Find Minimum in Rotated Sorted Array II LeetCode Solution

Problem – Find Minimum in Rotated Sorted Array II LeetCode Solution

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.

Follow up: This problem is similar to Find Minimum in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

Find Minimum in Rotated Sorted Array II LeetCode Solution in Python

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

Find Minimum in Rotated Sorted Array II LeetCode Solution in Java

public int findMin(int[] nums) {
	 int l = 0, r = nums.length-1;
	 while (l < r) {
		 int mid = (l + r) / 2;
		 if (nums[mid] < nums[r]) {
			 r = mid;
		 } else if (nums[mid] > nums[r]){
			 l = mid + 1;
		 } else {  
			 r--;  //nums[mid]=nums[r] no idea, but we can eliminate nums[r];
		 }
	 }
	 return nums[l];
}

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

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