LeetCode //C - 154. Find Minimum in Rotated Sorted Array II

154. Find Minimum in Rotated Sorted Array II

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, a1, a2, ..., an-1] 1 time results in the array a\[n-1, a0, a1, a2, ..., an-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 <= numsi <= 5000
• nums is sorted and rotated between 1 and n times.

From: LeetCode

Link: 154. Find Minimum in Rotated Sorted Array II

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Solution:

Ideas:

1. Initialization: Set two pointers, left and right, at the beginning and end of the array, respectively.

2. While Loop: Continue searching as long as left is less than right.

3. Middle Element: Calculate the middle position mid between left and right.

4. Decision Tree:

• If numsmid is greater than numsright, the minimum is in the right half (excluding mid), so move left to mid + 1.
• If numsmid is less than numsright, the minimum could be mid or to the left of mid, so move right to mid.
• If numsmid equals numsright, reduce right by one to gradually eliminate duplicates without skipping the minimum.

5. Conclusion: Once left equals right, the minimum element is found, as the search space is narrowed down to a single element.

Code:

int findMin(int* nums, int numsSize) {
    int left = 0, right = numsSize - 1;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] > nums[right]) {
            left = mid + 1;
        } else if (nums[mid] < nums[right]) {
            right = mid;
        } else { // nums[mid] == nums[right]
            right--;
        }
    }
    return nums[left];
}