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Find Minimum in Rotated Sorted Array⚓︎

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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 = [0,1,2,4,5,6,7] 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 in \(O(\log n)\) time.

Example 1:

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

Example 2:

  • 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:

  • 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.

Solution⚓︎

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class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = -1, right = nums.size();
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] >= nums[0]) left = mid;
            else right = mid;
        }
        return left == nums.size() - 1 ? nums[0] : nums[left + 1];
    }
};

Another way of writing:

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class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = -1, right = nums.size() - 1;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < nums[nums.size() - 1]) right = mid;
            else left = mid;
        }
        return nums[right];
    }
};