# Solving Leetcode Interviews in Seconds with AI: Partitioning Into Minimum Number Of Deci-Binary Numbers


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1689" 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 decimal number is called deci-binary if each of its digits is either 0 or 1 without any leading zeros. For example, 101 and 1100 are deci-binary, while 112 and 3001 are not. Given a string n that represents a positive decimal integer, return the minimum number of positive deci-binary numbers needed so that they sum up to n.   Example 1:  Input: n = "32" Output: 3 Explanation: 10 + 11 + 11 = 32  Example 2:  Input: n = "82734" Output: 8  Example 3:  Input: n = "27346209830709182346" Output: 9    Constraints:  1 <= n.length <= 105 n consists of only digits. n does not contain any leading zeros and represents a positive integer.  

	# Explanation
	Here's the breakdown of the solution:

*   **Key Idea:** The minimum number of deci-binary numbers needed is simply the largest digit in the input string. This is because we can always decompose the number such that each deci-binary number contributes 1 to each digit place up to the maximum digit.
*   **Example:** For "32", we need three deci-binary numbers. We can represent 32 as 10 + 11 + 11. The maximum digit is 3.
*   **Implementation:** Iterate through the string and find the maximum digit.

*   **Complexity:** O(N) time complexity, where N is the length of the string. O(1) space complexity.

	
	# Code
	```python
	def minPartitions(n: str) -> int:
    """
    Given a string n that represents a positive decimal integer,
    return the minimum number of positive deci-binary numbers needed so that they sum up to n.
    """
    max_digit = 0
    for digit in n:
        max_digit = max(max_digit, int(digit))
    return max_digit
	```