What is the time complexity of Partitioning Into Minimum Number Of Deci-Binary Numbers?

The optimal solution for Partitioning Into Minimum Number Of Deci-Binary Numbers runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only need to iterate through the digits once. The space complexity is O(1) since we are using a constant amount of space.

What is the brute force solution for Partitioning Into Minimum Number Of Deci-Binary Numbers?

The brute force approach for Partitioning Into Minimum Number Of Deci-Binary Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible deci-binary numbers and checking how many are needed to sum up to the given number. This is straightforward but inefficient as it tries to cover all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Partitioning Into Minimum Number Of Deci-Binary Numbers?

Partitioning Into Minimum Number Of Deci-Binary Numbers uses the following concepts: String, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Partitioning Into Minimum Number Of Deci-Binary Numbers?

Always think about the properties of the numbers involved — in this case, the digits. Start with a brute force approach to understand the problem before optimizing. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Partitioning Into Minimum Number Of Deci-Binary Numbers?

Overcomplicating the problem by trying to generate all combinations of deci-binary numbers. Not recognizing that the maximum digit gives the answer directly.

#1689

Partitioning Into Minimum Number Of Deci-Binary Numbers

Medium

StringGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach leverages the fact that the maximum digit in the number determines the minimum number of deci-binary numbers needed. Each deci-binary number can contribute a 1 to each digit position, so we only need as many as the highest digit.

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Algorithm

3 steps

1Step 1: Iterate through each character in the string.
2Step 2: Convert each character to an integer and keep track of the maximum value.
3Step 3: Return the maximum value found as it represents the minimum number of deci-binary numbers needed.

solution.py3 lines

1def min_deci_binary(n: str) -> int:
2    return max(int(digit) for digit in n)
3

ℹ

Complexity note: The time complexity is O(n) because we only need to iterate through the digits once. The space complexity is O(1) since we are using a constant amount of space.

1The maximum digit in the number directly determines the number of deci-binary numbers needed.
2Each deci-binary number can only contribute a 1 in each digit position.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.