#154: Find Minimum in Rotated Sorted Array II

Author: papilo-cloud

Modified_binary_search/find_min_II.py

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+# Suppose an array of length n sorted in ascending order is rotated between 1 and n times.
+# For example, the array nums = [0,1,4,4,5,6,7] might become:
+
+# [4,5,6,7,0,1,4] if it was rotated 4 times.
+# [0,1,4,4,5,6,7] if it was rotated 7 times.
+# Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array
+# [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
+
+# Given the sorted rotated array nums that may contain duplicates, return the minimum element
+# of this array.
+
+# You must decrease the overall operation steps as much as possible.
+
+
+# Example 1:
+
+# Input: nums = [1,3,5]
+# Output: 1
+# Example 2:
+
+# Input: nums = [2,2,2,0,1]
+# Output: 0
+ 
+
+# Constraints:
+
+# n == nums.length
+# 1 <= n <= 5000
+# -5000 <= nums[i] <= 5000
+
+
+from typing import List
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        l = 0
+        r = len(nums) - 1
+
+        while l < r:
+            m = (l+r)//2
+            if nums[m] == nums[r]:
+                r -= 1
+            elif nums[m] < nums[r]:
+                r = m
+            else:
+                l += 1
+        return nums[l]