Introduction

In this blog post, we will explore how to solve the LeetCode problem "2719" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. Return the number of good integers. Since the answer may be large, return it modulo 109 + 7. Note that digit_sum(x) denotes the sum of the digits of x. Example 1: 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. Example 2: Input: num1 = "1", num2 = "5", min_sum = 1, max_sum = 5 Output: 5 Explanation: The 5 integers whose sum of digits lies between 1 and 5 are 1,2,3,4, and 5. Thus, we return 5. Constraints: 1 <= num1 <= num2 <= 1022 1 <= min_sum <= max_sum <= 400

Explanation

Here's a breakdown of the solution:

• Dynamic Programming: The core idea is to use dynamic programming (specifically, memoization) to efficiently calculate the number of integers within a given range that satisfy the digit sum constraint. We break the problem down into smaller overlapping subproblems.

• Top-Down Approach with Memoization: A recursive function count_good explores all possible digits for each position in the number, keeping track of the current sum and the upper bound. The @lru_cache decorator stores results of previously computed subproblems to avoid redundant calculations.

• Helper function to count within range: We create a helper function count_range to compute the good integers within the range [0, num] for given min_sum, and max_sum. We finally compute count_range(num2, min_sum, max_sum) - count_range(num1 - 1, min_sum, max_sum). Note that num1 - 1 is converted to string before passed to count_range and is handled separately as count_range method would not accept a negative number.

• Time & Space Complexity: O(length * max_sum * 10), where length is the maximum length of num1 and num2. The space complexity is dominated by the memoization cache, also O(length * max_sum).

Code

    from functools import lru_cache

def solve():
    num1 = input()
    num2 = input()
    min_sum = int(input())
    max_sum = int(input())

    MOD = 10**9 + 7

    def count_range(num_str, min_sum, max_sum):
        n = len(num_str)

        @lru_cache(None)
        def count_good(index, current_sum, is_tight):
            if index == n:
                return min_sum <= current_sum <= max_sum

            count = 0
            upper_bound = int(num_str[index]) if is_tight else 9

            for digit in range(upper_bound + 1):
                new_sum = current_sum + digit
                new_is_tight = is_tight and (digit == int(num_str[index]))
                count = (count + count_good(index + 1, new_sum, new_is_tight)) % MOD

            return count

        return count_good(0, 0, True)

    # Handle num1 - 1.  If num1 is "1", then num1 - 1 = "0".
    # Otherwise, subtract 1 and convert to string.
    if num1 == "1":
        count_num1_minus_1 = 0
    else:
        num1_int = int(num1)
        count_num1_minus_1 = count_range(str(num1_int - 1), min_sum, max_sum)

    result = (count_range(num2, min_sum, max_sum) - count_num1_minus_1) % MOD
    return result

# Driver code (for local testing - remove in submission)
# num1 = "1"
# num2 = "12"
# min_sum = 1
# max_sum = 8
# print(solve(num1, num2, min_sum, max_sum))  # Output: 11

# num1 = "1"
# num2 = "5"
# min_sum = 1
# max_sum = 5
# print(solve(num1, num2, min_sum, max_sum))  # Output: 5

def solve_interactive():
    print(solve())

if __name__ == "__main__":
    solve_interactive()