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Partitioning Into Minimum Number Of Deci-Binary Numbers

Difficulty: Medium


Problem Description

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. A deci-binary number is defined as a number where each digit is either 0 or 1 without leading zeros.


Key Insights

  • Each digit of n contributes to the number of deci-binary numbers needed.
  • The maximum digit in n determines how many deci-binary numbers are required.
  • For each digit, we can create a deci-binary number that contributes to that digit without exceeding it.
  • The solution can be achieved in linear time by simply finding the maximum digit.

Space and Time Complexity

Time Complexity: O(n), where n is the length of the string n. Space Complexity: O(1), as we only use a constant amount of space for variables.


Solution

To solve the problem, we need to determine the maximum digit in the string n. This maximum digit indicates how many deci-binary numbers are necessary to sum to n, as each deci-binary number can only represent a single digit of either 0 or 1. The algorithm involves traversing through the string once to find the maximum digit, which gives us the result directly.


Code Solutions

def min_deci_binary_numbers(n: str) -> int:
    # Find the maximum digit in the string n
    max_digit = max(int(digit) for digit in n)
    return max_digit
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