package dev.bogwalk.batch3

import dev.bogwalk.util.maths.factorial

/**
 * Problem 34: Digit Factorials
 *
 * https://projecteuler.net/problem=34
 *
 * Goal: Find the sum of all numbers less than N that divide the sum of the factorial of their
 * digits (& therefore have minimum 2 digits).
 *
 * Constraints: 10 <= N <= 1e5
 *
 * Factorion: A natural number that equals the sum of the factorials of its digits.
 * The only non-single-digit factorions are: 145 and 40585.
 *
 * e.g.: N = 20
 *       qualifying numbers = {19}
 *       as 1! + 9! = 362_881, which % 19 = 0
 *       e.g. 18 does not work as 1! + 8! = 40321, which % 18 > 0
 *       sum = 19
 */

class DigitFactorials {
    // pre-calculation of all digit factorials to increase performance
    private val factorials = List(10) { it.factorial().intValueExact() }

    /**
     * HackerRank specific implementation that finds the sum of all numbers < [n] that are divisors
     * of the sum of the factorials of their digits.
     *
     * Increase this solution's efficiency by creating an array of upper constraint
     * size (plus 1) & looping once through all numbers then caching the result plus the
     * previously calculated sum, if it matches the necessary requirements.
     */
    fun sumOfDigitFactorialsHR(n: Int): Int {
        var sum = 0
        for (num in 10 until n) {
            val numSum = num.toString().sumOf { ch -> factorials[ch.digitToInt()] }
            if (numSum % num == 0) {
                sum += num
            }
        }
        return sum
    }

    /**
     * Project Euler specific implementation that finds the sum of all numbers that are factorions.
     *
     * The numbers cannot have more than 7 digits, as 9! * 8 returns only a 7-digit number.
     *
     * 9! * 7 returns 2_540_160, so the 1st digit of the 7-digit number cannot be greater than 2.
     */
    fun sumOfDigitFactorialsPE(): Int {
        var sum = 0
        for (num in 10 until 2_000_000) {
            val digits = num.toString().map { it.digitToInt() }
            var numSum = 0
            for (digit in digits) {
                numSum += factorials[digit]
                // prevents further unnecessary calculation
                if (numSum > num) break
            }
            if (numSum == num) sum += numSum
        }
        return sum
    }
}
package dev.bogwalk.batch3

import dev.bogwalk.util.maths.factorial

/**
 * Problem 34: Digit Factorials
 *
 * https://projecteuler.net/problem=34
 *
 * Goal: Find the sum of all numbers less than N that divide the sum of the factorial of their
 * digits (& therefore have minimum 2 digits).
 *
 * Constraints: 10 <= N <= 1e5
 *
 * Factorion: A natural number that equals the sum of the factorials of its digits.
 * The only non-single-digit factorions are: 145 and 40585.
 *
 * e.g.: N = 20
 *       qualifying numbers = {19}
 *       as 1! + 9! = 362_881, which % 19 = 0
 *       e.g. 18 does not work as 1! + 8! = 40321, which % 18 > 0
 *       sum = 19
 */

class DigitFactorials {
    // pre-calculation of all digit factorials to increase performance
    private val factorials = List(10) { it.factorial().intValueExact() }

    /**
     * HackerRank specific implementation that finds the sum of all numbers < [n] that are divisors
     * of the sum of the factorials of their digits.
     *
     * Increase this solution's efficiency by creating an array of upper constraint
     * size (plus 1) & looping once through all numbers then caching the result plus the
     * previously calculated sum, if it matches the necessary requirements.
     */
    fun sumOfDigitFactorialsHR(n: Int): Int {
        var sum = 0
        for (num in 10 until n) {
            val numSum = num.toString().sumOf { ch -> factorials[ch.digitToInt()] }
            if (numSum % num == 0) {
                sum += num
            }
        }
        return sum
    }

    /**
     * Project Euler specific implementation that finds the sum of all numbers that are factorions.
     *
     * The numbers cannot have more than 7 digits, as 9! * 8 returns only a 7-digit number.
     *
     * 9! * 7 returns 2_540_160, so the 1st digit of the 7-digit number cannot be greater than 2.
     */
    fun sumOfDigitFactorialsPE(): Int {
        var sum = 0
        for (num in 10 until 2_000_000) {
            val digits = num.toString().map { it.digitToInt() }
            var numSum = 0
            for (digit in digits) {
                numSum += factorials[digit]
                // prevents further unnecessary calculation
                if (numSum > num) break
            }
            if (numSum == num) sum += numSum
        }
        return sum
    }
}