4. Find Minimum in Rotated Sorted Array

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.

Approach

• First if condition means you are in a sorted window so no need to check further as the min would be the left one just find min with the solution and break
• You would have basically two sorted array inside of sort so we just need to see where we stand

class Solution {
    public int findMin(int[] nums) {
        int l = 0, r = nums.length - 1;
        int min = nums[0];
 
        while (l <= r) {
            if (nums[l] < nums[r]) {
                min = Math.min(min, nums[l]);
                break;
            }
 
            int m = l + (r - l) / 2;
            min = Math.min(min, nums[m]);
 
            if (nums[m] >= nums[l]) { // = req check [2,1]
                l = m + 1;
            } else {
                r = m - 1;
            }
        }
 
        return min;
    }
}