update solution: find-minimum-in-rotated-sorted-array

Author: dusunax

find-minimum-in-rotated-sorted-array/dusunax.py

@@ -1,15 +1,41 @@
 '''
 # 153. Find Minimum in Rotated Sorted Array
 
-## Time Complexity: O(n)
-- min() iterates through all elements to find the smallest one.
-
-## Space Complexity: O(1)
-- no extra space is used.
+> **why binary search works in a "rotated" sorted array?**  
+> rotated sorted array consists of **two sorted subarrays**, and the minimum value is the second sorted subarray's first element.  
+> so 👉 find the point that second sorted subarray starts.   
+>
+> - if nums[mid] > nums[right]? => the pivot point is in the right half.  
+> - if nums[mid] <= nums[right]? => the pivot point is in the left half.  
+> - loop until left and right are the same.
 '''
 class Solution:
-    def findMin(self, nums: List[int]) -> int:
+    '''
+    ## A. brute force(not a solution)
+    - TC: O(n)
+    - SC: O(1)
+    '''
+    def findMinBF(self, nums: List[int]) -> int:
         if len(nums) == 1:
             return nums[0]
 
-        return min(nums)
+        return min(nums) # check all elements
+
+    '''
+    ## B. binary search
+    - TC: O(log n)
+    - SC: O(1)
+    '''
+    def findMinBS(self, nums: List[int]) -> int:
+        left = 0
+        right = len(nums) - 1
+
+        while left < right:
+            mid = (left + right) // 2
+            
+            if nums[mid] > nums[right]:
+                left = mid + 1
+            else:
+                right = mid
+
+        return nums[left]