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 <= num2min_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 <= 10221 <= min_sum <= max_sum <= 400This method involves iterating over each integer between num1 and num2, calculating their digit sums, and counting those within the given range. Though simple, this approach considers each number individually and checks if it satisfies the conditions.
This solution in C involves iterating over each number from num1 to num2 and computing their digit sums. If the digit sum is within the specified range, the count is incremented. This code uses primitive handling for large numbers using strings since C does not inherently support very large integers.
C++
Java
Python
C#
JavaScript
The time complexity is O(N*M) where N is the range of numbers and M is the number of digits due to string manipulations, and the space complexity is O(1).
This approach uses dynamic programming concepts, specifically digit-based dynamic programming, to efficiently compute the number of good integers by understanding properties of digit sums and bounds. By breaking the number problem into subproblems per digit, we reduce unnecessary recalculation.
This dynamic programming solution for C++ leverages digit-based DP to efficiently count valid integers in a range without iterating through all possible numbers. This approach uses state parameters in a recursive function to track the digit position, the current digit sum, and whether the current number representation is 'tight' or bound to the actual number.
Java
The time complexity is O(D*sum_bound*TightStates), where D is the number of digits, and TightStates is typically 2, representing tight and non-tight bounds. The space complexity is O(D*sum_bound*TightStates) due to the DP storage.
| Approach | Complexity |
|---|---|
| Iterative Brute Force | The time complexity is O(N*M) where N is the range of numbers and M is the number of digits due to string manipulations, and the space complexity is O(1). |
| Dynamic Programming with Digit DP | The time complexity is O(D*sum_bound*TightStates), where D is the number of digits, and TightStates is typically 2, representing tight and non-tight bounds. The space complexity is O(D*sum_bound*TightStates) due to the DP storage. |
Find the Punishment Number of an Integer - Leetcode 2698 - Python • NeetCodeIO • 12,079 views views
Watch 9 more video solutions →Practice Count of Integers with our built-in code editor and test cases.
Practice on FleetCode