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

Output:

4

**Alternate Solution : **

It can be solved mathematically with a simple formula:

// 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; }

Output :

4 8573

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

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