FindMinimumInRotatedSortedArray153.kt

package 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.
 */
// this linear solution in O(N)
fun findMin(nums: IntArray): Int {
    var min  = Int.MAX_VALUE
    nums.forEach { num->
        min = Math.min(min,num)
    }
    return min
}

// this  solution in O(logN)
fun findMin2(nums: IntArray): Int {
    if(nums.size==1) return nums[0]
    var left =0
    var right = nums.size-1
    while (left < right)
    {
        var midPoint = left +(right-left) / 2
        if(midPoint>0&&nums[midPoint]<nums[midPoint-1])
        {
            return nums[midPoint]
        }else if( nums[midPoint] >=nums[left] && nums[midPoint] >nums[right])
        {
            left = midPoint+1
        }else
        {
            right=midPoint-1
        }
    }
    return nums[left]
}