A program was supposed to print an array of integers. The program forgot to print whitespaces and the array is printed as a string of digits and all we know is that all integers in the array were in the range [1, k] and there are no leading zeros in the array.
Given the string s and the integer k. There can be multiple ways to restore the array.
Return the number of possible array that can be printed as a string s using the mentioned program.
The number of ways could be very large so return it modulo 10^9 + 7
Example 1:
Input: s = "1000", k = 10000
Output: 1
Explanation: The only possible array is [1000]
Example 2:
Input: s = "1000", k = 10
Output: 0
Explanation: There cannot be an array that was printed this way and has all integer >= 1 and <= 10.
Example 3:
Input: s = "1317", k = 2000
Output: 8
Explanation: Possible arrays are [1317],[131,7],[13,17],[1,317],[13,1,7],[1,31,7],[1,3,17],[1,3,1,7]
Example 4:
Input: s = "2020", k = 30
Output: 1
Explanation: The only possible array is [20,20]. [2020] is invalid because 2020 > 30. [2,020] is ivalid because 020 contains leading zeros.
Example 5:
Input: s = "1234567890", k = 90
Output: 34
Constraints:
1 <= s.length <= 10^5.
s consists of only digits and doesn't contain leading zeros.
1 <= k <= 10^9.
class Solution:
    def numberOfArrays(self, s: str, k: int) -> int:
        MOD = 10 ** 9 + 7
        n = len(s)
        @lru_cache(None)
        def dp(i):
            if i == n:
                return 1
            if s[i] == '0':
                return 0
            num = 0
            ans = 0
            for j in range(i, n):
                num = num * 10 + ord(s[j]) - ord('0')
                if num > k:
                    break
                ans += dp(j+1)
            return ans % MOD
        
        return dp(0)

class Solution:
    def numberOfArrays(self, s: str, k: int) -> int:
        MOD = 10 ** 9 + 7
        n = len(s)
        @lru_cache(None)
        def dp(i):
            if i == n:
                return 1
            if s[i] == '0':
                return 0
            num = 0
            ans = 0
            for j in range(i, n):
                num = num * 10 + ord(s[j]) - ord('0')
                if num > k:
                    break
                ans += dp(j+1)
            return ans % MOD
        
        return dp(0)
