How do you solve Nth Magical Number on LeetCode?

Nth Magical Number (LeetCode #878) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log(n * min(a, b))) time complexity and O(1) space complexity. Key insight: Understanding the concept of LCM is crucial for optimizing the search for magical numbers.

What is the brute force solution for Nth Magical Number?

The brute force approach for Nth Magical Number is Brute Force, with O(n) time complexity and O(1) space complexity. We can check each number sequentially to see if it is magical, meaning it is divisible by either a or b. This approach is straightforward but inefficient for large n. The optimal Optimal Solution approach improves this to O(log(n * min(a, b))).

What algorithm or data structure is used to solve Nth Magical Number?

Nth Magical Number uses the following concepts: Math, Binary Search. The recommended patterns to study are: Binary Search, Mathematical Properties.

What are the interview tips for Nth Magical Number?

Always clarify the constraints and edge cases before diving into coding. Explain your thought process and reasoning behind each step as you code. Practice binary search problems to become comfortable with the technique.

What are common mistakes when solving Nth Magical Number?

Failing to consider edge cases where a and b are the same. Not using modulo operations correctly to handle large numbers.

#878

Nth Magical Number

Hard

MathBinary SearchBinary SearchMathematical Properties

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(log(n * min(a, b)))
Space	O(1)	O(1)

💡

Intuition

Time O(log(n * min(a, b)))Space O(1)

Instead of checking each number, we can use binary search to find the nth magical number more efficiently. We calculate the count of magical numbers up to a certain point and adjust our search range accordingly.

⚙️

Algorithm

5 steps

1Step 1: Calculate the least common multiple (LCM) of a and b.
2Step 2: Set the search range between 1 and n * min(a, b).
3Step 3: Use binary search to find the smallest number x such that the count of magical numbers up to x is at least n.
4Step 4: The count of magical numbers up to x is given by (x // a) + (x // b) - (x // lcm(a, b)).
5Step 5: Return the result modulo 10^9 + 7.

solution.py14 lines

1# Full working Python code
2import math
3
4def nthMagicalNumber(n, a, b):
5    mod = 10**9 + 7
6    lcm_ab = a * b // math.gcd(a, b)
7    left, right = 1, n * min(a, b)
8    while left < right:
9        mid = (left + right) // 2
10        if mid // a + mid // b - mid // lcm_ab < n:
11            left = mid + 1
12        else:
13            right = mid
14    return left % mod

ℹ

Complexity note: The time complexity is O(log(n * min(a, b))) due to the binary search over the range. The space complexity is O(1) since we are using a constant amount of space.

1Understanding the concept of LCM is crucial for optimizing the search for magical numbers.
2Binary search can significantly reduce the time complexity compared to a brute-force approach.

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