[LeetCode] Find Minimum in Rotated Sorted Array

As explained in the Solution tag, the key to solving this problem is to use invariants. We set two pointers: l for the left and r for the right. One key invariant is nums[l] > nums[r]. If this invariant does not hold, then we know the array has not been rotated and the minimum is justnums[l]. Otherwise, we reduce the search range by a half each time via comparisons betweennums[l], nums[r] and nums[(l + r) / 2], denoted as nums[mid]:

1. If nums[mid] > nums[r], then mid is in the first larger half and r is in the second smaller half. So the minimum will be right to mid, update l = mid + 1;
2. If nums[mid] <= nums[r], then the minimum is at least nums[mid] and elements right tomid cannot be the minimum. So update r = mid (note that it is not mid - 1 since midmay also be the index of the minimum).

When l == r or the invariant does not hold, we have found the answer, which is just nums[l].

Putting these togerther, we have the following codes.

---

C (0 ms)

int findMin(int* nums, int numsSize) {
    int l = 0, r = numsSize - 1;
    while (l < r && nums[l] > nums[r]) {
        int mid = (l & r) + ((l ^ r) >> 1);
        if (nums[mid] > nums[r]) l = mid + 1;
        else r = mid; 
    }
    return nums[l];
}

---

C++ (4 ms)

class Solution {
public:
    int findMin(vector<int>& nums) {
        int l = 0, r = nums.size() - 1;
        while (l < r && nums[l] > nums[r]) {
            int mid = (l & r) + ((l ^ r) >> 1); 
            if (nums[mid] > nums[r]) l = mid + 1;
            else r = mid;
        }
        return nums[l];
    }
};

---

Python (48 ms)

class Solution:
    # @param {integer[]} nums
    # @return {integer}
    def findMin(self, nums):
        l, r = 0, len(nums) - 1
        while l < r and nums[l] > nums[r]:
            mid = (l & r) + ((l ^ r) >> 1)
            if nums[mid] > nums[r]:
                l = mid + 1
            else:
                r = mid
        return nums[l]