Update search_a_2d_matrix.cpp

Hello there, I`d like to propose my own solution for this binary search problem as it reduces the problem to "704. Binary Search" problem.
Reduction is performed by considering matrix as whole 1D array of length m*n, elements in this "array" are stored in sorted order as well: access to the position inside a matrix can be done by 2 simple computations:
```Array[k] == Matrix[k / n][k % n]```,
where n is the number of columns in a matrix (We simply linearize our matrix)

cpp/neetcode_150/05_binary_search/search_a_2d_matrix.cpp

@@ -47,3 +47,41 @@ class Solution {
  return false;
  }
 };
+
+// Same approach with implicit array linearization
+// T(n) = O(log_2(mn)) = O(log_2(m) + log_2(n))
+class Solution {
+public:
+ bool searchMatrix(vector<vector<int>>& matrix, int target) {
+
+ int left = 0;
+ int m = matrix.size();
+ int n = matrix[0].size();
+ int right = m * n - 1;
+
+ while (left <= right){
+
+ int middle = right + (left - right) / 2;
+
+ // compute sub-indices using matrix structure
+ int row = middle / n;
+ int col = middle % n; 
+
+
+ //ordinary binary search
+ int middle_x = matrix[row][col];
+ if (target > middle_x){
+ left = ++middle;
+ } else if (target < middle_x){
+ right = --middle;
+ } else {
+ return true;
+ }
+
+
+ }
+
+ return false;
+
+ }
+};