docs(docs): add documentation page for 74. Search a 2D Matrix problem

apps/docs/content/(array)/74-search-a-2d-matrix.mdx

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+---
+title: '74. Search a 2D Matrix'
+description: 'You are given an m x n integer matrix matrix with the following two properties:'
+sidebar:
+  label: 'Search a 2D Matrix'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Binary Search</Badge>
+
+::::warning
+You must write a solution in O(log(m * n)) time complexity.
+::::
+
+### Example 1:
+- Input: `matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3`
+- Output: `true`
+
+### Example 2:
+- Input: `matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13`
+- Output: `false`
+
+### Constraints:
+
+- `m == matrix.length`
+- `n == matrix[i].length`
+- `1 <= m, n <= 100`
+- `-10^4 <= matrix[i][j], target <= 10^4`
+
+## Solution
+
+```py
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        ROWS, COLS = len(matrix), len(matrix[0])
+
+        top, bot = 0, ROWS - 1
+        while top <= bot:
+            row = (top + bot) // 2
+            if target > matrix[row][-1]:
+                top = row + 1
+            elif target < matrix[row][0]:
+                bot = row - 1
+            else:
+                break
+
+        if not (top <= bot):
+            return False
+
+        row = (top + bot) // 2
+        l, r = 0, COLS - 1
+        while l <= r:
+            m = (l + r) // 2
+            if target > matrix[row][m]:
+                l = m + 1
+            elif target < matrix[row][m]:
+                r = m - 1
+            else:
+                return True
+        return False
+```