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Find Minimum in Rotated Sorted Array LC #153 Medium Binary Search · Rotation Pivot
Problem

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. Given the rotated array nums (all unique elements), find and return the minimum element in O(log n) time.

Example 1
Input: nums = [3,4,5,1,2]
Output: 1
Original [1,2,3,4,5] rotated 3 times.
Example 2
Input: nums = [4,5,6,7,0,1,2]
Output: 0
Original [0,1,2,4,5,6,7] rotated 4 times.
Example 3
Input: nums = [11,13,15,17]
Output: 11
Not rotated — min is the first element.
Constraints: n == nums.length  |  1 ≤ n ≤ 5000  |  −5000 ≤ nums[i] ≤ 5000  |  All integers are unique.
Try Examples
Custom:
Rotated Array — Pivot Detector
State Variables
lo
—
hi
—
mid
—
nums[mid]
—
nums[hi]
—
Step Logic
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Algorithm
1
Init lo=0, hi=n-1. The minimum is somewhere in [lo..hi].
2
Compute mid = lo + (hi-lo)/2.
3
If nums[mid] > nums[hi]: drop is in right half → lo = mid+1.
4
Else: right half sorted or same → minimum is at mid or left → hi = mid.
5
When lo == hi: nums[lo] is the minimum.
Time
O(log n)
Space
O(1)
Why It Works

A rotated sorted array has exactly one inflection point — the rotation pivot — where values drop. Comparing nums[mid] to nums[hi] tells us which half contains the drop:

nums[mid] > nums[hi]: the left half is "normally sorted" but the big values are there. The drop must be to the right.

nums[mid] ≤ nums[hi]: the right half is sorted (or equal). The minimum is at mid or further left.

We never compare to nums[lo] because the pivot could be exactly at mid.