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Does this video make a mistake or am I misunderstanding the recursive call?

In the example, we have an array `[0,1,2,3,4,5,6,7,8]` with a length of 9 and I set a target of 8. We check the list is not empty (`true`) and then set our midpoint equal to `Math.floor(list.length / 2)` which will evaluate to 4 - correct?

4 is not equal to 8 and 4 is less than our target 8. As a result we call `recursiveBinarySearch(list, target)` again with two arguments - our new list and target of 8. Pasan says our new list is has 3 values, [6,7,8] but does our list not in fact have 4 values [5,6,7,8] as we remove elements [0,1,2,3,4]? I am a little confused and wonder if this is a mistake or I have misunderstood something.

I think both approaches still work. For example, Pasan explains we calculate the new length of [6,7,8] which is equal to 3 and subsequently divide 3 by 2 = 1.5. We then floor 1.5 which means is 1. Our `midpoint = 1`. This will select the element at index 1 which is [6, 7, 8].

With my approach I calculate the new length of [5, 6, 7, 8] which is equal to 4 and divide 4 by 2 = 2. There is nothing to floor in this example. Our `midpoint = 2`. This will select the element at index 2 which is still 7. [5, 6, 7, 8].

Have I missed something here?

Thank you for calling this out with such detail. I had the same issue, but you covered it excellently Kit Howlett

Hey Kit Howlett you are correct! The graphic running of the code clearly has the list `[5, 6, 7, 8]` as the first recursive call as shown by the highlighted numbers. The audio however says this list contains "6, 7, and 8".
The cause may be due to some version confusion between showing the code implementation (where the example data is 1-based `[1, 2, 3, 4, 5, 6, 7, 8]` — length 8) and the graphic example (where the example data is 0-based `[0, 1, 2, 3, 4, 5, 6, 7, 8]` — length 9).