Given a tank with capacity C liters which is completely filled in starting. Everyday tank is filled with l liters of water and in the case of overflow extra water is thrown out. Now on i-th day i liters of water is taken out for drinking. We need to find out the day at which tank will become empty the first time.

### Manual Synopsis

Input : Capacity = 5

l = 2

Output : 4

At the start of 1st day, water in tank = 5

and at the end of the 1st day = (5 – 1) = 4

At the start of 2nd day, water in tank = 4 + 2 = 6

but tank capacity is 5 so water = 5

and at the end of the 2nd day = (5 – 2) = 3

At the start of 3rd day, water in tank = 3 + 2 = 5

and at the end of the 3rd day = (5 – 3) = 2

At the start of 4th day, water in tank = 2 + 2 = 4

and at the end of the 4th day = (4 – 4) = 0

So final answer will be 4

### Code

```
# Python3 code to find number of days
# after which tank will become empty
# Utility method to get
# sum of first n numbers
def getCumulateSum(n):
return int((n * (n + 1)) / 2)
# Method returns minimum number of days
# after which tank will become empty
def minDaysToEmpty(C, l):
# if water filling is more than
# capacity then after C days only
# tank will become empty
if (C <= l) : return C
# initialize binary search variable
lo, hi = 0, 1e4
# loop until low is less than high
while (lo < hi):
mid = int((lo + hi) / 2)
# if cumulate sum is greater than (C - l)
# then search on left side
if (getCumulateSum(mid) >= (C - l)):
hi = mid
# if (C - l) is more then
# search on right side
else:
lo = mid + 1
# Final answer will be obtained by
# adding l to binary search result
return (l + lo)
# Driver code
C, l = 5, 2
print(minDaysToEmpty(C, l))
# This code is contributed by Smitha Dinesh Semwal.
```

We can see that tank will be full for starting (l + 1) days because water taken out is less than water being filled. After that, each day water in the tank will be decreased by 1 more liter and on (l + 1 + i)th day (C – (i)(i + 1) / 2) liter water will remain before taking drinking water.

Now we need to find a minimal day (l + 1 + K), in which even after filling the tank by l liters we have water less than l in tank i.e. on (l + 1 + K – 1)th day tank becomes empty so our goal is to find minimum K such that,

C – K(K + 1) / 2 <= l

We can solve above equation using binary search and then (l + K) will be our answer. Total time complexity of solution will be O(log C)

**Output:**

4

**Alternate Solution :**

It can be solved mathematically with a simple formula:

Let’s Assume C>L. Let d be the amount of days after the L*th* when the tank become empty.During that time, there will be **(d-1)**refills and **d** withdrawals.

Hence we need to solve this equation :

Sum of all withdrawals is a sum of arithmetic progression,therefore :

Discriminant = 1+8(C-L)>0,because C>L.

Skipping the negative root, we get the following formula:

Therefore, the final alwer is:

```
# Python3 code to find number of days
# after which tank will become empty
import math
# Method returns minimum number of days
# after which tank will become empty
def minDaysToEmpty(C, l):
if (l >= C): return C
eq_root = (math.sqrt(1 + 8 * (C - l)) - 1) / 2
return math.ceil(eq_root) + l
# Driver code
print(minDaysToEmpty(5, 2))
print(minDaysToEmpty(6514683, 4965))
# This code is contributed by Smitha Dinesh Semwal.
```

Thanks to **Andrey Khayrutdinov** for suggesting this solution.

This article is contributed by **Utkarsh Trivedi**. If you like Geekycodes and would like to contribute, you can also write an article using or mail your article to here.

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