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+# 2999. Count the Number of Powerful Integers
+
+"""
+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(object):
+ def numberOfPowerfulInt(self, start, finish, limit, s):
+ """
+ :type start: int
+ :type finish: int
+ :type limit: int
+ :type s: str
+ :rtype: int
+ """
+ cnt = [0] * 16
+
+ def get_cnt(n, s):
+ res, i, size = cnt[len(n) - 1], 0, len(n) - len(s)
+ while True:
+ res += (
+ n[i:] >= s
+ if i == size
+ else cnt[len(n) - i - 1] * (min(limit, int(n[i]) - 1) + (i > 0))
+ )
+ i += 1
+ if i > size or int(n[i - 1]) > limit:
+ break
+ return res
+
+ for i in range(len(s), 16):
+ cnt[i] = 1 if i == len(s) else cnt[i - 1] * (limit + 1)
+ return get_cnt(str(finish), s) - get_cnt(str(start - 1), s)
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.numberOfPowerfulInt(start=1, finish=6000, limit=4, s="124"))
+ print(obj.numberOfPowerfulInt(start=15, finish=215, limit=6, s="10"))
+ print(obj.numberOfPowerfulInt(start=1000, finish=2000, limit=4, s="3000"))