Is Count Special Integers easy or hard?

Count Special Integers is labeled Hard because it requires recognizing a digit DP or combinatorics counting pattern. Brute force approaches are straightforward but inefficient. The difficulty lies in converting the problem into counting permutations of unused digits under prefix constraints.

Count Special Integers Python or Java solution

Python and Java implementations typically follow the digit counting strategy: compute permutations for shorter lengths, then iterate through digits of n while tracking used digits. If a duplicate digit appears, the loop stops early. FleetCode provides full implementations in Python, Java, C++, C#, and JavaScript.

How to solve Count Special Integers in O(log n)?

Treat the number digit by digit. First count all valid numbers with fewer digits using permutations. Then iterate through each digit of n and count how many smaller unused digits can appear at that position, multiplying by permutations of remaining digits. Stop when a repeated digit appears. This digit-based counting runs proportional to the number of digits, which is O(log n).

What is the best approach for Count Special Integers?

The most efficient approach uses digit position counting with combinatorics, often described as a digit DP style solution. It counts valid numbers by analyzing each digit of n and calculating how many permutations of unused digits fit under the prefix constraint. This runs in O(d * 10) time where d is the number of digits in n, which is much faster than generating numbers directly.

What data structure is used in Count Special Integers?

Most solutions use a small boolean array or set to track which digits have already been used while processing the digits of n. The algorithm also relies on permutation calculations from combinatorics rather than heavy data structures.

What is the time complexity of Count Special Integers?

The optimal combinatorics-based solution runs in O(d * 10) time where d is the number of digits in n. Since d is at most 10 for typical integer constraints, the algorithm is effectively constant time. Space complexity is O(1) because only a small set or boolean array is used to track used digits.

Is Count Special Integers asked at Google, Amazon, or Meta?

Digit DP and combinatorics counting problems similar to Count Special Integers appear in interviews at companies like Google, Amazon, and Meta. Interviewers often test the ability to convert brute-force enumeration into a counting problem using permutations and prefix constraints.

Description Hints 2 Solution Video 19:55

2376. Count Special Integers

Hard45.2% acceptance1 company

We call a positive integer special if all of its digits are distinct.

Given a positive integer n, return the number of special integers that belong to the interval [1, n].

Example 1:

Input: n = 20
Output: 19
Explanation: All the integers from 1 to 20, except 11, are special. Thus, there are 19 special integers.

Example 2:

Input: n = 5
Output: 5
Explanation: All the integers from 1 to 5 are special.

Example 3:

Input: n = 135
Output: 110
Explanation: There are 110 integers from 1 to 135 that are special.
Some of the integers that are not special are: 22, 114, and 131.

Constraints:

1 <= n <= 2 * 10⁹

Asked at 1 company

Google

Topics

Dynamic Programming Math

Hints open one at a time

Hint 1

Try to think of dynamic programming.

Hint 2

Use the idea of digit dynamic programming to build the numbers, in addition to a bitmask that will tell which digits you have used so far on the number that you are building.

19:55

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution