# Solving Leetcode Interviews in Seconds with AI: Restore The Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1416" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the breakdown of the solution:

*   **Dynamic Programming:** We use dynamic programming to solve this problem. `dp[i]` stores the number of possible arrays that can be formed from the substring `s[i:]`.
*   **Iterative Approach:** We iterate from the end of the string `s` to the beginning. For each position `i`, we try all possible lengths of the number starting at that position.
*   **Range Check:** For each possible number, we check if it's within the valid range `[1, k]`. If it is, we add the number of ways to form the rest of the string (i.e., `dp[i + length]`) to `dp[i]`.

*   **Runtime Complexity:** O(n*l), where n is the length of s, and l is the average length of a valid number (worst case is O(n^2) when all numbers are length 1)
*   **Storage Complexity:** O(n)

	
	# Code
	```python
	def solve():
    s = input()
    k = int(input())
    n = len(s)
    dp = [0] * (n + 1)
    dp[n] = 1
    mod = 10**9 + 7

    for i in range(n - 1, -1, -1):
        if s[i] == '0':
            dp[i] = 0
            continue
        
        num_str = ""
        for j in range(i, n):
            num_str += s[j]
            num = int(num_str)
            if num >= 1 and num <= k:
                dp[i] = (dp[i] + dp[j + 1]) % mod
            elif num > k:
                break
    return dp[0]

print(solve())
	```