Problem Description

You are given two numeric strings num1 and num2 representing integers, and two integers max_sum and min_sum. Your task is to count how many integers x are "good" based on the following criteria:

1. x must be within the range [num1, num2] (inclusive), meaning num1 <= x <= num2
2. The sum of digits of x must be within the range [min_sum, max_sum] (inclusive), meaning min_sum <= digit_sum(x) <= max_sum

For example, if x = 123, then digit_sum(x) = 1 + 2 + 3 = 6.

Since the count of good integers can be very large, you need to return the result modulo 10^9 + 7.

The challenge involves efficiently counting all integers in a potentially huge range that satisfy the digit sum constraint, without iterating through each number individually. The solution uses digit dynamic programming to handle this efficiently by processing the numbers digit by digit from left to right, keeping track of the current digit sum and whether we've reached the upper bound of our range.

Intuition

When we need to count integers in a range [num1, num2] that satisfy certain digit-based conditions, directly iterating through all numbers would be inefficient, especially for large ranges. Instead, we can transform this into a classic digit DP problem.

The key insight is to convert the range problem into a difference of two simpler problems. Rather than counting good integers in [num1, num2], we can:

• Count good integers in [0, num2]
• Count good integers in [0, num1-1]
• Subtract the second from the first to get our answer

For counting integers up to a given number with digit sum constraints, we can build the number digit by digit from left to right. At each position, we need to track:

1. Current position (pos): Which digit we're currently placing
2. Current digit sum (s): Sum of all digits we've placed so far
3. Limit flag (limit): Whether we're still bounded by the upper limit number

The limit flag is crucial - if we've been placing digits that match the upper bound so far (e.g., if the upper bound is 523 and we've placed 5 and 2), then at the current position we can only place digits up to the corresponding digit in the upper bound (in this case, 3). However, if we've already placed a smaller digit earlier (like placing 4 instead of 5 at the first position), then we're free to place any digit 0-9 at subsequent positions.

This recursive approach with memoization allows us to efficiently explore all valid numbers without explicitly generating them. When we reach the end of the number (processed all positions), we simply check if our accumulated digit sum falls within [min_sum, max_sum].

The modulo operation 10^9 + 7 is applied during the computation to prevent integer overflow while maintaining the correct result.

Learn more about Math and Dynamic Programming patterns.

Solution Approach

The solution implements digit dynamic programming with memoization. Here's how the implementation works:

Main Function Structure: The count function handles the range calculation by computing the answer for [0, num2] and subtracting the answer for [0, num1-1].

Core DP Function - dfs(pos, s, limit):

• pos: Current digit position being processed (from left to right)
• s: Current sum of digits accumulated so far
• limit: Boolean flag indicating if we're still bounded by the upper limit

Base Case: When pos >= len(num), we've formed a complete number. We return 1 if the digit sum s is within [min_sum, max_sum], otherwise return 0.

Recursive Case:

1. Determine the upper bound for the current digit:

   • If limit is True: upper bound is int(num[pos]) (the digit at current position in the limit number)
   • If limit is False: upper bound is 9 (we can use any digit)

2. Try all possible digits from 0 to up:

   • For each digit i, recursively call dfs(pos + 1, s + i, limit and i == up)
   • The new limit is True only if both current limit is True AND we chose the maximum possible digit i == up

3. Sum all possibilities and apply modulo 10^9 + 7

Memoization with @cache: The @cache decorator stores results of dfs calls to avoid redundant calculations. This is crucial for efficiency as many subproblems repeat.

Computing the Final Answer:

num = num2
a = dfs(0, 0, True)  # Count good integers in [0, num2]
dfs.cache_clear()    # Clear cache before next computation
num = str(int(num1) - 1)  # Convert to num1-1
b = dfs(0, 0, True)  # Count good integers in [0, num1-1]
return (a - b) % mod  # Return the difference

The cache is cleared between computations because the num variable changes, and cached results from the first computation would be invalid for the second.

Time Complexity: O(10 × n × max_sum) where n is the length of the number string. For each position (n), each possible digit sum (up to max_sum), we might try up to 10 digits.

Space Complexity: O(n × max_sum) for the memoization cache storing unique states.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example: Count good integers where num1 = "10", num2 = "15", min_sum = 3, max_sum = 5

We need to find integers in range [10, 15] whose digit sums are in [3, 5].

Step 1: Transform the problem Instead of directly counting in [10, 15], we compute:

• Count of good integers in [0, 15]
• Count of good integers in [0, 9] (which is num1-1)
• Answer = first count - second count

Step 2: Count good integers in [0, 15] using dfs(pos, s, limit)

Let's trace through building numbers digit by digit up to "15":

Starting with dfs(0, 0, True):

• Position 0 (tens place), sum=0, limit=True
• Upper bound is '1' (first digit of "15")
• Try digit 0: leads to numbers 00-09
  • dfs(1, 0, False) - limit becomes False since 0 < 1
  • At position 1, can use any digit 0-9
  • Results: 03(sum=3✓), 04(sum=4✓), 05(sum=5✓) → 3 good numbers
• Try digit 1: leads to numbers 10-15
  • dfs(1, 1, True) - limit stays True since 1 == 1
  • At position 1, upper bound is '5' (second digit of "15")
  • Try digits 0-5:
    • 10: sum=1+0=1 ✗
    • 11: sum=1+1=2 ✗
    • 12: sum=1+2=3 ✓
    • 13: sum=1+3=4 ✓
    • 14: sum=1+4=5 ✓
    • 15: sum=1+5=6 ✗
  • Results: 3 good numbers

