Replace Circular Entries with Neighbor Sums

### Problem For every position in circular array `code`, compute a replacement from the original array. If `k` is positive, sum the next k entries; if negative, sum the preceding -k entries; if zero, use zero. Return all replacements in their original position order. ```cpp class Solution { public: vector<int> decrypt(vector<int>& code, int k); }; ``` ### Examples **Example 1:** ```text code = [3,1,6,2], k = 2 Output: [7,8,5,4] Explanation: The last two positions wrap around to include the beginning. ``` **Example 2:** ```text code = [4,2,7], k = -1 Output: [7,4,2] Explanation: Each replacement is its predecessor in the circular order. ``` ### Constraints - `n == code.length`, with `1 <= n <= 100`. - `1 <= code[i] <= 100`. - `-(n-1) <= k <= n-1`. ### Clarifying Questions All replacements are simultaneous: newly computed values must never supply a later sum. The current entry is not among its own neighbors. ```hint Inspect the structure Which original entries belong to each sum for positive, negative, or zero `k`? ``` ### Reasoning checkpoints - Handle the sign of k and circular wrapping. - Read only original entries when creating replacements. ### Optional extensions How could a sliding sum reduce the work from O(n*|k|) to O(n)?