feat(work): add solution for Find Minimum in Rotated Sorted Array

work/Medium/Array/153.find-minimum-in-rotated-sorted-array.py

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+"""
+153. Find Minimum in Rotated Sorted Array
+Difficulty: Medium
+https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/
+
+──────────────────────────────────────────────────
+
+Suppose an array of length n sorted in ascending order is rotated
+between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7]
+might become:
+
+        • [4,5,6,7,0,1,2] if it was rotated 4 times.
+
+        • [0,1,2,4,5,6,7] 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 of unique elements, return the
+minimum element of this array.
+
+You must write an algorithm that runs in O(log n) time.
+
+
+
+Example 1:
+
+Input: nums = [3,4,5,1,2]
+Output: 1
+Explanation: The original array was [1,2,3,4,5] rotated 3 times.
+
+Example 2:
+
+Input: nums = [4,5,6,7,0,1,2]
+Output: 0
+Explanation: The original array was [0,1,2,4,5,6,7] and it was
+rotated 4 times.
+
+Example 3:
+
+Input: nums = [11,13,15,17]
+Output: 11
+Explanation: The original array was [11,13,15,17] and it was rotated
+4 times.
+
+
+
+Constraints:
+
+        • n == nums.length
+
+        • 1 <= n <= 5000
+
+        • -5000 <= nums[i] <= 5000
+
+        • All the integers of nums are unique.
+
+        • nums is sorted and rotated between 1 and n times.
+"""
+
+
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        l, r = 0, len(nums) - 1
+        lowest_index = -1
+
+        while l <= r:
+            m = (l + r) // 2
+            if nums[m] <= nums[-1]:
+                lowest_index = m
+                r = m - 1
+            else:
+                l = m + 1
+
+        return nums[lowest_index]