153. Find Minimum in Rotated Sorted Array

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153. Find Minimum in Rotated Sorted Array

Description

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

  • [4,5,6,7,0,1,2] if it was rotated 4 times.
  • [0,1,2,4,5,6,7] 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 of unique elements, return the minimum element of this array.

You must write an algorithm that runs inO(log n) time.

Example 1:

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Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

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Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

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Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times.

Constraints:

  • n == nums.length
  • 1 <= n <= 5000
  • -5000 <= nums[i] <= 5000
  • All the integers of nums are unique .
  • nums is sorted and rotated between 1 and n times.

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = 0, right = nums.size(), end = nums[right - 1];
        // the checking range: [left, right)
        // nums[right, n) is always less than end
        // nums[0, left - 1] is always bigger than end
        while (left < right) {
            int mid = (left + right) / 2;
            if (nums[mid] > end) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return nums[right];
    }
};