Sum of Divisors - FreeCodeCamp #138 Daily Challenge

Problem Statement: Sum of Divisors

Given a positive integer $n$, return the sum of all its divisors (integers that divide $n$ with no remainder).

For example, for $n = 6$, the divisors are $1, 2, 3, 6$ and the sum is $12$.

Examples:

• sumOfDivisors(6) → $12$
• sumOfDivisors(13) → $14$
• sumOfDivisors(28) → $56$
• sumOfDivisors(84) → $224$
• sumOfDivisors(549) → $806$
• sumOfDivisors(9348) → $23520$

Analysis and Test Cases

To solve the problem, we just need to identify all divisors of $n$ (that is, all $i$ such that $n \% i = 0$) and sum them. Here are some cases:

These examples cover small, prime, composite, and large values to validate efficiency.

Solution and Explanation

Strategy

We use a direct approach: iterate from $1$ to $n$ and sum the values that divide $n$ exactly. It’s simple, clear, and sufficient for the problem’s input sizes.

Flowchart

flowchart TD
  A["Input: n"] --> B["Initialize sum = 0"]
  B --> C["Iterate i = 1 to n"]
  C --> D{"n % i == 0?"}
  D -- Yes --> E["sum += i"]
  D -- No --> C
  E --> C
  C --> F["End of loop"]
  F --> G["Return sum"]

JavaScript Code

function sumOfDivisors(n) {
  let sum = 0
  for (let i = 1; i <= n; i++) {
    if (n % i === 0)
      sum += i
  }
  return sum
}

Complexity

• Time: $O(n)$ (one iteration per number up to $n$).
• Space: $O(1)$ (only scalar variables).

Edge Cases and Considerations

• If $n = 1$, the only divisor is $1$ (the function should return $1$).
• The statement assumes $n$ is positive.
• For primes, the sum is $n + 1$.
• Negatives and zero are not considered.

Reflections and Learnings

• The modulo operator is key to identifying divisors.
• For large $n$, you can optimize by iterating up to $\sqrt{n}$ and summing both divisors, achieving $O(\sqrt{n})$.
• The current version is didactic and clear for beginners.

Resources

• Divisor (Wikipedia)
• Sum of Divisors - GeeksforGeeks