Follow upfor "Find Minimum in Rotated Sorted Array":

What ifduplicatesare allowed?Would this affect the run-time complexity? How and why?

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e.,

`0 1 2 4 5 6 7`

might become `4 5 6 7 0 1 2`

).
Find the minimum element.

The array may contain duplicates.

**Understand the problem:**

The problem is similar to the search in rotated array II. The key is to handle when the duplicates exist. The only case where we need to handle specially is when nums[lo] == nums[mid] == nums[hi]. Because in this case, we have no idea which half to move. In this case, we move lo ++ .

**Code (Java):**

public class Solution { public int findMin(int[] nums) { if (nums == null || nums.length == 0) { return 0; } int lo = 0; int hi = nums.length - 1; while (lo + 1 < hi) { int mid = lo + (hi - lo) / 2; if (nums[lo] == nums[mid] && nums[mid] == nums[hi]) { lo++; } else if (nums[lo] < nums[mid] && nums[mid] <= nums[hi]) { return nums[lo]; } else if (nums[lo] <= nums[mid] && nums[mid] > nums[hi]) { lo = mid + 1; } else if (nums[lo] >= nums[mid] && nums[mid] <= nums[hi]) { hi = mid; } } if (nums[lo] <= nums[hi]) { return nums[lo]; } else { return nums[hi]; } } }

**Update on 10/12/15:**

public class Solution { public int findMin(int[] nums) { if (nums == null || nums.length == 0) { return 0; } int lo = 0; int hi = nums.length - 1; while (lo + 1 < hi) { int mid = lo + (hi - lo) / 2; if (nums[lo] == nums[mid]) { lo++; continue; } if (nums[lo] < nums[hi]) { return nums[lo]; } else if (nums[lo] < nums[mid]) { lo = mid; } else if (nums[mid] <= nums[hi]) { hi = mid; } } return Math.min(nums[lo], nums[hi]); } }

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