Find First and Last Position of Element in Sorted Array

[Question] Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value. If target is not found in the array, return [-1, -1].You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

// TC: O(2log(N))
// SC: O(1)

func findFirstAndLastOccurance(_ arr: [Int], _ targetElement: Int) -> [Int] {
    
    func findFirstOccurance() ->Int {
        let arrayCount = arr.count
        var low = 0
        var high = arrayCount - 1
        var idx = -1
        
        while low <= high {
            let mid = (low + high) / 2
            if arr[mid] >= targetElement {
                high = mid - 1;
            } else{
                low = mid + 1;
            }
            if arr[mid] == targetElement {
                idx = mid
            }
        }
        return idx
    }
    
    func findLastOccurance() ->Int {
        let arrayCount = arr.count
        var low = 0
        var high = arrayCount - 1
        var idx = -1
        
        while low <= high {
            let mid = (low + high) / 2
            if arr[mid] <= targetElement {
                low = mid + 1;
            } else{
                high = mid - 1;
            }
            if arr[mid] == targetElement {
                idx = mid
            }
        }
        return idx
    }

    return [findFirstOccurance(), findLastOccurance()]
}

var inputBinArray = [5,7,7,8,8,10]
let targetElem = 8
let opFirstAndLast = findFirstAndLastOccurance(inputBinArray, targetElem)
print("opFirstAndLast --- ", opFirstAndLast)// 3, 4

We can also solve a Question like Count Occurrences of a number in Sorted Array

// Question 7 : Count occurrence in Sorted array.
let ipCountArray = [2, 4, 6, 8, 8, 8, 11, 13]
let opCountOccurance = findFirstAndLastOccurance(ipCountArray, 8)
print("opCountOccurance---",(opCountOccurance.last! - opCountOccurance.first!) + 1 ) // 3