Find Minimum in Rotated Sorted Array II

Description

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?

Code

class Solution {
public:
    int findMin(vector<int>& nums) {
        int l = 0, r = nums.size() - 1;
 
        while(l < r) {
            int m = l + (r - l) / 2;
            if(nums[m] < nums[r]) {
            // the right half is sorted
                r = m;
            } else if(nums[m] > nums[r]) {
            // the left half is sorted
                l = m + 1;
            } else if(nums[m] == nums[r]) {
            // we have no clue
                r--;
            }
        }
 
        return nums[l];
    }
};

和 Find Minimum in Rotated Sorted Array 的差別？

以 0, 1, 2, 4, 4, 4, 4 為例：

可發現當 m < r 時，pivot 會在左半邊，因此 r = mid。當 m > r 時，pivot 會在右半邊，因此 l = mid + 1。剩下一種 case 就是當 m == r 時，因為可以重複，因此有可能出現 4, 4, 4, 4, 0, 4, 4 這種 case，pivot 就不一定在左半邊。

l        mid      r
0, 1, 2, 4, 4, 4, 4   l < m <= r

4, 0, 1, 2, 4, 4, 4   l > m < r
4, 4, 0, 1, 2, 4, 4   l > m < r
4, 4, 4, 0, 1, 2, 4   l > m < r

4, 4, 4, 4, 0, 1, 2   l <= m > r
2, 4, 4, 4, 4, 0, 1   l < m > r
1, 2, 4, 4, 4, 4, 0   l < m > r

Source

• Find Minimum in Rotated Sorted Array II - LeetCode