#2376

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Count Special Integers

Count the number of special integers in the interval [1, n] where digits of each integer are distinct.

Problem Statement

You are given a positive integer n, and your task is to return the number of special integers between 1 and n. A special integer is defined as one where all of its digits are distinct. For example, the number 123 is special, but 112 is not, since the digit 1 repeats.

Your solution must handle values of n up to 2 * 10^9, which means a brute force approach may not be efficient enough. Instead, you can use dynamic programming with state transitions to efficiently count the number of special integers in the given range.

Examples

Example 1

Input: n = 20

Output: 19

All the integers from 1 to 20, except 11, are special. Thus, there are 19 special integers.

Example 2

Input: n = 5

Output: 5

All the integers from 1 to 5 are special.

Example 3

Input: n = 135

Output: 110

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

Solution Approach

State Transition Dynamic Programming

The problem can be solved using dynamic programming with a state transition approach. For each digit position, keep track of whether the digit has been used, and use transitions to count valid numbers while ensuring that digits are distinct.

Digit Counting

Count the possible valid digits for each position while avoiding repetitions. This helps to narrow down the search space and avoid brute-forcing all possible values, leading to a more efficient solution.

Recursive with Memoization

Implement a recursive function that checks for valid numbers and memoizes the results to avoid redundant calculations. This approach is especially useful when exploring different digit combinations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity of the solution depend on the depth of the recursion and the number of digits in n. With proper memoization and dynamic programming, the solution can run efficiently even for large values of n.

What Interviewers Usually Probe

Evaluate the candidate's understanding of dynamic programming and recursion.
Assess if the candidate can handle large input sizes efficiently.
Test the candidate's ability to reduce a brute-force approach to a more optimized solution using state transitions.

Common Pitfalls or Variants

Common pitfalls

Forgetting to memoize intermediate results, leading to redundant calculations.
Incorrectly handling leading zeros or invalid digits.
Failing to consider all digit positions, which can result in incorrect counts.

Follow-up variants

Extend the problem to count special integers within a range [m, n].
Count the number of special integers that can be formed by rearranging the digits of n.
Modify the problem to include additional constraints, such as excluding certain digits.

FAQ

What is a special integer in the Count Special Integers problem?

A special integer is a number where all digits are distinct. For example, 123 is special, but 112 is not because the digit 1 repeats.

How do state transitions help solve this problem?

State transitions allow us to efficiently track used digits and count valid numbers without brute-forcing all possible integers.

Can I use a brute force approach for this problem?

A brute force approach may not be efficient enough, especially for large values of n. A dynamic programming approach is recommended.

What is the time complexity of the solution?

The time complexity depends on the number of digits in n and the efficiency of memoization. It can be significantly reduced with dynamic programming.

What are the key challenges in solving this problem?

The main challenges include handling large input sizes efficiently and ensuring that digits are distinct without missing any possibilities.

terminal

Solution

Solution 1: State Compression + Digit DP

This problem essentially asks for the number of numbers in the given range $[l, ..r]$ that satisfy certain conditions. The conditions are related to the composition of the numbers rather than their size, so we can use the concept of Digit DP to solve it. In Digit DP, the size of the number has little impact on the complexity.

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class Solution:
    def countSpecialNumbers(self, n: int) -> int:
        @cache
        def dfs(i: int, mask: int, lead: bool, limit: bool) -> int:
            if i >= len(s):
                return int(lead ^ 1)
            up = int(s[i]) if limit else 9
            ans = 0
            for j in range(up + 1):
                if mask >> j & 1:
                    continue
                if lead and j == 0:
                    ans += dfs(i + 1, mask, True, limit and j == up)
                else:
                    ans += dfs(i + 1, mask | 1 << j, False, limit and j == up)
            return ans

        s = str(n)
        return dfs(0, 0, True, True)

Continue Practicing

#2400 number of ways to reach a position after exactly k steps #2338 count the number of ideal arrays #2310 sum of numbers with units digit k