# 1416. Restore The Array
###### tags: `leetcode`
## Description
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 s 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, return the number of the possible arrays that can be printed as s using the mentioned program. Since the answer may be very large, return it modulo 109 + 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]

- Constraints:

>1 <= s.length <= 105
s consists of only digits and does not contain leading zeros.
1 <= k <= 109

## Solution
- The problem can be solved by using dynamic programming, keeping the substring count for each iteration
- In order to fasten up, get the degree for the `k` and for each time, we can add up the value in the degree without checking the availability
- If the substring starts with `0`, the answer for the substring should be 0

if (s[i] == '0') dp[i] = 0;

- Add for all the value that is inside the degree range

for (int j = 1; j < kVal && i + j < s.length(); j++) dp[i] = (dp[i] + dp[i + j]) % 1000000007;

- For the one that is just met, calculate the value and do the comparison. For the whole string, the combination would be 1

if (i + kVal < s.length())
{
    if (stol(s.substr(i, kVal)) <= k) dp[i] = (dp[i] + dp[i + kVal]) % 1000000007;
}
else if (stol(s.substr(i)) <= k) dp[i]++;