Given an array of digits which is sorted in non-decreasing order. You can write numbers using each digits[i] as many times as we want. For example, if digits = ['1','3','5'], we may write numbers such as '13', '551', and '1351315'.
Return the number of positive integers that can be generated that are less than or equal to a given integer n.
Example 1:
Input: digits = ["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: digits = ["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 array.
Example 3:
Input: digits = ["7"], n = 8 Output: 1
Constraints:
1 <= digits.length <= 9digits[i].length == 1digits[i] is a digit from '1' to '9'.digits are unique.digits is sorted in non-decreasing order.1 <= n <= 109class Solution {
public:
int atMostNGivenDigitSet(vector<string>& digits, int n) {
vector<int> d;
int ch = digits.size();
// convert to vector of ints instead of string
for(int i = 0; i < ch; i++)
d.push_back(stoi(digits[i]));
int p = log10(n), res=0;
// all combs with less digits than n's (ie, <p digits)
for(int i = 1; i <= p; i++){
res+=pow(ch, i);
}
cout << res << endl;
// combs with p digits but less than n
string bound = to_string(n);
for (int i = 0; i <= p; i++){
bool eq = false;
int k = (bound[i] - '0');
cout << k << " -> ";
// find that digit in vector `d`
int gti = -1;
for (int idx = 0; idx < ch; idx++){ // doesnt affect complexity -> Atmost 9 runs
gti = d[idx]<k ? idx : gti;
if (d[idx]==k) eq = true;
}
int temp = (gti+1)*pow(ch, p-i);
res += temp>0 ? temp:0;
cout << gti+1 << " " << temp << endl;
if(!eq) break;
}
// check if n can be formed with given digits... If yes, +1
int pdt = 1;
for(int i = 0; i <= p; i++){
auto it = find(d.begin(), d.end(), (bound[i] - '0'));
if (it == d.end()) //not found
pdt = 0;
}
res+=pdt;
cout << endl << res<< endl;
return (res);
}
};