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.

**Examples:**

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

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)

## C++

// C/C++ code to find number of days after which // tank will become empty #include <bits/stdc++.h> using namespace std; // Utility method to get sum of first n numbers int getCumulateSum(int n) { return (n * (n + 1)) / 2; } // Method returns minimum number of days after // which tank will become empty int minDaysToEmpty(int C, int 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 int lo = 0; int hi = 1e4; int mid; // loop until low is less than high while (lo < hi) { mid = (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 to test above methods int main() { int C = 5; int l = 2; cout << minDaysToEmpty(C, l) << endl; return 0; }

## Java

// Java code to find number of days after which // tank will become empty public class Tank_Empty { // Utility method to get sum of first n numbers static int getCumulateSum(int n) { return (n * (n + 1)) / 2; } // Method returns minimum number of days after // which tank will become empty static int minDaysToEmpty(int C, int 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 int lo = 0; int hi = (int)1e4; int mid; // loop until low is less than high while (lo < hi) { mid = (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 to test above methods public static void main(String args[]) { int C = 5; int l = 2; System.out.println(minDaysToEmpty(C, l)); } } // This code is contributed by Sumit Ghosh

## Python3

# 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.

## C#

// C# code to find number // of days after which // tank will become empty using System; class GFG { // Utility method to get // sum of first n numbers static int getCumulateSum(int n) { return (n * (n + 1)) / 2; } // Method returns minimum // number of days after // which tank will become empty static int minDaysToEmpty(int C, int 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 int lo = 0; int hi = (int)1e4; int mid; // loop until low is // less than high while (lo < hi) { mid = (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 static public void Main () { int C = 5; int l = 2; Console.WriteLine(minDaysToEmpty(C, l)); } } // This code is contributed by ajit

**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:

## C++

// C/C++ code to find number of days after which // tank will become empty #include <bits/stdc++.h> using namespace std; // Method returns minimum number of days after // which tank will become empty int minDaysToEmpty(int C, int l) { if (l >= C) return C; double eq_root = (std::sqrt(1+8*(C-l)) - 1) / 2; return std::ceil(eq_root) + l; } // Driver code to test above methods int main() { cout << minDaysToEmpty(5, 2) << endl; cout << minDaysToEmpty(6514683, 4965) << endl; return 0; }

## Java

// Java code to find number of days // after which tank will become empty import java.lang.*; class GFG { // Method returns minimum number of days // after which tank will become empty static int minDaysToEmpty(int C, int l) { if (l >= C) return C; double eq_root = (Math.sqrt(1 + 8 * (C - l)) - 1) / 2; return (int)(Math.ceil(eq_root) + l); } // Driver code public static void main(String[] args) { System.out.println(minDaysToEmpty(5, 2)); System.out.println(minDaysToEmpty(6514683, 4965)); } } // This code is contributed by Smitha Dinesh Semwal.

## Python3

# 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.

## C#

// C# code to find number // of days after which // tank will become empty using System; class GFG { // Method returns minimum // number of days after // which tank will become empty static int minDaysToEmpty(int C, int l) { if (l >= C) return C; double eq_root = (Math.Sqrt(1 + 8 * (C - l)) - 1) / 2; return (int)(Math.Ceiling(eq_root) + l); } // Driver code static public void Main () { Console.WriteLine(minDaysToEmpty(5, 2)); Console.WriteLine(minDaysToEmpty(6514683, 4965)); } } // This code is contributed by ajit

## PHP

<?php // PHP code to find number // of days after which // tank will become empty // Method returns minimum // number of days after // which tank will become empty function minDaysToEmpty($C, $l) { if ($l >= $C) return $C; $eq_root = (int)sqrt(1 + 8 * ($C - $l) - 1) / 2; return ceil($eq_root) + $l; } // Driver code echo minDaysToEmpty(5, 2), "\n"; echo minDaysToEmpty(6514683, 4965), "\n"; // This code is contributed // by akt_mit ?>

**Output :**

4 8573

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

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