My solution for 1460

problems/1460/jeremymanning.md

@@ -1,11 +1,38 @@
 # [Problem 1460: Make Two Arrays Equal by Reversing Subarrays](https://leetcode.com/problems/make-two-arrays-equal-by-reversing-subarrays/description/?envType=daily-question)
 
 ## Initial thoughts (stream-of-consciousness)
+- There are two ways to solve this:
+    - We could sort both arrays and then compare each corresponding element in turn.  This would take $O(n \log n)$ time.
+    - We could count the number of occurances of each unique element, store in a hash table, and ensure that the keys/values all match.  This would take $O(n)$ time (i.e., faster).  So let's go with this approach.
 
 ## Refining the problem, round 2 thoughts
+- No special cases to deal with...
 
 ## Attempted solution(s)
 ```python
-class Solution:  # paste your code here!
-    ...
+class Solution:
+    def canBeEqual(self, target: List[int], arr: List[int]) -> bool:
+        target_count = {}
+        arr_count = {}
+        for t, a in zip(target, arr):
+            if t in target_count:
+                target_count[t] += 1
+            else:
+                target_count[t] = 1
+
+            if a in arr_count:
+                arr_count[a] += 1
+            else:
+                arr_count[a] = 1
+
+        for t, t_count in target_count.items():
+            if (t not in arr_count) or (target_count[t] != arr_count[t]):
+                return False
+
+        return True
 ```
+- Given test cases pass; submitting...
+
+![Screenshot 2024-08-02 at 8 12 24 PM](https://github.com/user-attachments/assets/0d1fd1fe-1937-4d81-8a2e-d55f3ce42dab)
+
+Solved!