Nth Fibonacci Number - FreeCodeCamp #145 Daily Challenge

2026-01-02 1 min

Table of Contents

Problem statement

Given a positive integer n, return the nth number in the Fibonacci sequence. The sequence starts like this (1-based indices):

1 → 0 2 → 1 3 → 1 4 → 2 5 → 3 6 → 5

Reference examples:

nthFibonacci(4) → 2
nthFibonacci(10) → 34
nthFibonacci(15) → 377
nthFibonacci(40) → 63245986
nthFibonacci(75) → 1304969544928657

Analysis

Input n is interpreted as 1-based: n = 1 corresponds to 0, n = 2 to 1. The goal is to return the value at position n.

Requirements and considerations:

Accept n as a positive integer.
Keep the solution efficient for moderate n (e.g. up to tens of thousands if using BigInt).
Handle JavaScript numeric precision limits for very large results.

Solution (Approach)

The simplest space-efficient approach is iterative: keep the two latest values and advance until position n. Time complexity is O(n) and space complexity is O(1).

Implementation (JavaScript, 1-based)

function nthFibonacci(n) {
  if (n <= 0)
    throw new Error('n must be a positive integer (1-based)')
  if (n === 1)
    return 0
  if (n === 2)
    return 1

  let a = 0
  let b = 1
  for (let i = 3; i <= n; i++) {
    const next = a + b
    a = b
    b = next
  }
  return b
}

// Example usage
console.log(nthFibonacci(4)) // 2
console.log(nthFibonacci(10)) // 34

Complexity

Time complexity: O(n).
Space complexity: O(1).

Edge cases and notes

n <= 0: throw an error (invalid input).
n = 1 and n = 2: base cases return 0 and 1 respectively.
Precision: use BigInt if the result may exceed Number.MAX_SAFE_INTEGER.
Faster alternatives (O(log n)) include matrix exponentiation or fast doubling; these are useful for extremely large n.

Conclusion

The iterative solution is clear, efficient and adequate for most practical uses. For applications requiring exact results for very large indices, use BigInt or logarithmic-time algorithms.

Resources

freecodecamp daily-challenge