Update 17-00153-find-minimum-in-rotated-sorted-array.py

Author: leetlab11

05-Binary Search/17-00153-find-minimum-in-rotated-sorted-array.py

@@ -1,4 +1,47 @@
+# 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.
+
+# --------------------------------------------------------------------------------------------------------------------------------------
+
+# array is rotated but still sorted, so we can use the sorted property.
+# there are 2 sorted arrays- left and right
+# find m by taking the midpoint of l and m(floor division)
+# when we are in left array, try to search right as that might have smaller elements;
+# so when m >= l that means we are in left array; so l = m+1(we are trying to move right)
+# when we are in right array, try to search left as the left side in right array will have smaller elements
+# so when m < l that means we are in right array; so r = m-1(we are trying to move left)
+# when we are in an array that is sorted, this algorithm won't work, so search for the leftmost element and compare with res
+# if that l is smaller than res, that is the o/p and the loop will break
 
 class Solution:
     def findMin(self, nums: List[int]) -> int:
@@ -19,3 +62,28 @@ def findMin(self, nums: List[int]) -> int:
                 r = m - 1
 
         return res
+
+Time Complexity = O(logn)
+Space Complexity = O(1)
+
+
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+
+