Given a sorted and rotated array numbers containing unique elements, find and return the minimum element in this array.
Rotating an array [a[0], a[1], a[2], ..., a[n-1]] once results in [a[n-1], a[0], a[1], a[2], ..., a[n-2]]. Rotating it a second time results in [a[n-2], a[n-1], a[0], a[1], ..., a[n-3]].
Develop an algorithm that runs in O(log n) time complexity.
numbers: number[]: An array of integersInput: numbers = [1,2,3,4]Output: 1Explanation: The original array [1,2,3,4] was rotated 0 times
Input: numbers = [3,4,1,2]Output: 1Explanation: The original array [1,2,3,4] was rotated 2 times
Input: numbers = [6,7,8,-5,-4,2]Output: -5Explanation: The original array [-5,-4,2,6,7,8] was rotated 3 times.
numbers.length <= 1000numbers[i] <= 10,000Given a sorted and rotated array numbers containing unique elements, find and return the minimum element in this array.
Rotating an array [a[0], a[1], a[2], ..., a[n-1]] once results in [a[n-1], a[0], a[1], a[2], ..., a[n-2]]. Rotating it a second time results in [a[n-2], a[n-1], a[0], a[1], ..., a[n-3]].
Develop an algorithm that runs in O(log n) time complexity.
numbers: number[]: An array of integersInput: numbers = [1,2,3,4]Output: 1Explanation: The original array [1,2,3,4] was rotated 0 times
Input: numbers = [3,4,1,2]Output: 1Explanation: The original array [1,2,3,4] was rotated 2 times
Input: numbers = [6,7,8,-5,-4,2]Output: -5Explanation: The original array [-5,-4,2,6,7,8] was rotated 3 times.
numbers.length <= 1000numbers[i] <= 10,000console.log() statements will appear here.