Intuition:
The array is originally sorted but rotated at some pivot. Because of this rotation, one half of the array will always be sorted, and the minimum element lies at the point where the order breaks. We can use binary search to narrow down the location of this minimum efficiently.

Instead of scanning every element, we compare the middle element with the rightmost element to decide which half contains the minimum value.

Step-by-step thinking:

Initialize two pointers: left at the start and right at the end of the array.
If the array is empty, return 0.
While left is less than right:

Calculate the middle index mid.
If nums[mid] is greater than nums[right], the minimum must be in the right half.
Move left to mid + 1.
Otherwise, the minimum is in the left half (including mid).
Move right to mid.

When the loop ends, left points to the minimum element.
Return nums[left].

This works because each comparison eliminates half of the remaining search space, just like a standard binary search.

Time Complexity: O(log n)
Space Complexity: O(1)

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74. Find Minimum in Rotated Sorted Array

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