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# Problem

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

## example 1

 ``````1 2 `````` ``````Input: [1,3,5,6], 5 Output: 2 ``````

## example 2

 ``````1 2 `````` ``````Input: [1,3,5,6], 2 Output: 1 ``````

## example 3

 ``````1 2 `````` ``````Input: [1,3,5,6], 7 Output: 4 ``````

## example 4

 ``````1 2 `````` ``````Input: [1,3,5,6], 0 Output: 0 ``````

# Solution

Using binary search with additional :

• return `0` if target less then `nums[0]`
• return `len(nums)` if target grater then `nums[len(nums)-1]`
• where usually return `-1` ( not exist element) – return `left`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 `````` ``````func searchInsert(nums []int, target int) int { left := 0 right:=len(nums)-1 if target>nums[right] { return len(nums) } if target < nums[left] { return 0 } for left<=right { med := (left+right)/2 if nums[med] == target { return med } if nums[med]>target { right=med-1 continue } if nums[med]