154

154. Find Minimum in Rotated Sorted Array II (Hard)

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?

 

Companies:
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Related Topics:
Array, Binary Search

Similar Questions:

• Find Minimum in Rotated Sorted Array (Medium)

Solution 1. Binary Search (L < R)

// OJ: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/
// Author: github.com/lzl124631x
// Time: average O(logN), worst O(N)
// Space: O(1)
class Solution {
public:
    int findMin(vector<int>& A) {
        int L = 0, R = A.size() - 1;
        while (L < R) {
            int M = (L + R) / 2;
            if (A[M] > A[R]) L = M + 1;
            else if (A[M] < A[R]) R = M;
            else --R; // since `A[L]` might be the answer and `A[M]` is the same as `A[R]`, it's safe to --R.
        }
        return A[L];
    }
};