Find the maximum number x for which the sum of x natural numbers don’t exceeds the litres of water available.
Given that the first bucket has a capacity of 1 litre, second bucket has capacity of 2 litres and so on.
So we just need the maximum number x so that 1+2+3+....+x <=n, where n is the litres of the water available. Thus, problem is a searching problem, we are just searching for the required x.
First approach is we can use linear search to find the value of x but it will exceed the time limit. So, we have to think of a more efficient searching approach.
Next approach is to use the binary search as it is very efficient and very well within the time limit of the problem.
Author’s solution can be found here.
Why can’t we just use the formula for Sum of first n natural numbers? (n*(n+1))/2 = N
This was mine, I don’t know why I was getting TLE.
I mean, I’m okay with getting a wrong answer (I just re-read the question and saw the line, “The bucket will be considered filled if it has atleast 1 litre in it.”) But why am I getting a TLE? Is the sqrt() function that heavy/slow?
for t in range(input()):
It is because the problem is not that easy. The time limit of the problem is very less. So you can’t use the linear approach. The problem was designed that way.
I based my solution on this editorial but I got a WA response.
Can someone point out the mistake in my code.
And I am trying to understand the approach used in the following solutions…
singh_8 and sayan009 but cant get it…
See in line 22, you should be checking for <=N (you did <N instead).
- in line 22’s “if” satisfies, i.e mid * (mid-1) / 2 <= N, then the result var should be set to mid-1 (you did mid instead).
Hope it helps.
I used the formula for sum of first n natural numbers and got a WA,
but a simple linear approach for summing up the first n natural numbers gave AC, instead of TLE.
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You can use the formula (n*(n+1))/2=N and convert them into a quadratic equation to solve the problem. You can check my solution for this problem link text
Here i use sheree dharacharya formula to find the solution. And it accepted in one go.