Locate Any Strict Peak in a Matrix

### Problem No side-adjacent entries of rectangular `mat` are equal. A peak exceeds every existing up, down, left, and right neighbor; outside neighbors have value -1. Return coordinates `[row,column]` of any peak. Use `O(m log n)` or `O(n log m)` time for m rows and n columns. ```cpp class Solution { public: vector<int> findPeakGrid(vector<vector<int>>& mat); }; ``` ### Examples **Example 1:** ```text mat = [[2,8],[5,3]] One valid output: [0,1] Explanation: 8 is a peak, and [1,0] at value 5 is another permitted answer. ``` **Example 2:** ```text mat = [[4,7,2],[3,9,5]] One valid output: [1,1] Explanation: 9 exceeds all existing side neighbors and the outside value -1. ``` ### Constraints - `m == mat.length`, `n == mat[i].length`. - `1 <= m,n <= 500`. - `1 <= mat[i][j] <= 10^5`. - No side-adjacent cells have equal values. - Required time: `O(m log n)` or `O(n log m)`. ### Clarifying Questions Any qualifying peak is accepted; there can be several. Diagonal neighbors do not affect the condition. ```hint Inspect the structure Which neighboring cells matter, what boundary values apply, and may different valid coordinates be accepted? ``` ### What a Strong Answer Covers - Show that the returned coordinates exceed every relevant side-adjacent value. - Justify the required time bound while accepting any qualifying peak and handling one-row or one-column inputs. ### Follow-up Questions When rows greatly outnumber columns, how could choosing the other split direction reduce work?