Is Count Fancy Numbers in a Range easy or hard?

Count Fancy Numbers in a Range is categorized as a Hard problem because it requires recognizing the Digit DP pattern. Understanding how to model the digit states and manage the tight bound is the main challenge.

Count Fancy Numbers in a Range Python/Java solution

Most implementations define a recursive digit DP function with parameters such as position, tight flag, and rule-specific state. Memoization stores intermediate results. The same logic translates directly across Python, Java, C++, Go, and TypeScript.

What is the best approach for Count Fancy Numbers in a Range?

Digit Dynamic Programming is the most efficient approach. Instead of iterating through every number, it builds valid numbers digit by digit while respecting the upper bound and the fancy-number constraints. This reduces the complexity to roughly O(d * states * 10), where d is the number of digits.

Is Count Fancy Numbers in a Range asked at Google/Amazon/Meta?

Digit DP problems that count numbers with digit constraints are common in interviews at companies like Google, Amazon, and Meta. Variants often involve counting numbers without repeated digits, limiting digit sums, or enforcing adjacency rules.

What data structure is used in Count Fancy Numbers in a Range?

The solution primarily uses recursion with memoization (a DP table or hash map) to store computed states. The digits of the bound are typically stored in an array or string to process them sequentially during the DP transitions.

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

The optimal Digit DP solution runs in O(d * states * 10) time, where d is the number of digits in the bound. Each DP state explores digits 0–9 while memoization avoids recomputation. Space complexity is O(d * states) for the DP cache.

How to solve Count Fancy Numbers in a Range in O(d * states * 10)?

Use a digit DP function that processes the number from the most significant digit to the least. Maintain state variables such as the current index, whether the prefix is tight with the bound, and any rule-specific properties required for a fancy number. Compute count(high) and subtract count(low − 1) to get the final answer.

Description Hints 4 Solution Video 22:37

3869. Count Fancy Numbers in a Range

Hard26.6% acceptance

You are given two integers l and r.

An integer is called good if its digits form a strictly monotone sequence, meaning the digits are strictly increasing or strictly decreasing. All single-digit integers are considered good.

An integer is called fancy if it is good, or if the sum of its digits is good.

Return an integer representing the number of fancy integers in the range [l, r] (inclusive).

A sequence is said to be strictly increasing if each element is strictly greater than its previous one (if exists).

A sequence is said to be strictly decreasing if each element is strictly less than its previous one (if exists).

Example 1:

Input: l = 8, r = 10

Output: 3

Explanation:

8 and 9 are single-digit integers, so they are good and therefore fancy.
10 has digits [1, 0], which form a strictly decreasing sequence, so 10 is good and thus fancy.

Therefore, the answer is 3.

Example 2:

Input: l = 12340, r = 12341

Output: 1

Explanation:

12340
- 12340 is not good because [1, 2, 3, 4, 0] is not strictly monotone.
- The digit sum is 1 + 2 + 3 + 4 + 0 = 10.
- 10 is good as it has digits [1, 0], which is strictly decreasing. Therefore, 12340 is fancy.
12341
- 12341 is not good because [1, 2, 3, 4, 1] is not strictly monotone.
- The digit sum is 1 + 2 + 3 + 4 + 1 = 11.
- 11 is not good as it has digits [1, 1], which is not strictly monotone. Therefore, 12341 is not fancy.

Therefore, the answer is 1.

Example 3:

Input: l = 123456788, r = 123456788

Output: 0

Explanation:

123456788 is not good because its digits are not strictly monotone.
The digit sum is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 8 = 44.
44 is not good as it has digits [4, 4], which is not strictly monotone. Therefore, 123456788 is not fancy.

Therefore, the answer is 0.

Constraints:

1 <= l <= r <= 10¹⁵

Topics

Dynamic Programming Math

Hints open one at a time

Hint 1

Precompute goods: all strictly increasing numbers made from digits 1-9 and all strictly decreasing numbers made from digits 0-9, remove duplicates.

Hint 2

Precompute good_sums: all integers in [1, 144] whose decimal digits are strictly increasing or strictly decreasing (single-digit counts).

Hint 3

Implement digit DP cnt(bound) that counts numbers <= bound whose digit sum is in good_sums.

Hint 4

Answer = cnt(r) - cnt(l - 1) + count(goods in [l, r]) - count(g in goods where l <= g <= r and sum(g) in good_sums).

22:37

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1class Solution:
2    def countFancy(self, l: int, r: int) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

8
10

Input

12340
12341

Input

123456788
123456788

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