How do you solve Numbers At Most N Given Digit Set on LeetCode?

Numbers At Most N Given Digit Set (LeetCode #902) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding how to count combinations efficiently is crucial.

What is the brute force solution for Numbers At Most N Given Digit Set?

The brute force approach for Numbers At Most N Given Digit Set is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible numbers using the given digits and checking how many of them are less than or equal to n. This is straightforward but inefficient, especially as n increases. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Numbers At Most N Given Digit Set?

Numbers At Most N Given Digit Set uses the following concepts: Array, Math, String, Binary Search, Dynamic Programming. The recommended patterns to study are: Combinatorial Counting, Dynamic Programming.

What are the interview tips for Numbers At Most N Given Digit Set?

Always start with a brute force approach to understand the problem better. Discuss your thought process and explore edge cases during the interview. Practice counting problems to become comfortable with combinatorial logic.

What are common mistakes when solving Numbers At Most N Given Digit Set?

Generating all combinations without considering the constraints, leading to timeouts. Not handling edge cases where n is smaller than the smallest digit.

#902

Numbers At Most N Given Digit Set

Hard

ArrayMathStringBinary SearchDynamic ProgrammingCombinatorial CountingDynamic Programming

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 properties of the digits and the structure of the number n to count valid combinations without generating them. We use mathematical counting based on the length of n and the digits available.

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Algorithm

3 steps

1Step 1: Count all numbers with fewer digits than n using the formula len(digits) ^ i for i from 1 to len(n) - 1.
2Step 2: For numbers with the same number of digits as n, iterate through each digit of n and count how many valid digits are less than the current digit.
3Step 3: If a digit matches, continue checking the next digit; if a digit is smaller, add the count of all combinations that can be formed with the remaining digits.

solution.py20 lines

1def atMostNGivenDigitSet(digits, n):
2    count = 0
3    n_str = str(n)
4    length_n = len(n_str)
5    
6    # Count numbers with fewer digits than n
7    count += sum(len(digits) ** i for i in range(1, length_n))
8    
9    # Count numbers with the same number of digits as n
10    for i in range(length_n):
11        found_smaller = False
12        for d in digits:
13            if d < n_str[i]:
14                count += len(digits) ** (length_n - i - 1)
15            elif d == n_str[i]:
16                found_smaller = True
17                break
18        if not found_smaller:
19            break
20    return count

ℹ

Complexity note: The time complexity is O(n) because we iterate through the digits of n, and for each digit, we check against the digits array, which is constant in size.

1Understanding how to count combinations efficiently is crucial.
2Recognizing the constraints of the problem helps in optimizing the approach.

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