findMinimumInRotatedSortedArray.ts

/*

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.

*/

function findMin(nums: number[]): number {
	let left = 0;
	let right = nums.length - 1;
	let mid = Math.floor((left + right) / 2);

	while (left < right) {
		if (nums[left] < nums[right]) {
			return nums[left];
		}

    // If mid is greater than right, the minimum is on the right
    // If mid is less than right, the minimum is on the left
		if (nums[mid] > nums[right]) {
			left = mid + 1;
		} else {
			right = mid;
		}

		mid = Math.floor((left + right) / 2);
	}

	return nums[left];
}

function findMax(nums: number[]): number {
  let left = 0;
  let right = nums.length - 1;
  let mid = Math.floor((left + right) / 2);

  while (left < right) {
    if (nums[left] < nums[right]) {
      return nums[right];
    }

    // If mid is greater than left, the maximum is on the right
    // If mid is less than left, the maximum is on the left
    if (nums[mid] >= nums[left]) {
      left = mid;
    } else {
      right = mid - 1;
    }

    mid = Math.floor((left + right) / 2);
  }

  return nums[left];
}

console.log(findMin([17, 13, 15, 16]));