach 4: Pure Recursion (Not Recommended)

Time Complexity: O(2^n) - Exponential!
Space Complexity: O(n) - Recursion stack

class Solution {
public:
    int fib(int n) {
        if(n <= 0) return 0;
        if(n == 1) return 1;
        return fib(n - 1) + fib(n - 2);
    }
};

Why not recommended:

Extremely slow for large n
Recalculates same values multiple times
Only shown for educational purposes

Approach 5: Matrix Exponentiation (Advanced)

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

Uses matrix exponentiation for logarithmic time complexity:

class Solution {
public:
    int fib(int n) {
        if(n <= 0) return 0;
        if(n == 1) return 1;
        
        // Matrix: [F(n+1) F(n)  ] = [1 1]^n
        //         [F(n)   F(n-1)]   [1 0]
        vector<vector<int>> base = \{\{1, 1\}, \{1, 0\}\};
        vector<vector<int>> result = matrixPower(base, n);
        
        return result[0][1];
    }
    
private:
    vector<vector<int>> matrixPower(vector<vector<int>>& m, int n) {
        if(n == 1) return m;
        
        vector<vector<int>> half = matrixPower(m, n / 2);
        vector<vector<int>> result = matrixMultiply(half, half);
        
        if(n % 2 == 1) {
            result = matrixMultiply(result, m);
        }
        
        return result;
    }
    
    vector<vector<int>> matrixMultiply(vector<vector<int>>& a, vector<vector<int>>& b) {
        return {
            {a[0][0]*b[0][0] + a[0][1]*b[1][0], a[0][0]*b[0][1] + a[0][1]*b[1][1]},
            {a[1][0]*b[0][0] + a[1][1]*b[1][0], a[1][0]*b[0][1] + a[1][1]*b[1][1]}
        };
    }
};

Pros:

O(log n) time complexity
Efficient for very large n

Cons:

More complex implementation
Overkill for small n (n ≤ 30)

Complexity Analysis

Approach	Time	Space	Pros	Cons
DP with Cache	O(n)	O(n)	Simple, clear	O(n) space
Space-Optimized	O(n)	O(1)	Optimal space	Can’t access history
Recursive + Memo	O(n)	O(n)	Intuitive	Stack overhead
Pure Recursion	O(2^n)	O(n)	Simple	Extremely slow
Matrix Exponentiation	O(log n)	O(1)	Very fast	Complex

Implementation Details

Cache Initialization

vector<int> cache(n + 1, 0);

Creates an array of size n + 1 initialized to 0. This allows indexing from 0 to n.

Base Case Handling

if(n <= 0) return 0;
if(n == 1) return 1;

Early returns for base cases avoid unnecessary computation and array access.

Loop Construction

for(int i = 2; i <= n; i++) {
    cache[i] = cache[i - 1] + cache[i - 2];
}

Builds Fibonacci numbers sequentially from F(2) to F(n).

Common Mistakes

Off-by-one errors: Using i < n instead of i <= n
Array bounds: Not allocating n + 1 elements
Base case order: Checking n == 1 before n <= 0
Uninitialized cache: Not setting cache[0] and cache[1]
Wrong return value: Returning cache[n-1] instead of cache[n]

Optimization Tips

Space Optimization: Use two variables instead of array for O(1) space
Early Returns: Handle base cases immediately
Memoization: Cache results to avoid recalculation (already done in DP)
Matrix Exponentiation: Use for very large n (though n ≤ 30 here)

70. Climbing Stairs - Same recurrence relation
746. Min Cost Climbing Stairs - Fibonacci with costs
1137. N-th Tribonacci Number - Three-term recurrence
509. Fibonacci Number - This problem

Real-World Applications

Algorithm Analysis: Fibonacci heap data structure
Nature: Modeling growth patterns (pinecones, sunflowers)
Art: Golden ratio and aesthetic proportions
Computer Science: Dynamic programming examples
Mathematics: Number theory and sequences

Pattern Recognition

This problem demonstrates the “Classic DP Pattern”:

Identify base cases
Define recurrence relation
Build solution bottom-up or top-down
Optimize space if possible

Fibonacci Sequence Properties

Golden Ratio: As n increases, F(n+1)/F(n) approaches φ ≈ 1.618
Binet’s Formula: Closed-form solution using golden ratio
Pisano Period: Fibonacci modulo m has a repeating cycle
Sum Property: Sum of first n Fibonacci numbers = F(n+2) - 1

Why DP is Preferred

Avoids Recalculation: Pure recursion recalculates F(3) multiple times
Efficient: O(n) time vs O(2^n) for naive recursion
Simple: Easy to understand and implement
Space Trade-off: Can optimize to O(1) space easily

This problem is a perfect introduction to dynamic programming, demonstrating how memoization can transform exponential time complexity into linear time.

Robina Li

[Easy] 509. Fibonacci Number

[Easy] 509. Fibonacci Number

Examples

Constraints

Clarification Questions

Interview Deduction Process (10 minutes)

Step 1: Brute-Force Approach (2 minutes)

Step 2: Semi-Optimized Approach (3 minutes)

Step 3: Optimized Solution (5 minutes)

Solution: Dynamic Programming (Bottom-Up)

Solution: DP with Cache Array

How the Algorithm Works

Step-by-Step Example: `n = 5`

Visual Representation

Key Insights

Algorithm Breakdown

Edge Cases

Alternative Approaches

Approach 2: Space-Optimized Iterative (O(1) Space)

Approach 3: Recursive with Memoization

Approach 4: Pure Recursion (Not Recommended)

Approach 5: Matrix Exponentiation (Advanced)

Complexity Analysis

Implementation Details

Cache Initialization

Base Case Handling

Loop Construction

Common Mistakes

Optimization Tips

Real-World Applications

Pattern Recognition

Fibonacci Sequence Properties

Why DP is Preferred

Related Posts

Recent Posts

[Easy] 509. Fibonacci Number

Examples

Constraints

Clarification Questions

Interview Deduction Process (10 minutes)

Step 1: Brute-Force Approach (2 minutes)

Step 2: Semi-Optimized Approach (3 minutes)

Step 3: Optimized Solution (5 minutes)

Solution: Dynamic Programming (Bottom-Up)

Solution: DP with Cache Array

How the Algorithm Works

Step-by-Step Example: n = 5

Visual Representation

Key Insights

Algorithm Breakdown

Edge Cases

Alternative Approaches

Approach 2: Space-Optimized Iterative (O(1) Space)

Approach 3: Recursive with Memoization

Approach 4: Pure Recursion (Not Recommended)

Approach 5: Matrix Exponentiation (Advanced)

Complexity Analysis

Implementation Details

Cache Initialization

Base Case Handling

Loop Construction

Common Mistakes

Optimization Tips

Related Problems

Real-World Applications

Pattern Recognition

Fibonacci Sequence Properties

Why DP is Preferred

Related Posts

Recent Posts

Step-by-Step Example: `n = 5`