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+Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
+
+*   Integers in each row are sorted from left to right.
+*   The first integer of each row is greater than the last integer of the previous row.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/10/05/mat.jpg)
+
+    Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
+    Output: true
+    
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2020/10/05/mat2.jpg)
+
+    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`
+*   `-104 <= matrix[i][j], target <= 104`
+
+https://leetcode.com/problems/search-a-2d-matrix/
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diff --git a/0074_search-a-2d-matrix/python3/solution.py b/0074_search-a-2d-matrix/python3/solution.py
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+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        '''
+        Two binary searches, first across rows and then
+        across columns to quickly find the target
+        '''
+        
+        mid = -1
+        start, end = 0, len(matrix) - 1
+        
+        # Figure out which row the number belongs to
+        while start <= end:
+            mid = (start + end) // 2
+            
+            if target > matrix[mid][-1]:
+                start = mid + 1
+            elif target < matrix[mid][0]:
+                end = mid - 1
+            else:
+                break
+                
+        # If we never managed to find a row, that means number
+        # doesn't exist
+        if not start <= end:
+            return False
+        
+        # The row to search
+        row = matrix[mid]
+        start, end = 0, len(row) - 1
+        
+        while start <= end:
+            mid = (start + end) // 2
+            
+            if target == row[mid]:
+                return True
+            elif target > row[mid]:
+                start = mid + 1
+            else:
+                end = mid - 1
+        
+        return False
+        
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