Water Tank

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 Lth when the tank become empty.During that time, there will be (d-1)refills and d withdrawals.
Hence we need to solve this equation :
C+(d-1)*L=\sum_{i=1}^{d} Withdrawal_i

Sum of all withdrawals is a sum of arithmetic progression,therefore :
C+(d-1)*L=\frac {L+1+L+d}{2}*2
2C +2dL - 2L=2dL+d +d^2
d^2+d-2(C-L)=0

Discriminant = 1+8(C-L)>0,because C>L.
Skipping the negative root, we get the following formula:
d=-1 +\frac { \sqrt{1-8(C-L)} }{2}
Therefore, the final alwer is:
min,days=L +ceil\left (  \frac { \sqrt{1-8(C-L)}-1 }{2}\right )

# 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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