Implement sum(1)(2)(3)(4)..(n) in JavaScript

Implement sum(1)(2)(3)(4)..(n) in JavaScript

This is a famous Amazon interview question. Here is how to implement it.

The Problem

sum(1)(2)(3)(4)(); // 10
sum(1)(2)(3); // function (can continue chaining)

Solution 1: Terminate with Empty Call

function sum(a) {
  return function (b) {
    if (b !== undefined) return sum(a + b);
    return a;
  };
}
sum(1)(2)(3)(4)(); // 10

Each call returns a function. When called with no argument, it returns the accumulated sum.

Solution 2: Using toString

function sum(n) {
  const fn = (m) => sum(n + m);
  fn.toString = () => n;
  return fn;
}
sum(1)(2)(3) + 0; // 6 (JavaScript calls toString)
console.log(sum(1)(2)(3)); // 6 (console.log calls toString)

Solution 3: Using valueOf

function sum(n) {
  const fn = (m) => sum(n + m);
  fn.valueOf = () => n;
  return fn;
}
Number(sum(1)(2)(3)); // 6

How It Works (Solution 1)

1. sum(1) returns a function with a = 1.
2. Calling it with 2 returns sum(1 + 2) = sum(3).
3. Calling it with 3 returns sum(3 + 3) = sum(6).
4. Calling it with 4 returns sum(6 + 4) = sum(10).
5. Calling it with no argument () returns 10.

The Takeaway

sum(1)(2)(3)(4)..(n): each call returns a function that accumulates the sum. Terminate with an empty call (check b !== undefined) or use toString/valueOf for automatic conversion. This tests currying and closures.