LeetCode #153 — MEDIUM

Find Minimum in Rotated Sorted Array

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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.

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[3,4,5,1,2]

Example 2

[4,5,6,7,0,1,2]

Example 3

[11,13,15,17]

Core Insight

What unlocks the optimal approach

Array was originally in ascending order. Now that the array is rotated, there would be a point in the array where there is a small deflection from the increasing sequence. eg. The array would be something like [4, 5, 6, 7, 0, 1, 2].
You can divide the search space into two and see which direction to go. Can you think of an algorithm which has O(logN) search complexity?
<ol> <li>All the elements to the left of inflection point > first element of the array.</li> <li>All the elements to the right of inflection point < first element of the array.</li> <ol>

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #153: Find Minimum in Rotated Sorted Array
class Solution {
    public int findMin(int[] nums) {
        int n = nums.length;
        if (nums[0] <= nums[n - 1]) {
            return nums[0];
        }
        int left = 0, right = n - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (nums[0] <= nums[mid]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return nums[left];
    }
}

// Accepted solution for LeetCode #153: Find Minimum in Rotated Sorted Array
func findMin(nums []int) int {
	n := len(nums)
	if nums[0] <= nums[n-1] {
		return nums[0]
	}
	left, right := 0, n-1
	for left < right {
		mid := (left + right) >> 1
		if nums[0] <= nums[mid] {
			left = mid + 1
		} else {
			right = mid
		}
	}
	return nums[left]
}

# Accepted solution for LeetCode #153: Find Minimum in Rotated Sorted Array
class Solution:
    def findMin(self, nums: List[int]) -> int:
        if nums[0] <= nums[-1]:
            return nums[0]
        left, right = 0, len(nums) - 1
        while left < right:
            mid = (left + right) >> 1
            if nums[0] <= nums[mid]:
                left = mid + 1
            else:
                right = mid
        return nums[left]

// Accepted solution for LeetCode #153: Find Minimum in Rotated Sorted Array
impl Solution {
    pub fn find_min(nums: Vec<i32>) -> i32 {
        let mut left = 0;
        let mut right = nums.len() - 1;
        while left < right {
            let mid = left + (right - left) / 2;
            if nums[mid] > nums[right] {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        nums[left]
    }
}

// Accepted solution for LeetCode #153: Find Minimum in Rotated Sorted Array
function findMin(nums: number[]): number {
    let left = 0;
    let right = nums.length - 1;
    while (left < right) {
        const mid = (left + right) >>> 1;
        if (nums[mid] > nums[right]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return nums[left];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.