Merge pull request #13 from alberthu16/day18

Day 18: search matrix pt2 in ruby!!

Author: Albert Hu

day18/README.md

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+question of the day: https://leetcode.com/problems/search-a-2d-matrix-ii/#/description
+
+Write an efficient algorithm that searches for a value in an `m` x `n`
+matrix. This matrix has the following properties:
+
+Integers in each row are sorted in ascending from left to right.
+
+Integers in each column are sorted in ascending from top to bottom.
+
+For example,
+
+Consider the following matrix:
+
+```ruby
+[
+  [1,   4,  7, 11, 15],
+  [2,   5,  8, 12, 19],
+  [3,   6,  9, 16, 22],
+  [10, 13, 14, 17, 24],
+  [18, 21, 23, 26, 30]
+]
+```
+
+Given *target* = `5`, return `true`.  
+Given *target* = `20`, return `false`.  
+
+## Ideas
+
+Brute force solution: just iterate through every coordinate in the
+matrix and compare value by value to see if any match. Return `true`
+when a match is found, or `false` if the whole matrix has been searched
+without finding a match. This is an `O(rows*columns)` runtime solution,
+but doesn't take advantage of the existing ordering.
+
+Now, unlike yesterday's problem where in addition to the elements
+in each row being ordered in increasing order, the same property held
+across rows. The matices in this problem are not like that. The
+elements in each column are also increasing order now. This means I can't
+do a binary search across all the elements in the matrix as easily.
+
+Something I can still do right off the bat, is a binary search within
+each row. For example, in the example matrix, if I'm going to look for 
+`13`, I can start in the middle of the first row, then search the right half
+(since `7` < `13`), then search the right half again (since `11` < `13`) and
+finally give up on the row once I see `15` is the last option. I'd
+move down to the next row and repeat the binary search. This solution
+is a slight improvement, especially on larger matrices. It'd have a
+`O(rows*log(columns))` runtime. However, I'm still not using the
+increasing-down-columns property to my advantage.
+
+At this moment, I can't think of a better way to solve this problem.
+
+I'll just implement the binary search by rows solution for now.
+
+## Code
+
+[Ruby](./searchMatrix2.rb)
+
+## 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.
+
+@apengwin says there's a known `O(m + n)` solution to this problem and
+that I should "Think about using the sorted-ness of the matrix to my
+advantage". 
+
+

day18/searchMatrix2.rb

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+def searchMatrix(matrix, target)
+  # the downside to this is that I do a binary search
+  # on every row before I 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
+
+  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()