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Find Minimum in Rotated Sorted Array II⚓︎

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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.

Solution⚓︎

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class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = -1, right = nums.size() - 1;
        while (left + 1 < right && nums[right] == nums[0]) right--;
        if (nums[0] <= nums[right]) return nums[0];

        while (left + 1 < right) {
            int mid = left + right >> 1;
            if (nums[mid] >= nums[0]) left = mid;
            else right = mid;
        }

        return left == nums.size() - 1 ? nums[0] : nums[left + 1];
    }
};
  • Time complexity (worst): \(O(n)\);
  • Space complexity: \(O(1)\).