Problem Description

A deci-binary number is a positive integer where every digit is either 0 or 1, with no leading zeros. Examples of deci-binary numbers include 101 and 1100, while 112 and 3001 are not deci-binary because they contain digits other than 0 or 1.

Given a string n representing a positive decimal integer, you need to find the minimum number of positive deci-binary numbers that sum up to n.

For example, if n = "32", you could express it as the sum of deci-binary numbers like:

• 11 + 11 + 10 = 32 (using 3 deci-binary numbers)

The key insight is that when adding deci-binary numbers column by column, each column can contribute at most 1 from each deci-binary number. Therefore, if a digit in position i of the target number is d, you need at least d deci-binary numbers to contribute to that position.

Since you need enough deci-binary numbers to satisfy the digit with the maximum value, the answer is simply the maximum digit in the string n. This is why the solution return int(max(n)) works - it finds the largest digit in the string representation of the number.

Intuition

Let's think about how deci-binary numbers work when we add them together. Since each deci-binary number only contains digits 0 or 1, when we add multiple deci-binary numbers in a column, each number can contribute at most 1 to that column's sum.

Consider the number 32:

• The tens place has digit 3
• The ones place has digit 2

To make the digit 3 in the tens place, we need at least 3 deci-binary numbers, each contributing 1 to that position. Similarly, to make the digit 2 in the ones place, we need at least 2 deci-binary numbers contributing 1 to that position.

We can visualize this addition:

  1 1  (first deci-binary)
  1 1  (second deci-binary)
+ 1 0  (third deci-binary)
-----
  3 2

The key realization is that we need enough deci-binary numbers to satisfy the position with the largest digit. In this case, the digit 3 requires 3 deci-binary numbers. Once we have 3 numbers, we can arrange their bits to also satisfy the digit 2 (and any other smaller digits).

This pattern holds for any number: if the maximum digit in the number is d, we need exactly d deci-binary numbers. Each of these numbers will have a 1 in positions where we need to increment that digit, and 0 elsewhere.

Therefore, the minimum number of deci-binary numbers needed equals the maximum digit in the given number, which explains why max(n) gives us the answer.

Learn more about Greedy patterns.

Solution Approach

The solution is remarkably elegant and simple once we understand the underlying pattern. We need to find the maximum digit in the string representation of the number.

Implementation Steps:

1. Convert the string to individual digits: Since the input n is already a string, each character represents a digit.

2. Find the maximum digit: We can use Python's built-in max() function directly on the string. When applied to a string, max() compares characters lexicographically, which for single digits ('0' through '9') gives us the largest digit.

3. Convert to integer: The max() function returns a character (e.g., '7'), so we convert it to an integer using int().

The complete implementation:

class Solution:
    def minPartitions(self, n: str) -> int:
        return int(max(n))

Why this works:

• max(n) iterates through all characters in the string and returns the largest one
• For a string like "432", max("432") returns '4'
• int('4') converts it to the integer 4

Time Complexity: O(n) where n is the length of the input string, as we need to examine each digit once to find the maximum.

Space Complexity: O(1) as we only use a constant amount of extra space regardless of input size.

This approach leverages the mathematical insight that the minimum number of deci-binary numbers needed equals the maximum digit value, eliminating the need for complex decomposition algorithms or dynamic programming.

Example Walkthrough

Let's walk through the solution with the example n = "827":

Step 1: Identify each digit

• Position 0 (hundreds): digit is 8
• Position 1 (tens): digit is 2
• Position 2 (ones): digit is 7

Step 2: Determine minimum deci-binary numbers needed

• To form digit 8 in the hundreds place, we need 8 deci-binary numbers each contributing 1
• To form digit 2 in the tens place, we need 2 deci-binary numbers each contributing 1
• To form digit 7 in the ones place, we need 7 deci-binary numbers each contributing 1

Step 3: Find the maximum requirement

• The maximum digit is 8, so we need at least 8 deci-binary numbers

Step 4: Verify with decomposition We can construct 8 deci-binary numbers that sum to 827:

  1 0 1  (= 101)
  1 0 1  (= 101)
  1 0 1  (= 101)
  1 0 1  (= 101)
  1 0 1  (= 101)
  1 0 1  (= 101)
  1 0 1  (= 101)
+ 1 1 0  (= 110)
-------
  8 2 7  (= 827)

Notice how:

• 8 numbers contribute 1 to the hundreds place (giving us 8)
• Only 2 of them contribute 1 to the tens place (giving us 2)
• 7 of them contribute 1 to the ones place (giving us 7)

Step 5: Apply the solution

def minPartitions(self, n: str) -> int:
    return int(max(n))  # max("827") = '8', int('8') = 8

The answer is 8, which matches our maximum digit. This confirms that we always need exactly as many deci-binary numbers as the value of the largest digit in our target number.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string. This is because the max() function needs to iterate through each character in the string to find the maximum digit, requiring a single pass through all characters.

The space complexity is O(1). The max() function returns a single character (the maximum digit), and int() converts it to an integer. No additional data structures are created that scale with the input size - only a constant amount of extra space is used regardless of the string length.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting to Actually Construct the Deci-Binary Numbers

Many developers initially try to build the actual deci-binary numbers that sum to n, leading to unnecessarily complex solutions with arrays and loops.

Incorrect Approach:

def minPartitions(self, n: str) -> int:
    result = []
    remaining = int(n)
    while remaining > 0:
        deci_binary = ""
        temp = remaining
        for digit in str(temp):
            if int(digit) > 0:
                deci_binary += "1"
                remaining -= 10 ** (len(str(temp)) - len(deci_binary))
            else:
                deci_binary += "0"
        result.append(int(deci_binary))
    return len(result)

Why it's wrong: This approach is overly complicated and may not even produce correct results. The problem only asks for the COUNT, not the actual numbers.

2. Converting to Integer First

Some might convert the entire string to an integer before processing:

Problematic Approach:

def minPartitions(self, n: str) -> int:
    num = int(n)  # Unnecessary conversion
    return int(max(str(num)))

Why it's inefficient: This adds an unnecessary conversion step. Since the input is already a string, converting to int and back to string wastes time and could cause issues with very large numbers.

3. Using a Loop to Find Maximum

Writing manual loops when built-in functions suffice:

Overly Complex:

def minPartitions(self, n: str) -> int:
    max_digit = 0
    for char in n:
        max_digit = max(max_digit, int(char))
    return max_digit

Why it's suboptimal: While this works correctly, it's more verbose than necessary. Python's max() function handles this elegantly in one line.

4. Misunderstanding the Return Type

Forgetting to convert the character to integer:

Wrong:

def minPartitions(self, n: str) -> int:
    return max(n)  # Returns '9' instead of 9 for "932"

Solution: Always remember to wrap with int() since max() on a string returns a character.

5. Edge Case Paranoia

Adding unnecessary checks for edge cases that don't exist according to problem constraints:

Unnecessary:

def minPartitions(self, n: str) -> int:
    if not n or n == "0":  # Problem states n is positive
        return 0
    if len(n) == 1:  # Unnecessary special case
        return int(n)
    return int(max(n))

Why it's wrong: The problem guarantees a positive integer, so these checks just add complexity without value.

Correct Solution:

The elegant one-liner leverages Python's built-in functions and the mathematical insight:

def minPartitions(self, n: str) -> int:
    return int(max(n))