Problem:
Score : 300 points

Problem Statement

To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:

1 yen (the currency of Japan)

6 yen, 6^2(=36) yen, 6^3(=216) yen, ...

9 yen, 9^2(=81) yen, 9^3(=729) yen, ...

At least how many operations are required to withdraw exactly N yen in total?

It is not allowed to re-deposit the money you withdrew.

Constraints

1 \leq N \leq 100000

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If at least x operations are required to withdraw exactly N yen in total, print x.

Sample Input 1

127

Sample Output 1

4

By withdrawing 1 yen, 9 yen, 36(=6^2) yen and 81(=9^2) yen, we can withdraw 127 yen in four operations.

Sample Input 2

3

Sample Output 2

3

By withdrawing 1 yen three times, we can withdraw 3 yen in three operations.

Sample Input 3

44852

Sample Output 3

16

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();

        System.out.println(solve2(N));
    }

    private static int solve2(int N) {
        int[] X = new int[N+1];

        X[1] = 1;
        for (int i=2; i<=N; i++) {
            int min = Integer.MAX_VALUE;

            int power6 = 6;
            while (power6 < i) {
                min = Math.min(min, X[i-power6]+1);

                power6 *= 6;
            }

            int power9 = 9;
            while (power9 < i) {
                min = Math.min(min, X[i-power9]+1);

                power9 *= 9;
            }

            for (int j=1; j<10 && j < i; j++) {
                min = Math.min(min, X[i-j]+j);
            }

            X[i] = min;
        }

        return X[N];
    }

    private static int solve(int N) {
        int x = 0;

        while (N > 0) {
            int max9 = power(N, 9);
            int max6 = power(N, 6);

            System.err.println("N = " + N + ", max9 = " + max9 + ", max6 = " + max6 + ", x = " + x);
            if (max9 > max6 && max9 > 0) {
                N -= max9;
                x++;
            } else if (max6 > max9 && max6 > 0) {
                N -= max6;
                x++;
            } else {
                x += N;
                N = 0;
            }
        }

        return x;
    }

    private static int power(int max, int base) {
        if (max < base) return -1;

        int cur = base;
        while (cur*base < max) {
            cur *= base;
        }

        return cur;
    }
}