Create 581_shortest_unsorted_continuous_subarray.md

Author: Joseph Luce

leetcode/easy/581_shortest_unsorted_continuous_subarray.md

@@ -0,0 +1,42 @@
+# 581. Shortest Unsorted Continuous Subarray
+
+## Best solution
+- Runtime: O(N)
+- Space: O(1)
+- N = Number of elements in array
+
+The idea is quite simple. Start from left to right, find the first unsorted index, then from right to left, find the last unsorted index.
+From then, you have three subarrays, the left sub-array, the middle sub-array, and the right sub-array. 
+The middle sub-array is obviously your known set of unsorted numbers but you do not know if your middle sub-array can be further extended to the left or the right.
+For example, [1,2,3,4,1,2,1,2,3], the first pass through will only identify index 3 and index 6 as the middle sub-array. 
+However, you need a second pass through to identify index 1 to index 8 by checking if the numbers of the left or right sub-arrays are in the middle sub-array.
+
+```
+class Solution:
+    def findUnsortedSubarray(self, nums: List[int]) -> int:
+        first = last = None
+        for index, n in enumerate(nums[:-1]):
+            if n > nums[index+1]:
+                first = index
+                last = index+1
+                break
+        for index, n in reversed(list(enumerate(nums))):
+            if index > 0 and n < nums[index-1]:
+                last = index
+                break
+        if first is not None:
+            MIN, MAX = min(nums[first:last+1]), max(nums[first:last+1])
+            # check outer sub-arrays for similar numbers
+            for index, n in enumerate(nums[:first]):
+                if n in range(MIN+1, MAX+1):
+                    first = index
+                    break
+            for index, n in reversed(list(enumerate(nums))):
+                if index <= last: # don't iterate into known unsorted array
+                    break
+                if n in range(MIN, MAX):
+                    last = index
+                    break
+            return last-first+1
+        return 0
+```