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.

We use a `modified` binary search to find the minimum element:-

- At every point, compare (‘i’, ‘mid’) with (‘mid+1’, ‘j’). Whichever section has the lower value, let’s move that direction.
if ( (a < c && a < d) || (b < c && b < d) ) // go left else // go right

- Return the last number remaining in the recursion (when i === j)

/** * @param {number[]} nums * @return {number} */ var findMin = function(nums) { return findMinRecursive(nums, 0, nums.length-1) }; var findMinRecursive = function(nums, i, j) { // one or two digits if ( i === j ) return nums[i]; let mid = i + Math.floor((j-i)/2); let a = nums[i]; let b = nums[mid]; let c = nums[mid+1]; let d = nums[j]; if ( (a < c && a < d) || (b < c && b < d) ) { // go left return findMinRecursive(nums, i, mid); } else { // go right return findMinRecursive(nums, mid+1, j); } } |