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; ` `} ` |

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## 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 ` |

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## 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. ` |

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## 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 ` |

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**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; ` `} ` |

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## 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. ` |

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## 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. ` |

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## 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 ` |

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## 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 ` `?> ` |

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**Output :**

4 8573

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

This article is contributed by **Utkarsh Trivedi**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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