# Number of days after which tank will become empty

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
Recommended Practice

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  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

## Javascript

 

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 : 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:  ## C++

 // C/C++ code to find number of days after which // tank will become empty #include  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

 = $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 ?>

## Javascript

 

Output :

4
8573

Thanks to Andrey Khayrutdinov for suggesting this solution.
If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.