How do you solve Find the K-Beauty of a Number on LeetCode?

Find the K-Beauty of a Number (LeetCode #2269) 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: Substrings can be efficiently generated by using string slicing.

What is the time complexity of Find the K-Beauty of a Number?

The optimal solution for Find the K-Beauty of a Number runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only loop through the string once to check each substring of length k, making it efficient.

What is the brute force solution for Find the K-Beauty of a Number?

The brute force approach for Find the K-Beauty of a Number is Brute Force, with O(n²) time complexity and O(1) space complexity. We will generate all possible substrings of length k from the number converted to a string and check if each substring is a divisor of the original number. This is straightforward but can be inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the K-Beauty of a Number?

Find the K-Beauty of a Number uses the following concepts: Math, String, Sliding Window. The recommended patterns to study are: String Manipulation, Sliding Window.

What are the interview tips for Find the K-Beauty of a Number?

Always clarify the constraints and edge cases before jumping into coding. Think about the efficiency of your solution and discuss it with the interviewer. Practice writing clean and concise code, as readability is important.

What are common mistakes when solving Find the K-Beauty of a Number?

Forgetting to handle the case where the substring is '0'. Not converting the substring back to an integer before checking divisibility.

#2269

Find the K-Beauty of a Number

Easy

MathStringSliding WindowString ManipulationSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

We can still use the same approach but optimize the way we check for divisibility by ensuring we only convert substrings once and handle leading zeros effectively.

⚙️

Algorithm

5 steps

1Step 1: Convert the number 'num' to a string.
2Step 2: Initialize a counter for valid divisors.
3Step 3: Loop through the string and extract substrings of length k.
4Step 4: Convert each substring to an integer and check if it's a divisor of 'num', skipping '0'.
5Step 5: Return the count of valid divisors.

solution.py16 lines

1# Full working Python code
2
3def k_beauty(num: int, k: int) -> int:
4    num_str = str(num)
5    count = 0
6    for i in range(len(num_str) - k + 1):
7        substring = num_str[i:i + k]
8        if substring != '0':  # Skip if substring is '0'
9            divisor = int(substring)
10            if num % divisor == 0:
11                count += 1
12    return count
13
14# Example usage
15print(k_beauty(240, 2))  # Output: 2
16

ℹ

Complexity note: The time complexity is O(n) because we only loop through the string once to check each substring of length k, making it efficient.

1Substrings can be efficiently generated by using string slicing.
2Divisibility checks are straightforward but must handle edge cases like '0'.

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