# Solving Leetcode Interviews in Seconds with AI: Number of Ways to Separate Numbers


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1977" 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
	> You wrote down many positive integers in a string called num. However, you realized that you forgot to add commas to seperate the different numbers. You remember that the list of integers was non-decreasing and that no integer had leading zeros. Return the number of possible lists of integers that you could have written down to get the string num. Since the answer may be large, return it modulo 109 + 7.   Example 1:  Input: num = "327" Output: 2 Explanation: You could have written down the numbers: 3, 27 327  Example 2:  Input: num = "094" Output: 0 Explanation: No numbers can have leading zeros and all numbers must be positive.  Example 3:  Input: num = "0" Output: 0 Explanation: No numbers can have leading zeros and all numbers must be positive.    Constraints:  1 <= num.length <= 3500 num consists of digits '0' through '9'.  

	# Explanation
	Here's the solution:

*   **Dynamic Programming:** Use dynamic programming to store the number of valid lists ending at each prefix of the input string. The state `dp[i]` represents the number of non-decreasing lists that can be formed using the first `i` characters of `num`.
*   **Iteration:** Iterate through all possible prefixes and consider all possible last numbers in the list. The last number must be a valid integer (no leading zeros) and greater than or equal to the previous number in the list.
*   **Comparison:** Compare the current potential last number with the previous number (if there is one). If the current number is smaller, then we can't use it. Otherwise, add the number of ways to form a list ending with the previous number to the current count.

*   **Runtime Complexity:** O(N^2)
*   **Storage Complexity:** O(N^2)

	
	# Code
	```python
	def solve():    num = input()
    n = len(num)
    mod = 10**9 + 7

    if n > 1 and num[0] == '0':
        print(0)
        return

    if n == 1 and num[0] == '0':
        print(0)
        return

    dp = [[0] * (n + 1) for _ in range(n + 1)]
    dp[0][0] = 1

    for i in range(1, n + 1):
        for j in range(1, i + 1):
            # num[i-j:i] is the current number
            current_num_str = num[i-j:i]

            if current_num_str[0] == '0' and len(current_num_str) > 1:
                continue

            current_num = int(current_num_str)

            if i - j == 0:
                dp[i][j] = dp[i][j] + dp[i - j][0]
                dp[i][j] %= mod
                continue

            for k in range(1, i - j + 1):
                prev_num_str = num[i - j - k:i - j]

                if prev_num_str[0] == '0' and len(prev_num_str) > 1:
                    continue

                prev_num = int(prev_num_str)

                if len(current_num_str) > len(prev_num_str):
                    dp[i][j] = (dp[i][j] + dp[i-j][k]) % mod
                elif len(current_num_str) < len(prev_num_str):
                    continue
                else:
                    if current_num >= prev_num:
                         dp[i][j] = (dp[i][j] + dp[i-j][k]) % mod

    ans = 0
    for j in range(1, n + 1):
        ans = (ans + dp[n][j]) % mod
    print(ans)


solve()
	```