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In this post, you will find the solution for the **Find Minimum in Rotated Sorted Array II** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Find Minimum in Rotated Sorted Array II– LeetCode Problem

Find Minimum in Rotated Sorted Array II– LeetCode Problem

**Problem:**

Suppose an array of length `n`

sorted in ascending order is **rotated** between `1`

and `n`

times. For example, the array `nums = [0,1,4,4,5,6,7]`

might become:

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

if it was rotated`4`

times.`[0,1,4,4,5,6,7]`

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`

that may contain **duplicates**, return *the minimum element of this array*.

You must decrease the overall operation steps as much as possible.

**Example 1:**

Input:nums = [1,3,5]Output:1

**Example 2:**

Input:nums = [2,2,2,0,1]Output:0

**Constraints:**

`n == nums.length`

`1 <= n <= 5000`

`-5000 <= nums[i] <= 5000`

`nums`

is sorted and rotated between`1`

and`n`

times.

Find Minimum in Rotated Sorted Array II– LeetCode Solutions

Find Minimum in Rotated Sorted Array II Solution in C++:

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

Find Minimum in Rotated Sorted Array II Solution in Java:

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

Find Minimum in Rotated Sorted Array II Solution in Python:

class Solution: def findMin(self, nums: List[int]) -> int: l = 0 r = len(nums) - 1 while l < r: m = (l + r) // 2 if nums[m] == nums[r]: r -= 1 elif nums[m] < nums[r]: r = m else: l = m + 1 return nums[l]