<\/span><\/h2>\nTo efficiently find the minimum value in a rotated sorted array, we can adapt the binary search algorithm. This approach leverages the fact that the array is still partially sorted, even after the rotation.<\/p>\n
1. We start by setting a left pointer to the array’s first element and a right pointer to the last element.
\n2. While the left pointer is not equal to the right pointer, we calculate the middle index using the formula `mid = left + (right – left) \/\/ 2`.
\n3. We compare the middle element with the rightmost element:
\n a. If the middle element is greater than the rightmost element, then the minimum value must be present somewhere to the right of the middle. We update the left pointer to `mid + 1`.
\n b. If the middle element is smaller than the rightmost element, then the minimum value must be present somewhere to the left of the middle or it may be the middle element itself. We update the right pointer to `mid`.
\n4. The search space is halved in each iteration, and eventually, when the left and right pointers meet, we have found the minimum value.<\/p>\n
The minimum value in a rotated sorted array can be found by using a binary search approach like the one described above.<\/b><\/p>\n