How do you solve Ugly Number III on LeetCode?

Ugly Number III (LeetCode #1201) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log(2 * 10^9)) time complexity and O(1) space complexity. Key insight: Understanding the inclusion-exclusion principle is crucial for counting ugly numbers efficiently.

What is the time complexity of Ugly Number III?

The optimal solution for Ugly Number III runs in O(log(2 * 10^9)) time and uses O(1) space. The time complexity is O(log(2 * 10^9)) due to the binary search, and the space complexity is O(1) since we only use a few variables.

What is the brute force solution for Ugly Number III?

The brute force approach for Ugly Number III is Brute Force, with O(n²) time complexity and O(1) space complexity. We can find ugly numbers by checking each number sequentially to see if it is divisible by a, b, or c. This approach is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(log(2 * 10^9)).

What algorithm or data structure is used to solve Ugly Number III?

Ugly Number III uses the following concepts: Math, Binary Search, Combinatorics, Number Theory. The recommended patterns to study are: Binary Search, Inclusion-Exclusion Principle.

What are the interview tips for Ugly Number III?

Always clarify the problem and constraints before jumping into coding. Explain your thought process and approach as you code to demonstrate your understanding. Practice counting problems using inclusion-exclusion to build intuition for similar problems.

What are common mistakes when solving Ugly Number III?

Students often forget to handle the inclusion-exclusion principle correctly, leading to inaccurate counts. Not considering the limits of the binary search can lead to infinite loops or incorrect results.

#1201

Ugly Number III

Medium

MathBinary SearchCombinatoricsNumber TheoryBinary SearchInclusion-Exclusion Principle

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log(2 * 10^9))
Space	O(1)	O(1)

💡

Intuition

Time O(log(2 * 10^9))Space O(1)

Instead of checking each number, we can use binary search to efficiently find the nth ugly number by counting how many ugly numbers exist up to a certain point using the inclusion-exclusion principle.

⚙️

Algorithm

3 steps

1Step 1: Define a function f(k) that counts how many ugly numbers are ≤ k using the inclusion-exclusion principle.
2Step 2: Use binary search between 1 and a large enough number (2 * 10^9) to find the smallest number x such that f(x) = n.
3Step 3: Return x as the nth ugly number.

solution.py23 lines

1# Full working Python code
2
3from math import gcd
4
5def lcm(x, y):
6    return x * y // gcd(x, y)
7
8def countUglyNumbers(k, a, b, c):
9    ab = lcm(a, b)
10    ac = lcm(a, c)
11    bc = lcm(b, c)
12    abc = lcm(ab, c)
13    return (k // a) + (k // b) + (k // c) - (k // ab) - (k // ac) - (k // bc) + (k // abc)
14
15def nthUglyNumber(n, a, b, c):
16    left, right = 1, 2 * 10**9
17    while left < right:
18        mid = (left + right) // 2
19        if countUglyNumbers(mid, a, b, c) < n:
20            left = mid + 1
21        else:
22            right = mid
23    return left

ℹ

Complexity note: The time complexity is O(log(2 * 10^9)) due to the binary search, and the space complexity is O(1) since we only use a few variables.

1Understanding the inclusion-exclusion principle is crucial for counting ugly numbers efficiently.
2Binary search allows us to quickly hone in on the nth ugly number without generating all previous numbers.

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