153. Find Minimum In Rotated Sorted Array¶

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153. Find Minimum in Rotated Sorted Array

Medium

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.

Solution¶

class Solution {
    public int findMin(int[] arr) {
        int n = arr.length;
        int low = 0, high = n - 1, ans = Integer.MAX_VALUE;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[low] <= arr[high]) {
                ans = Math.min(ans, arr[low]);
                break;
            }
            else if (arr[low] <= arr[mid]) {
                ans = Math.min(ans, arr[low]);
                low = mid + 1;
            }
            else {
                ans = Math.min(ans, arr[mid]);
                high = mid - 1;
            }
        }
        return ans; 
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here