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Partitioning Into Minimum Number Of Deci-Binary Numbers

Number: 1807

Difficulty: Medium

Paid? No

Companies: Amazon, Apple, Nutanix

Problem Description

Given a string n representing a positive decimal integer, determine the minimum number of positive deci-binary numbers needed such that their sum equals n. A deci-binary number is one where each digit is either 0 or 1, with no leading zeros.

Key Insights

  • The minimum number of deci-binary numbers required is equal to the maximum digit in the string n.
  • Each deci-binary number can contribute at most 1 at each digit position.
  • Therefore, for a digit d in n, you need at least d deci-binary numbers to sum up to that particular digit.

Space and Time Complexity

Time Complexity: O(n), where n is the length of the input string, because we iterate through each digit once. Space Complexity: O(1), using a constant amount of extra space.

Solution

The idea behind the solution is based on the observation that for any digit in the number, you must have at least that many deci-binary numbers, since each deci-binary number can only contribute a 1 at that digit's place. Thus, the answer is the maximum digit in the input string n. We simply iterate over the string, convert each character to an integer, and track the maximum value found.

Code Solutions

# Function to find the minimum number of deci-binary numbers required
def minPartitions(n: str) -> int:
    # Initialize the maximum digit seen so far to 0
    max_digit = 0
    # Iterate through each character in the string n
    for char in n:
        # Convert the character to an integer
        digit = int(char)
        # Update max_digit if the current digit is greater than max_digit
        if digit > max_digit:
            max_digit = digit
        # Early exit: if max_digit is 9, we cannot get a higher value.
        if max_digit == 9:
            return 9
    # Return the maximum digit found
    return max_digit

# Example usage:
# print(minPartitions("82734"))
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