Leetcode #2999: Count the Number of Powerful Integers
In this guide, we solve Leetcode #2999 Count the Number of Powerful Integers in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given three integers start, finish, and limit. You are also given a 0-indexed string s representing a positive integer.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Math, String, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: start = 1, finish = 6000, limit = 4, s = "124"
Output: 5
Explanation: 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.
Python Solution
class Solution:
def numberOfPowerfulInt(self, start: int, finish: int, limit: int, s: str) -> int:
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
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.