## 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.

## Solution

## Explanation

Okay so the array is sorted

We have to find out where the given number would be inserted

this sounds like a binary search problem

Well brute force algorithm would be the easiest But it won’t be efficient

Binary search would give us `logn`

time

We can do some border cases before even starting our binary algorithm
ex: if target is less than first number from the array return 0
ex: if target is greater than last number from the array return `last index + 1`

I use this logic to get the `mid`

`const mid = Math.floor((high-low)/2 + low)`

I have used recursion here
But, you can use simple `while`

loop for the binary search
I believe that should give you a better runtime

### Ask questions like

Will the array contain negative numbers

Will there be duplicate the the array What would be the result in case of duplicate numbers

How big would be the array Will it fit in memory

## Sample test cases

## Highlights

**Runtime:** 92 ms, faster than 14.27% of JavaScript online submissions for Search Insert Position.

**Memory Usage:** 34 MB, less than 63.58% of JavaScript online submissions for Search Insert Position.

### Disclaimer

This may not be the optimal solution. And that’s okay. The purpose here is to practice problem solving and have fun with algorithms. I am constantly learning new optimized solutions for these problems.

Please comment below if you have a better solution. Let’s learn algorithms and data structures together.