Find Minimum in Rotated Sorted Array

Problem Statement

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

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.

Examples

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] 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 (no rotation effect).

Intuition

In a rotated sorted array:

• One half is always sorted
• The minimum is at the rotation point (where sorted order breaks)
• If nums[mid] > nums[right]: minimum is in right half (rotation point is right)
• If nums[mid] ≤ nums[right]: minimum is in left half or at mid

Key insight: Compare mid with right endpoint to determine which half contains the minimum.

Pattern Recognition

This is Modified Binary Search with these characteristics:

• Sorted array with rotation
• Find specific property (minimum value)
• O(log n) requirement
• Decision based on comparing mid with boundaries

Approach

1. Initialize: left = 0, right = n - 1

2. Binary Search:

   • If nums[left] < nums[right]: Array not rotated, return nums[left]
   • Calculate mid = left + (right - left) / 2
   • If nums[mid] > nums[right]:
     • Rotation point is in right half
     • left = mid + 1
   • Else:
     • Minimum could be at mid or in left half
     • right = mid

3. Return: nums[left] (minimum element)

Edge Cases

• Array not rotated (already sorted)
• Single element
• Two elements
• Rotated by n times (back to original)

Complexity Analysis

• Time Complexity: O(log n)
  • Binary search, eliminating half each iteration
• Space Complexity: O(1)
  • Only using pointers