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// 2999. Count the Number of Powerful Integers
#include "leetcode.h"
/*
* you are given three integers 'start', 'finish', and 'limit'. you are also
* given a 0-indexed string 's' representing a positive integer. a positive
* integer 'x' is called powerful if it ends with 's' and each digit in 'x' is
* at most 'limit'. return the total number of powerful integers in the range
* '[start .. finish]'. a string 'x' is a suffix of a string 'y' if and only if
* 'x' is a substring of 'y' that starts from some index in 'y' and extends to
* the index 'y.length - 1'.
*/
class Solution {
private:
long long cnt[16];
long long get_cnt(char l, const string &n, const string &s) {
long long res = cnt[n.size() - 1], i = 0, size = n.size() - s.size();
do
res += i == size ? n.substr(i) >= s
: cnt[n.size() - i - 1] * (min(l, n[i]) - '1' + (i > 0));
while (i < size && n[i++] < l);
return res;
}
public:
long long numberOfPowerfulInt(long long start, long long finish, int limit,
string s) {
for (int i = s.size(); i < 16; ++i)
cnt[i] = i == s.size() ? 1 : cnt[i - 1] * (limit + 1);
return get_cnt(limit + '1', to_string(finish), s) -
get_cnt(limit + '1', to_string(start - 1), s);
}
};
int main() {
Solution obj;
printf("%lld\n", obj.numberOfPowerfulInt(1, 6000, 4, "124")); // expect: 5
printf("%lld\n", obj.numberOfPowerfulInt(15, 215, 6, "")); // expect: 2
printf("%lld\n", obj.numberOfPowerfulInt(1000, 2000, 4, "3000")); // expect: 0
}
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