154. Find Minimum in Rotated Sorted Array II

Problem

Suppose an array sorted in ascending order 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.

Example 1:

Input: [1,3,5]
Output: 1

Example 2:

Input: [2,2,2,0,1]
Output: 0

Note:

• This is a follow up problem to Find Minimum in Rotated Sorted Array.
• Would allow duplicates affect the run-time complexity? How and why?

Solution (Java)

class Solution {
    public int findMin(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        return find(0, nums.length - 1, nums);
    }

    private int find(int left, int right, int[] nums) {
        if (left + 1 >= right) {
            return Math.min(nums[left], nums[right]);
        }
        int mid = left + (right - left) / 2;
        if (nums[left] == nums[right] && nums[left] == nums[mid]) {
            return Math.min(find(left, mid, nums), find(mid, right, nums));
        }
        if (nums[left] >= nums[right]) {
            if (nums[mid] >= nums[left]) {
                return find(mid, right, nums);
            } else {
                return find(left, mid, nums);
            }
        } else {
            return find(left, mid, nums);
        }
    }
}

Solution (Javascript)

/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function(nums) {
  var left = 0;
  var right = nums.length - 1;
  var mid = 0;
  while (left < right) {
    mid = Math.floor((left + right) / 2);
    if (nums[mid] > nums[right]) {
      left = mid + 1;
    } else if (nums[mid] < nums[right]) {
      right = mid;
    } else {
      right--;
    }
  }
  return nums[left];
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).