Peak index in a mountain array

Go

Binary search

Time complexity: $O(log(n))$ - n is a length of an array
Auxiliary space: $O(1)$ - constant amount of space

func peakIndexInMountainArray(nums []int) int {
    left, right := 0, len(nums)

    var mid int

    for left <= right {
        mid = left + (right - left) / 2
        
        if nums[mid] > nums[mid + 1] && nums[mid] > nums[mid - 1]{
            return mid
        }

        if nums[mid] > nums[mid + 1] {
            right = mid - 1
        } else { 
            left = mid + 1
        }
    }
    
    return mid
}

Array iteration

Time complexity: $O(n)$ - n is a length of an array
Auxiliary space: $O(1)$ - constant amount of space

func peakIndexInMountainArray(nums []int) int {

    for i, value := range nums {
        if i == 0 {
            continue
        } else if value > nums[i-1] && value > nums[i+1] {
            return i
        }
    }

    return 0
}

Typescript

Max number (using standard library)

Time complexity: $O(log(n))$ - n is a length of an array
Auxiliary space: $O(1)$ - constant amount of space

function peakIndexInMountainArray(nums: number[]): number {
  return nums.indexOf(Math.max(...nums));
}

thoughtbook

Peak index in a mountain array

Go

Binary search

Array iteration

Typescript

Max number (using standard library)