154.FindMinimumInRotatedSortedArrayIi.cpp

/*
 * @lc app=leetcode id=154 lang=cpp
 *
 * [154] Find Minimum in Rotated Sorted Array II
 *
 * https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/description/
 *
 * algorithms
 * Hard (42.00%)
 * Likes:    1464
 * Dislikes: 271
 * Total Accepted:    238.2K
 * Total Submissions: 566.3K
 * Testcase Example:  '[1,3,5]'
 *
 * 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.
 *
 *
 * 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 is the same as Find Minimum in Rotated Sorted Array but with
 * duplicates. Would allow duplicates affect the run-time complexity? How and
 * why?
 */

// @lc code=start
class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = 0;
        int right = nums.size() - 1;
        while (left < right) {
            // move the left boundary if its equal to right boundary
            while (left < right && nums[left] == nums[right]) {
                left++;
            }

            // if the array is already sorted or there's only 1 element, break
            if (left == right || nums[left] < nums[right]) {
                break;
            }

            int mid = (left + right) / 2;
            if (nums[mid] >= nums[left]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        return nums[left];
    }
};
// @lc code=end