Recover Countersignature Number with airSlate SignNow

Recover countersignature number

hey everyone welcome back and let's write some more neat code today so today let's solve the problem missing number and this is another problem from the blind 75 list a list of commonly code questions we've been tracking this is missing number in the binary category you'll see why in a moment and the link to this sheet will be in the description if you do want to take a look we have solved most of the problems in this list so far so let's continue so the thing i like about this problem is that there's so many different solutions to this problem even though it's an easy problem i think it's a good problem to just understand a lot of the different solutions involved with this problem you can definitely learn a lot so we're given an array of numbers containing n distinct numbers in the range 0 to n even though this range actually has n plus 1 numbers inside it we're only choosing n distinct numbers from this range and we want to return the only number that's missing from this array so we want to find the missing number from this range that that we did not pick there are many different ways to solve this problem the follow-up is can we implement a solution with o of one extra space and o of n run time complexity that's going to be the challenge and actually there's two different ways we can arrive at such a solution i'll show you both of them because they're both pretty interesting okay so in the first example we're given three distinct numbers because the length of this is three so the range that we're looking at is from zero to three right there's four different numbers in this range zero one two and three but we've only chosen three of them in our input we want to find the one that's missing so we can kind of just go through this input right three okay the missing number is not three zero the missing number is not zero one the missing number is not one so the only remaining number is two so two is the missing number that was a pretty easy way to solve the problem what exactly did we do right now well you could have you could think about this list of numbers as basically a hash set or hash map or whatever you want and we're basically iterating through the list of numbers and then crossing out the ones that are already uh and contained in the input list so basically we used o of n extra memory with a hash set or map and o of n time complexity because we had to iterate through the input array so that's actually the most simple solution right using hashing to do this problem but it does take of n extra memory the follow-up was can we do this in o of one...