902. Numbers At Most N Given Digit Set

We have a sorted set of digits D, a non-empty subset of {‘1′,’2′,’3′,’4′,’5′,’6′,’7′,’8′,’9’}. (Note that ‘0’ is not included.)

Now, we write numbers using these digits, using each digit as many times as we want. For example, if D = {‘1′,’3′,’5′}, we may write numbers such as ’13’, ‘551’, ‘1351315’.

Return the number of positive integers that can be written (using the digits of D) that are less than or equal to N.

Example 1:
Input: D = ["1","3","5","7"], N = 100 Output: 20 Explanation: The 20 numbers that can be written are: 1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.
Example 2:

Input: D = ["1","4","9"], N = 1000000000 Output: 29523 Explanation: We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers, 81 four digit numbers, 243 five digit numbers, 729 six digit numbers, 2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers. In total, this is 29523 integers that can be written using the digits of D.
Note:
1. D is a subset of digits '1'-'9' in sorted order.
2. 1 <= N <= 10^9

My Solution(88ms 22.3MB)

public class Solution {
    public int AtMostNGivenDigitSet (string[] D, int N) {
        int ret = 0;
        bool donotIgnoreMatch = true;
        bool currentMatchCheck = false;

        // Convert D string[] to int[]
        int[] newD = new int[D.Length];
        for (int i = 0; i < D.Length; i++) {
            newD[i] = int.Parse (D[i]);
        }

        //Convert int N to string and get Length;
        string NString = N.ToString ();
        int NLength = NString.Length;
        for (int i = 0; i < NLength; i++) {
            int num = int.Parse (NString[i].ToString ());
            int availableDigit = 0;
            if (donotIgnoreMatch) {
                currentMatchCheck = false;
                foreach (int d in newD) {
                    if (d < num) {
                        availableDigit += 1;
                    } else if (d == num) {
                        availableDigit += 1;
                        currentMatchCheck = true;
                    } else {
                        break;
                    }
                }
                donotIgnoreMatch = currentMatchCheck;
            } else {
                availableDigit = newD.Length;
            }

            if (i == 0 && availableDigit == 0) {
                continue;
            }

            // Calculate posibility in current digit
            ret += availableDigit * (int) Math.Pow (newD.Length, NLength - i - 1);

        }
        return ret;
    }
}

No.1 Solution(88ms)

public class Solution {
    public int AtMostNGivenDigitSet(string[] D, int N) {
        string s = N.ToString();
        int NLength = s.Length;
        int L = D.Length;

        int res = 0;

        for (int i = 1; i < NLength; ++i)
            res += (int)Math.Pow(L, i);

        for(int i = 0; i<s.Length; ++i)
        {
            char curCh = s[i];
            int curLength = NLength - i - 1;

            bool meetEqual = false;
            for(int j = 0; j<D.Length; ++j)
            {
                if (D[j][0] < curCh)
                    res += (int)Math.Pow(L, curLength);
                else if (D[j][0] == curCh)
                {
                    meetEqual = true;
                    break;
                }
            }

            if (!meetEqual) return res;
        }

        return res + 1;
    }
}

Things to improve

Even the speed is the same, the memory usage is something I should pay more attention on. Since we are not reach enough to get TB of memory.

Leave a Reply

Your email address will not be published. Required fields are marked *