You are given an integer array digitSum of length n.
An array arr of length n is considered valid if:
0 <= arr[i] <= 5000arr[i] equals digitSum[i].Return an integer denoting the number of distinct valid arrays. Since the answer may be large, return it modulo 109 + 7.
An array is said to be non-decreasing if each element is greater than or equal to the previous element, if it exists.
Example 1:
Input: digitSum = [25,1]
Output: 6
Explanation:
Numbers whose sum of digits is 25 are 799, 889, 898, 979, 988, and 997.
The only number whose sum of digits is 1 that can appear after these values while keeping the array non-decreasing is 1000.
Thus, the valid arrays are [799, 1000], [889, 1000], [898, 1000], [979, 1000], [988, 1000], and [997, 1000].
Hence, the answer is 6.
Example 2:
Input: digitSum = [1]
Output: 4
Explanation:
The valid arrays are [1], [10], [100], and [1000].
Thus, the answer is 4.
Example 3:
Input: digitSum = [2,49,23]
Output: 0
Explanation:
There is no integer in the range [0, 5000] whose sum of digits is 49. Thus, the answer is 0.
Constraints:
1 <= digitSum.length <= 10000 <= digitSum[i] <= 50Problem Overview: You need to count how many arrays satisfy two constraints: the array is non‑decreasing and the total digit sum equals a given target. Each element contributes to the total sum while also respecting the ordering constraint. The challenge is handling both conditions efficiently without enumerating every possible array.
Approach 1: Brute Force Enumeration (Exponential Time, O(10^n) time, O(n) space)
The most direct idea is backtracking. Build the array position by position and only allow values greater than or equal to the previous value to maintain the non‑decreasing property. Track the running digit sum and stop exploring a branch once the sum exceeds the target. While this approach correctly generates all valid arrays, the branching factor (up to 10 choices per position) leads to exponential growth. It becomes impractical even for moderate n. This method mainly helps verify correctness on small inputs.
Approach 2: Dynamic Programming by Length, Last Value, and Sum (O(n * S * 10) time, O(S * 10) space)
Dynamic programming avoids recomputation by storing intermediate states. Define dp[i][d][s] as the number of ways to build an array of length i where the last digit is d and the total sum is s. For the next element, you can only choose digits d' such that d' ≥ d to maintain non‑decreasing order. Transition by iterating possible next digits and updating the sum. This converts the exponential search into polynomial time. The approach is a classic application of dynamic programming with constrained transitions.
Approach 3: Prefix Sum Optimized DP (O(n * S * 10) time, O(S * 10) space)
The DP transition repeatedly sums ranges of previous states because any digit ≥ the previous one is allowed. Instead of iterating through all possible digits each time, maintain prefix sums over the digit dimension. This allows computing transitions in constant time per state. The optimization significantly reduces constant factors and is typically the implementation used in competitive solutions. Techniques like cumulative sums and state compression often appear in problems combining ordering constraints with totals, similar to patterns in prefix sum or counting problems.
Approach 4: Combinatorics Perspective (Problem‑specific, ~O(S * 10))
Non‑decreasing arrays can be viewed as choosing digits with repetition while maintaining sorted order. That turns the array into a multiset of digits. If the only constraint were length, this would reduce to a stars‑and‑bars counting problem. The digit sum restriction introduces an additional constraint, which can still be handled using combinatorial DP or generating functions. This interpretation connects the problem with classic combinatorics counting techniques.
Recommended for interviews: The dynamic programming solution with prefix optimization is what interviewers usually expect. Starting with brute force shows you understand the constraints. Moving to DP demonstrates the ability to recognize overlapping subproblems and reduce exponential search to polynomial time.
Solutions for this problem are being prepared.
Try solving it yourself| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Backtracking | O(10^n) | O(n) | Small inputs or validating correctness of optimized solutions |
| Dynamic Programming (Length, Last Digit, Sum) | O(n * S * 10) | O(S * 10) | General solution that handles ordering and sum constraints efficiently |
| Prefix Sum Optimized DP | O(n * S * 10) | O(S * 10) | Preferred implementation when DP transitions involve digit ranges |
| Combinatorics / Counting | ~O(S * 10) | O(S) | When the problem can be reframed as multiset counting with sum constraints |
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