How do you solve Count Fancy Numbers in a Range on LeetCode?

Count Fancy Numbers in a Range (LeetCode #3869) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Single-digit numbers are always good.

What is the time complexity of Count Fancy Numbers in a Range?

The optimal solution for Count Fancy Numbers in a Range runs in O(n) time and uses O(n) space. Precomputation allows for constant-time checks against good numbers, making the overall complexity linear.

What is the brute force solution for Count Fancy Numbers in a Range?

The brute force approach for Count Fancy Numbers in a Range is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each number in the range [l, r] to see if it's good or if the sum of its digits is good. This is straightforward but inefficient for large ranges. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Fancy Numbers in a Range?

Count Fancy Numbers in a Range uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Fancy Numbers in a Range?

Explain your thought process clearly. Use examples to illustrate your points. Practice coding under time constraints.

What are common mistakes when solving Count Fancy Numbers in a Range?

Not considering single-digit numbers as good. Failing to precompute good sums.

#3869

Count Fancy Numbers in a Range

Hard

MathDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Precompute all good numbers and their sums to quickly determine if a number is fancy without checking each one individually.

⚙️

Algorithm

3 steps

1Step 1: Precompute all good numbers (strictly increasing and decreasing) up to a certain limit.
2Step 2: Precompute good sums for numbers from 1 to 144.
3Step 3: For each number in the range [l, r], check against precomputed good numbers and sums.

solution.py16 lines

1def precompute_goods():
2    goods = set()
3    for i in range(1, 10):
4        goods.add(i)
5    for i in range(1, 10):
6        for j in range(i + 1, 10):
7            goods.add(int(str(i) + str(j)))
8    return goods
9
10def count_fancy(l, r):
11    goods = precompute_goods()
12    count = 0
13    for i in range(l, r + 1):
14        if i in goods or sum(int(d) for d in str(i)) in goods:
15            count += 1
16    return count

ℹ

Complexity note: Precomputation allows for constant-time checks against good numbers, making the overall complexity linear.

1Single-digit numbers are always good.
2The sum of digits can also be good, which adds complexity.

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