Last updated on October 5th, 2024 at 09:09 pm

Here, We see ** Find Minimum in Rotated Sorted Array LeetCode Solution**. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

*List of all LeetCode Solution*

*List of all LeetCode Solution*

## Topics

Array, Binary Search

## Companies

Microsoft

## Level of Question

Medium

**Find Minimum in Rotated Sorted Array LeetCode Solution**

## Table of Contents

**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.

**1. Find Minimum in Rotated Sorted Array LeetCode Solution C++**

class Solution { public: int findMin(vector<int>& nums) { int n = nums.size(); int low=0, high=n-1; while(low<high){ if(nums[low] <= nums[high]) return nums[low]; int mid = low + (high-low)/2; if(nums[low] > nums[mid]){ high=mid; } else if(nums[mid] > nums[high]) { low=mid+1; } } if(nums[low] <= nums[high]) return nums[low]; return -1; } };

**2. Find Minimum in Rotated Sorted Array LeetCode Solution Java**

class Solution { public int findMin(int[] nums) { int l = 0; int r = nums.length - 1; while (l < r) { final int m = (l + r) / 2; if (nums[m] < nums[r]) r = m; else l = m + 1; } return nums[l]; } }

**3. Find Minimum in Rotated Sorted Array Solution JavaScript**

var findMin = function(nums) { let left = 0; let right = nums.length - 1; while (left < right) { let mid = left + Math.floor((right - left) / 2); if (nums[mid] > nums[right]) { left = mid + 1; } else { right = mid; } } return nums[left]; };

**4. Find Minimum in Rotated Sorted Array Solution Python**

class Solution(object): def findMin(self, nums): left = 0 right = len(nums) - 1 while left < right: mid = left + (right - left) / 2 if nums[mid] > nums[right]: left = mid + 1 else: right = mid return nums[left]