Day 18: finish explanation

Author: thatguyintech

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