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79. Find Minimum in Rotated Sorted Array

mediumAsked at Plaid

Find the minimum element in a rotated sorted array. Plaid asks this because finding the rotation pivot (e.g., the day a circular billing window restarts) is the same primitive.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Plaid loops.

  • Glassdoor (2025)— Plaid SWE II OA.
  • LeetCode Discuss (2026)— Plaid rotation-pivot finder.

Problem

Suppose an array of length n sorted in ascending order is rotated between 1 and n 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.

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.

Examples

Example 1

Input
nums = [3,4,5,1,2]
Output
1

Example 2

Input
nums = [4,5,6,7,0,1,2]
Output
0

Example 3

Input
nums = [11,13,15,17]
Output
11

Approaches

1. Linear scan

Walk and track min.

Time
O(n)
Space
O(1)
function findMin(nums) { return Math.min(...nums); }

Tradeoff: Misses the O(log n) requirement.

2. Binary search for the pivot

Compare nums[mid] to nums[hi]. If nums[mid] > nums[hi], pivot is to the right of mid. Else, mid is in the same sorted half as hi, so pivot is at mid or to the left.

Time
O(log n)
Space
O(1)
function findMin(nums) {
  let lo = 0, hi = nums.length - 1;
  while (lo < hi) {
    const mid = (lo + hi) >> 1;
    if (nums[mid] > nums[hi]) lo = mid + 1;
    else hi = mid;
  }
  return nums[lo];
}

Tradeoff: Logarithmic. Comparing to nums[hi] (not nums[lo]) is the standard convention; comparing to nums[lo] requires an extra check for the un-rotated case.

Plaid-specific tips

Plaid grades this on whether you compare to nums[hi]. Bonus signal: walk through the un-rotated case — if no rotation, nums is fully sorted and nums[mid] <= nums[hi] always, so hi keeps shrinking and we land at index 0. Connect to billing-cycle pivots where the 'minimum' is the start of the current cycle.

Common mistakes

  • Comparing to nums[lo] — fails on un-rotated arrays without extra handling.
  • Using lo <= hi as the loop condition — leads to off-by-one because hi = mid (not mid - 1).
  • Forgetting that the array might not be rotated at all.

Follow-up questions

An interviewer at Plaid may pivot to one of these next:

  • With duplicates (LC 154) — O(n) worst case.
  • Find the rotation count.
  • Search for a target after finding the pivot (LC 33).

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Output

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FAQ

Why compare to nums[hi]?

It captures both rotated and un-rotated cases in one comparison. Comparing to nums[lo] would need a separate check for the un-rotated case.

Why hi = mid (not mid - 1)?

Because mid itself could be the minimum. Setting hi = mid keeps it as a candidate; lo = mid + 1 only when we've proven mid is not the answer.

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