Leetcode #2719: Count of Integers
In this guide, we solve Leetcode #2719 Count of 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 two numeric strings num1 and num2 and two integers max_sum and min_sum. We denote an integer x to be good if: num1 <= x <= num2 min_sum <= digit_sum(x) <= max_sum.
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: num1 = "1", num2 = "12", min_sum = 1, max_sum = 8
Output: 11
Explanation: There are 11 integers whose sum of digits lies between 1 and 8 are 1,2,3,4,5,6,7,8,10,11, and 12. Thus, we return 11.
Python Solution
class Solution:
def count(self, num1: str, num2: str, min_sum: int, max_sum: int) -> int:
def dfs(pos: int, s: int, limit: bool) -> int:
if pos >= len(num):
return int(min_sum <= s <= max_sum)
up = int(num[pos]) if limit else 9
return (
sum(dfs(pos + 1, s + i, limit and i == up) for i in range(up + 1)) % mod
)
mod = 10**9 + 7
num = num2
a = dfs(0, 0, True)
dfs.cache_clear()
num = str(int(num1) - 1)
b = dfs(0, 0, True)
return (a - b) % mod
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.