Project Euler : fix solution for problem 20 (Factorial digit sum).

Author: lvlte

Project-Euler/Problem020.js

@@ -1,21 +1,39 @@
-/*
-Factorial digit sum
+/**
+ * Problem 20 - Factorial digit sum
+ *
+ * @see {@link https://projecteuler.net/problem=20}
+ *
+ * n! means n × (n − 1) × ... × 3 × 2 × 1
+ *
+ * For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
+ * and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27
+ *
+ * Find the sum of the digits in the number 100!
+ */
 
-n! means n × (n − 1) × ... × 3 × 2 × 1
+const factorialDigitSum = function (n = 100) {
 
-For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
-and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
+  // Consider each digit*10^exp separately, right-to-left ([units, tens, ...]).
+  let digits = [1];
 
-Find the sum of the digits in the number 100!
-*/
+  for (let x=2; x<=n; x++) {
+    let carry = 0;
+    for (let exp=0; exp<digits.length; exp++) {
+      const prod = digits[exp]*x + carry;
+      carry = Math.floor(prod/10);
+      digits[exp] = prod % 10;
+    }
+    while (carry > 0) {
+      digits.push(carry%10);
+      carry = Math.floor(carry/10);
+    }
+  }
 
-const findFactorialDigitSum = (num) => {
-  let result = 0
-  const stringifiedNumber = factorize(num).toLocaleString('fullwide', { useGrouping: false })
-  stringifiedNumber.split('').map(num => { result += Number(num) })
-  return result
+  // (digits are reversed but we only want the sum so it doesn't matter)
+
+  return digits.reduce((prev, current) => prev + current, 0)
 }
 
-const factorize = (num) => num === 0 ? 1 : num * factorize(num - 1)
+console.log('Factorial digit sum of 100! :', factorialDigitSum())
 
-console.log(findFactorialDigitSum(100))
+module.exports = factorialDigitSum