How do you solve Count Special Integers on LeetCode?

Count Special Integers (LeetCode #2376) 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: Distinct digits are crucial for a number to be special.

What is the brute force solution for Count Special Integers?

The brute force approach for Count Special Integers is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check each integer from 1 to n to see if it has all distinct digits. This is straightforward but inefficient for large n. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Special Integers?

Count Special Integers uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Bitmasking.

What are the interview tips for Count Special Integers?

Explain your thought process clearly while coding. Consider edge cases like small n values (1-10). Practice writing recursive functions with memoization.

What are common mistakes when solving Count Special Integers?

Not considering leading zeros in digit combinations. Forgetting to reset the bitmask when exploring new digits.

#2376

Count Special Integers

Hard

MathDynamic ProgrammingDynamic ProgrammingBitmasking

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)

We can use digit dynamic programming to count special integers more efficiently by building numbers with distinct digits and using a bitmask to track used digits.

⚙️

Algorithm

4 steps

1Step 1: Convert n to a string to analyze its digits.
2Step 2: Use a recursive function with memoization to count valid numbers with distinct digits, considering the current position, the used digits, and whether we are still bound by n.
3Step 3: Count all valid numbers with fewer digits than n.
4Step 4: Count valid numbers with the same number of digits as n, ensuring they do not exceed n.

solution.py28 lines

1# Full working Python code
2
3def count_special_integers(n):
4    s = str(n)
5    length = len(s)
6    dp = [[-1] * (1 << 10) for _ in range(length + 1)]
7
8    def dfs(pos, mask, tight):
9        if pos == length:
10            return 1
11        if dp[pos][mask] != -1:
12            return dp[pos][mask]
13        limit = int(s[pos]) if tight else 9
14        count = 0
15        for digit in range(0, limit + 1):
16            if mask & (1 << digit) == 0:
17                count += dfs(pos + 1, mask | (1 << digit), tight and (digit == limit))
18        dp[pos][mask] = count
19        return count
20
21    total = 0
22    for i in range(1, length):
23        total += 9 * (9 ** (i - 1))  # Count numbers with fewer digits
24    total += dfs(0, 0, True)  # Count numbers with the same number of digits
25    return total
26
27# Example usage
28print(count_special_integers(20))

ℹ

Complexity note: The time complexity is O(n) due to the digit dynamic programming approach, which efficiently counts valid numbers without generating them all.

1Distinct digits are crucial for a number to be special.
2Using dynamic programming can significantly reduce computation time.

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