#2999

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Count the Number of Powerful Integers

Count the number of powerful integers in a given range by applying state transition dynamic programming with constraints on digit values.

math string dynamic programming

Problem Statement

You are given three integers start, finish, and limit, along with a string s representing a positive integer. A positive integer x is considered powerful if it ends with the string s as a suffix, and each of its digits is less than or equal to the limit. Your task is to determine how many powerful integers exist within the range [start, finish].

The string s serves as a suffix for any integer x you are counting. For example, if s = "124", any valid integer x must end in "124" and have all its digits <= the limit. Your goal is to count how many such integers lie within the range from start to finish.

Examples

Example 1

Input: start = 1, finish = 6000, limit = 4, s = "124"

Output: 5

The powerful integers in the range [1..6000] are 124, 1124, 2124, 3124, and, 4124. All these integers have each digit <= 4, and "124" as a suffix. Note that 5124 is not a powerful integer because the first digit is 5 which is greater than 4. It can be shown that there are only 5 powerful integers in this range.

Example 2

Input: start = 15, finish = 215, limit = 6, s = "10"

Output: 2

The powerful integers in the range [15..215] are 110 and 210. All these integers have each digit <= 6, and "10" as a suffix. It can be shown that there are only 2 powerful integers in this range.

Example 3

Input: start = 1000, finish = 2000, limit = 4, s = "3000"

Output: 0

All integers in the range [1000..2000] are smaller than 3000, hence "3000" cannot be a suffix of any integer in this range.

Constraints

1 <= start <= finish <= 1015
1 <= limit <= 9
1 <= s.length <= floor(log10(finish)) + 1
s only consists of numeric digits which are at most limit.
s does not have leading zeros.

Solution Approach

State Transition Dynamic Programming

Use a digit-based dynamic programming approach to count numbers in the range [1, x] that satisfy the suffix condition and digit limit. The idea is to transition through states based on whether the number is still below the limit and ends with the specified suffix.

Handle Large Range Efficiently

Due to the large possible value of finish (up to 10^15), we need to use logarithmic time complexity to handle the range efficiently. This ensures the solution can scale for large inputs while keeping space usage manageable.

Edge Case Considerations

Pay attention to edge cases where the suffix is larger than the upper bound of the range or where the limit on digits precludes any valid numbers. These scenarios can be handled using early termination or checks before applying the dynamic programming solution.

Complexity Analysis

Metric	Value
Time	O(\log(\textit{finish}))
Space	O(\log(\textit{finish}))

The time complexity of the solution is O(log(finish)) due to the use of digit dynamic programming, which reduces the problem size by focusing on the digits of the numbers. The space complexity is O(log(finish)) as well, since we only need to store information about the digits in the range up to the finish value.

What Interviewers Usually Probe

Check if the candidate can recognize the need for digit DP in problems involving number ranges and suffixes.
Assess whether the candidate can handle edge cases and constraints efficiently, especially when dealing with large inputs.
Look for understanding of how dynamic programming can be adapted to count numbers with specific suffixes.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the constraint on digits may lead to incorrect solutions that count numbers outside the valid range.
Forgetting to account for all possible numbers with the given suffix in the range could result in an incomplete solution.
Failing to optimize for large inputs, leading to time or space inefficiency when handling upper bounds of the range.

Follow-up variants

Change the limit on the digits to create a different set of valid numbers.
Modify the suffix string to test if the solution can handle varying patterns and lengths.
Increase the range size to test the scalability of the solution with larger upper bounds.

FAQ

What is a powerful integer in the context of this problem?

A powerful integer ends with a specified suffix and has digits that are all less than or equal to a given limit.

How can I count powerful integers in a range efficiently?

You can use digit dynamic programming (digit DP) to count valid numbers in logarithmic time by transitioning through states based on digits and suffix conditions.

What is the significance of the suffix in this problem?

The suffix condition means that the integer must end with the string s, and this constraint is crucial for counting valid powerful integers.

Why is the time complexity O(log(finish))?

The time complexity is O(log(finish)) because we only need to consider the digits of the numbers within the given range, and this reduces the problem size significantly.

Can this approach handle very large values of finish?

Yes, the digit DP approach is specifically designed to handle large values of finish efficiently by reducing the problem to a manageable size based on the digits.

terminal

Solution

Solution 1: Digit DP

This problem is essentially about finding the count of numbers in the given range $[l, .., r]$ that satisfy the conditions. The count depends on the number of digits and the value of each digit. We can solve this problem using the Digit DP approach, where the size of the number has minimal impact on the complexity.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

class Solution:
    def numberOfPowerfulInt(self, start: int, finish: int, limit: int, s: str) -> int:
        @cache
        def dfs(pos: int, lim: int) -> int:
            if len(t) < n:
                return 0
            if len(t) - pos == n:
                return int(s <= t[pos:]) if lim else 1
            up = min(int(t[pos]) if lim else 9, limit)
            ans = 0
            for i in range(up + 1):
                ans += dfs(pos + 1, lim and i == int(t[pos]))
            return ans

        n = len(s)
        t = str(start - 1)
        a = dfs(0, True)
        dfs.cache_clear()
        t = str(finish)
        b = dfs(0, True)
        return b - a

Continue Practicing

#2851 string transformation #3260 find the largest palindrome divisible by k #2719 count of integers