Here, We see Find Minimum in Rotated Sorted Array II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.
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Find Minimum in Rotated Sorted Array II 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 = [0,1,4,4,5,6,7]
might become:
[4,5,6,7,0,1,4]
if it was rotated4
times.[0,1,4,4,5,6,7]
if it was rotated7
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
Find Minimum in Rotated Sorted Array II 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;
}
};
Code language: PHP (php)
Find Minimum in Rotated Sorted Array II 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];
}
}
Code language: PHP (php)
Find Minimum in Rotated Sorted Array II LeetCode Solution JavaScript
var findMin = function(nums) {
var left = 0,
right = nums.length - 1
while (left < right){
var mid = Math.floor((left + right)/2)
if (nums[mid] > nums[right]) left = mid + 1
else right = mid
}
return nums[left]
};
Code language: JavaScript (javascript)
Find Minimum in Rotated Sorted Array II Solution Python
class Solution:
def findMin(self, num):
first, last = 0, len(num) - 1
while first < last:
midpoint = (first + last) // 2
if num[midpoint] > num[last]:
first = midpoint + 1
else:
last = midpoint
return num[first]
Code language: HTML, XML (xml)