Find the Smallest Value After One Digit-Sum Replacement

# Find the Smallest Value After One Digit-Sum Replacement ### Problem You are given a positive integer array `nums`. Replace each element with the sum of the decimal digits of that original element, exactly once. Return the minimum of the resulting values. A digit sum may itself have multiple digits; do not keep summing its digits. ### C++ Interface ```cpp class Solution { public: int minElement(vector<int>& nums); }; ``` ### Examples #### Example 1 **Input:** `nums = [248,90,10000]` **Output:** `1` **Explanation:** The replacements are `[14,9,1]`, whose minimum is `1`. #### Example 2 **Input:** `nums = [88,97,799]` **Output:** `16` **Explanation:** The replacements are `[16,16,25]`, whose minimum is `16`. The values `16` and `25` are not replaced again. ### Constraints - `1 <= nums.size() <= 100` - `1 <= nums[i] <= 10^4` ### Clarifying Questions - Which number's digits are summed? The original value at each position. - Must the result be a single digit? No, there is only one replacement per element. - Is the transformed array returned? No, return only its minimum value. ### What a Strong Answer Covers - Correct decimal digit sums for all original elements, including internal or trailing zeros. - Exactly one replacement per element, even when the resulting sum has several digits. - The returned minimum, behavior on a one-element array, and time and space bounds under the given value limit. ### Follow-up Questions - How would repeatedly summing until one digit remained change the task? - Is storing the whole transformed array necessary to return its minimum? ```hint Handle one original number Work out its decimal digit sum independently of the other elements, then compare that one sum with the best result so far. ```