# 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: Please solve it on “PRACTICE ” first, before moving on to the solution.

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

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:

Therefore, the final alwer is:

## 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  ?>

Output :

4
8573


Thanks to Andrey Khayrutdinov for suggesting this solution.

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