Problem Link: https://codeforces.com/edu/course/2/lesson/6/2/practice/contest/283932/problem/A
Accepted Code
Not Accepted Code
The only change in the two code is the value being assigned to l, when i normally implement binary search i use it like in the not accepted one but why in this problem its giving wrong answer?
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Lets have this set were — is not fit and + is fit - — — — + + + l = -1 and r = 7 int the first time in while loop: m =3 and l became 4 you notice that l is in a fit position and r in a fit position and that is not acceptable , we need only r to be in + positions and l to be in — position. Then , the while loop continue working and your code retrun 5 not 4 because of that we set l to m not to m+1
While implementing the BS the way you have done you need to use: if(isfit(l)) cout<<l<<endl; else cout<<r<<endl; because you are storing the ans in r, when r = mid and l = mid + 1 and l satisfies the condition it will not get updated into r because of the condition "while(r-l>1)" so at the end you have to put an extra check.
This will get AC: https://ide.geeksforgeeks.org/online-cpp-compiler/c8c64c48-58eb-424c-9bf5-5eb5752e69a0
The way BS is implemented in this code is better because you have to print only that variable which is storing the ans: https://ide.geeksforgeeks.org/online-cpp-compiler/fce3d222-6532-49d1-8fcb-67c65c2fa39a
First code finds point where isfit changes 'sign', if isfit(l) before running BS was = false you don't need to check it anymore, answer is always r. Second code is kind of mix with no clear idea behind it. It should work after changing to
while r>l, but 1st is just a lot better concept.btw
mid = (l+r)/2saves some keystrokes.