Total for [0, 15]: 3 + 3 = 6 good numbers

Step 3: Count good integers in [0, 9]

Starting with dfs(0, 0, True):

• Position 0 (ones place), sum=0, limit=True
• Upper bound is '9' (the only digit of "9")
• Try digits 0-9:
  • 0: sum=0 ✗
  • 1: sum=1 ✗
  • 2: sum=2 ✗
  • 3: sum=3 ✓
  • 4: sum=4 ✓
  • 5: sum=5 ✓
  • 6: sum=6 ✗
  • 7: sum=7 ✗
  • 8: sum=8 ✗
  • 9: sum=9 ✗

Total for [0, 9]: 3 good numbers

Step 4: Calculate final answer Answer = 6 - 3 = 3

The good integers in [10, 15] are: 12 (sum=3), 13 (sum=4), 14 (sum=5)

Key Observations:

1. The limit flag controls whether we can use any digit (0-9) or are restricted by the upper bound
2. Once we place a digit smaller than the bound, all subsequent positions become unrestricted
3. Memoization prevents recalculating the same (pos, s, limit) states multiple times
4. The digit sum constraint is only checked when we've built a complete number (base case)

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * S * d) where n is the length of the longer number between num1 and num2, S is the maximum possible sum of digits (which is 9 * n), and d is the number of possible digits (10).

The DFS function explores states defined by:

• pos: position in the number (0 to n-1)
• s: current sum of digits (0 to 9*n)
• limit: boolean flag

For each state, we iterate through at most 10 digits (0-9). Due to memoization with @cache, each unique state is computed only once. The total number of unique states is n * (9n + 1) * 2 = O(n²). Since we iterate through up to 10 digits for each state, the overall time complexity is O(10 * n²) = O(n²).

However, more precisely, the time complexity is O(n * S) where S represents the range of possible sums, which is bounded by 9n. Since we call the function twice (once for num2 and once for num1 - 1), the complexity remains O(n * S).

Space Complexity: O(n * S) where n is the length of the number and S is the maximum possible sum.

The space is used for:

• The memoization cache storing results for each unique (pos, s, limit) combination
• The recursion stack depth which is at most n

The cache stores at most n * (9n + 1) * 2 states, which simplifies to O(n²) or more precisely O(n * S) space. The recursion stack adds O(n) space, but this is dominated by the cache size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Handling of num1 - 1 Leading to Edge Cases

The Pitfall: When computing str(int(num1) - 1), there's a critical edge case when num1 = "0". This results in int("0") - 1 = -1, which becomes "-1" as a string. The digit DP function isn't designed to handle negative numbers, leading to incorrect results or unexpected behavior.

Example:

num1 = "0"
num1_minus_one = str(int(num1) - 1)  # Results in "-1"
# The DP function will process "-1" incorrectly

Solution: Add a special check for when num1 = "0":

if num1 == "0":
    # Special case: when num1 is 0, we don't need to subtract anything
    count_up_to_num1_minus_one = 0
else:
    num1_minus_one = str(int(num1) - 1)
    count_up_to_num1_minus_one = digit_dp(0, 0, True, num1_minus_one)

2. Cache Pollution Between Different Problem Instances

The Pitfall: The @cache decorator maintains state between different calls to the count method. If the Solution class instance is reused for multiple test cases, the cache from previous computations can interfere with new ones, especially since the parameters min_sum and max_sum are used inside the cached function but aren't part of the cache key.

Example:

sol = Solution()
# First call with min_sum=5, max_sum=10
result1 = sol.count("1", "100", 5, 10)
# Second call with different min_sum/max_sum
result2 = sol.count("1", "100", 3, 8)  # May use cached values with wrong sum bounds!

Solution: Either include all relevant parameters in the function signature or clear the cache at the beginning of each count call:

def count(self, num1: str, num2: str, min_sum: int, max_sum: int) -> int:
    from functools import cache
  
    MOD = 10**9 + 7
  
    @cache
    def digit_dp(position: int, digit_sum: int, is_limit: bool, 
                 current_number: str, min_s: int, max_s: int) -> int:
        # Include min_s and max_s as parameters
        if position >= len(current_number):
            return 1 if min_s <= digit_sum <= max_s else 0
        # ... rest of the function
  
    # Pass min_sum and max_sum explicitly
    count_up_to_num2 = digit_dp(0, 0, True, num2, min_sum, max_sum)

3. Integer Overflow in Modulo Operation

The Pitfall: When computing (count_up_to_num2 - count_up_to_num1_minus_one) % MOD, if count_up_to_num1_minus_one > count_up_to_num2, the subtraction results in a negative number. In Python, this isn't typically an issue as the modulo operation handles negatives correctly, but in other languages or when porting this solution, it can cause problems.

Example:

# If count_up_to_num2 = 5 and count_up_to_num1_minus_one = 8
result = (5 - 8) % MOD  # In Python: gives correct positive result
# In some other languages: might give negative result

Solution: Ensure the result is always positive by adding MOD before taking modulo:

return ((count_up_to_num2 - count_up_to_num1_minus_one) % MOD + MOD) % MOD

Or more elegantly:

return (count_up_to_num2 - count_up_to_num1_minus_one + MOD) % MOD