How do you solve Count the Number of Powerful Integers on LeetCode?

Count the Number of Powerful Integers (LeetCode #2999) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(d^2) time complexity and O(d) space complexity. Key insight: Powerful integers must end with the specified suffix.

What is the time complexity of Count the Number of Powerful Integers?

The optimal solution for Count the Number of Powerful Integers runs in O(d^2) time and uses O(d) space. The time complexity is O(d^2) where d is the number of digits in the number being processed. This is efficient compared to the brute force method.

What is the brute force solution for Count the Number of Powerful Integers?

The brute force approach for Count the Number of Powerful Integers is Brute Force, with O(n * m) time complexity and O(1) space complexity. The brute-force approach checks every integer in the range from start to finish to see if it meets the criteria of being a powerful integer. This is straightforward but inefficient for large ranges. The optimal Optimal Solution approach improves this to O(d^2).

What algorithm or data structure is used to solve Count the Number of Powerful Integers?

Count the Number of Powerful Integers uses the following concepts: Math, String, Dynamic Programming. The recommended patterns to study are: Digit Dynamic Programming, Backtracking.

What are the interview tips for Count the Number of Powerful Integers?

Clearly explain your thought process while coding. Consider edge cases, such as when start is less than the suffix. Practice digit DP problems to become familiar with the approach.

What are common mistakes when solving Count the Number of Powerful Integers?

Forgetting to check all digits against the limit. Not handling leading zeros correctly.

#2999

Count the Number of Powerful Integers

Hard

MathStringDynamic ProgrammingDigit Dynamic ProgrammingBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n * m)	O(d^2)
Space	O(1)	O(d)

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Intuition

Time O(d^2)Space O(d)

The optimal solution uses a digit dynamic programming (DP) approach to efficiently count the powerful integers without checking each one individually. This method leverages the properties of numbers and their digits to reduce the search space.

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Algorithm

3 steps

1Step 1: Define a DP function that counts valid integers up to a given number x.
2Step 2: Use recursion with memoization to explore digit placements while ensuring they meet the suffix and digit limit conditions.
3Step 3: Calculate the count of powerful integers in the range [1, finish] and subtract those in [1, start-1].

solution.py11 lines

1def countPowerfulIntegers(start, finish, limit, s):
2    def dp(pos, is_tight, has_suffix):
3        if pos == len(s):
4            return 1 if has_suffix else 0
5        count = 0
6        limit_digit = int(s[pos]) if is_tight else limit
7        for digit in range(0, limit_digit + 1):
8            count += dp(pos + 1, is_tight and (digit == limit_digit), has_suffix or (pos == len(s) - 1 and digit == int(s[-1])))
9        return count
10    return dp(0, True, False)
11    total_count = countPowerfulIntegers(finish) - countPowerfulIntegers(start - 1)

ℹ

Complexity note: The time complexity is O(d^2) where d is the number of digits in the number being processed. This is efficient compared to the brute force method.

1Powerful integers must end with the specified suffix.
2Each digit must be within the specified limit.

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