Day 18: finish ruby solution

Author: thatguyintech

day18/README.md

@@ -56,4 +56,11 @@ I'll just implement the binary search by rows solution for now.
 
 [Ruby](./searchMatrix2.rb)
 
-## Follow up
+## Follow up
+
+I can optimize this approach even further by checking which dimension
+is longer (rows or columns) and doing binary search along the larger
+valued dimension to maximize efficiency.
+
+I'll update this soluiton once I think of an even more optimal approach.
+

day18/searchMatrix2.rb

@@ -0,0 +1,58 @@
+def searchMatrix(matrix, target)
+  # the downside to this is that I do a binary search
+  # on every row before I return
+  return matrix.map {|row| binarySearch(row, target)}.any? {|result| result == true}
+end
+
+###########
+# Helpers #
+###########
+
+def binarySearch(row, target)
+  low, high = 0, row.size
+
+  while low < high  
+    mid = (low + high) / 2
+    if row[mid] == target
+      return true
+    elsif target > row[mid]
+      low = mid + 1
+    else
+      high = mid
+    end
+  end
+
+  return false
+end
+
+#########
+# Tests #
+#########
+
+class AssertionError < RuntimeError
+end
+
+def assert &block
+    raise AssertionError unless yield
+end
+
+def tests
+  matrix = [[1,2,3],
+            [4,5,6],
+            [7,8,9]]
+  assert { searchMatrix(matrix, 1) }
+  assert { searchMatrix(matrix, 2) }
+  assert { searchMatrix(matrix, 5) }
+  assert { searchMatrix(matrix, 6) }
+  assert { searchMatrix(matrix, 8) }
+  assert { searchMatrix(matrix, 9) }
+
+  assert { searchMatrix([[]], 0) == false }
+  assert { searchMatrix([[0]], 1) == false }
+  assert { searchMatrix([[1, 2]], 0) == false }
+  assert { searchMatrix([[1],[2],[3]], 0) == false }
+  assert { searchMatrix([[1],[2],[3]], 1) == true }
+  assert { searchMatrix([[1],[2],[3]], 3) == true }
+end
+
+tests()