diff --git a/median-of-two-sorted-arrays/README.md b/median-of-two-sorted-arrays/README.md
new file mode 100644
index 0000000..e49c3c4
--- /dev/null
+++ b/median-of-two-sorted-arrays/README.md
@@ -0,0 +1,31 @@
+# median-of-two-sorted-arrays
+
+## Two Pointers
+
+When building a merge sort algorithm, the last step is to merge two sorted arrays - just like with this problem (at least for the brute force soluton).
+One interesting thing about this is that we can keep track of how far into the final sorted array we are during this 'merge' step.
+
+For example, merging `[1, 2, 3]` and `[2, 7, 9]`, we can determine which index of the final array we are at.
+
+```
+[1] => 1
+[1, 2] => 2
+[1, 2, 2] => 3
+[1, 2, 2, 3] => 4
+[1, 2, 2, 3, 7] => 5
+[1, 2, 2, 3, 7, 9] => 6
+```
+
+Given this, we can also determine at what point we have 'reached' the middle of the final array.
+Given this, we can extract the middle value(s) from the final array, iterating just like we do during a merge sort, but only until we reach the middle.
+
+In the example above, we reeach the middle at index 3 - and given half the final array's sie is 3 - that's the index we want to stop at.
+
+The actual implementation is with two pointers - the index we're currently iterating over in each array.
+We keep track of the last two values we've seen, and when we reach the middle, we can determine if we need to return one or two values.
+
+> For a real implementation, you could setup two different algorithms, one for even-size final arrays, and one for odd. This would remove a couple instructions fo the loop, but given that the Big O is the same - it's not worth it for Leetcode.
+
+## Binary Search
+
+TODO: Writeup
\ No newline at end of file
diff --git a/median-of-two-sorted-arrays/src/main.rs b/median-of-two-sorted-arrays/src/bin/two-pointers.rs
similarity index 100%
rename from median-of-two-sorted-arrays/src/main.rs
rename to median-of-two-sorted-arrays/src/bin/two-pointers.rs