Can you explain your code

You are awesome!

Can this be any simpler !!! :)

You are awesome! Thank you! :)

It's work but why u use -t+1 as in every case at last t will be 1

splendid :)

Wow. I knew I was overcomplicating things. You did in 5 lines what took me about 50. Very elegant. Beautiful.

I came up with the same solution in C++ and I was getting time outs on several test cases

Does it pass all test cases ? I doubt

really very nice logic

Simply amazing man...and here i was stuck trying to solve it in such a complex way...Kudos !

this is awesome!!

Getting timeout for TC - 4, 5, 6, 8

It is not Big-O (log n). It will not pass all test case

stupid man!!!! its giving wrong answer.

my solution was identical save in another progamming language.

That is amazing... I always wish to come up with a solution like this..

awesome

my solution is the same. you didnt miss anything people with more complex solutions are using a hammer and looking at it like a nail. thus missing the simpler solution.

what about t=10

You are really good. #Respect.

Great Solution

You are great!

I did the exact same thing in python. Glad someone else was on the same wavelength.

It did take me a second try since initially, I did a serial subtraction of t, but that was orders slower and forced me to think about the number pattern.

A bit easy for a difficulty of 30, but I'm not complaining!

Awesome :)

Your solution is good, works in many languages, but int Java 7 it times out for some test cases, so in that language a faster solution is needed. Seems to me that the timeout period may have been too short because your solution is O(log(n)).

I made the same approach.

:-)

static long strangeCounter(long t) { /* * Write your code here. */ // System.out.println("value of t"+t); // long rem=t%3; // long count=t/3; long stable=0; long temp=3; for(long i=1;i<=t;i=i+temp){ if(i==1){ continue; } else{ temp=temp*2; }

     stable=i;
    }
    stable=t-stable;
    if(t==1){
        temp=3;
    }
    else{
    temp=temp-stable;
        }
   return temp;
}

SIMPLY AWESOME

The solution is working but I can't catch the logic for the solution

That was my first solution, but it times out in C#

That is awesome !!!

some testcases went timed out !

This is much more efficient way:

function strangeCounter($t) {
    
    $scnt = 3;
    
    while ($t+2 >= $scnt)
        $scnt *= 2;
    
    $scnt /= 2;
    
    return $scnt-($t - $scnt+2);
    
}

never thought of doing it this way. My approach was a lot longer;

WOW

In Python :

t=int(input())
b=3
summ=4
if t>3:
    while summ<=t:
        a=summ
        b=b*2
        summ=a+b
    print(summ-t)
else:
    print(4-t)

Genius level intellect ! Coming to a conclusion to there is a pattern is fine. But to execute it with such ease, nice

Thanks You, I got this solution. I didn't think deep enough

static long strangeCounter(long t) {
    int i = 1;
    int rate = 3;
    while (i + rate <= t) {
        i = i + rate;
        rate = rate * 2;
    }
    return rate - t + i;
}

After Improve my code from your code

static long strangeCounter(long t) {
    int rate = 3;
    while (t > rate) {
        t = t - rate;
        rate = rate * 2;
    }
    return rate - t + 1;
}

I still cannot pass the time cases

Amazing!

C++ solution :

#include<iostream>
using namespace std;

int main()
{
   long bound=3, t; cin >> t;
   while(t>bound)
   {
     t-=bound;
     bound*=2;
   }
   cout << (bound-t+1);
}

your approach to solving this problem is very simple and catchy...each time the loop executes, the value of t decreases...leading to reverse pattern of respective times in stop watch and in actual as soon as the loop ends...

your logic is so awesome

can you explain how this solution is in log n

How can you think this approch.

genius...

Its very optimize. good

how u guys think like this??am still not getting it even after seeing this solution?? :( :,(

Here is my JS approach, very similar:

    let timer = 3;

    while(t > timer){
        timer = timer * 2 + 3;
    }
    return timer - (t-1);

Had the same solution but in C#. For some reason I still get runtime errors. Switched to Python using my same solution and it passed just fine. :/

It didn't work because of timeOut problem in few test cases. WHY ???

How you think like this man

Hey c, c++ guys out there ! Use long long int to store values of integers related to t if you are going through the above code logic.

It can be done in O(1) like this

long strangeCounter(long t) {

long n = ceil(log((t+3)/3.0)/log(2));
return 6*pow(2, n-1) - t -2; 

}

it does not pass all testcases in cpp

that's awesomesauce!!

It took me a lot of time to figure out how to calculate the first value of the colum. Just that it should be log2((t-1)/3 + 1).

This was really helpful. Thanks!

its not o(1) its log(n)

but too much math

my solution is passing 7 test but its fail for 4 tests ..in those input were ig no so i hv used long bt its not working

Yes it is a nice problem..Good work to brain and simple if we understand the logic...

Could you please explain

You can just calculate the answer without the while loop:

   val cycleLength = (Math.pow(2, log((t.toLong + 2) / 3, 2)) * 3).toLong;

   val answer = cycleLength + cycleLength - t - 2;

Note that the log function, here, truncates to only the integer part (i.e., returns the floor of the log base 2).

At the start of each cycle, the value is 3 times some power of 2, and that value is t plus 2. Then each successive value decrements the starting value by one, until it reaches one, then a new cycle starts.

cool

Don't post solutions here.

Such a Genius. Would like to get advice from you about improving logical skills.

it fails.

bro can you give me any idea how devlop my algorithmic skill.

One of the most accurate hints.

You can get a one-liner closed form expression, ugly as hell, but confirming the previous comment.

result = 6 * Pow(2, Floor(Log2((t+2)/3))) - 2 - t