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# Number of days after which tank will become empty

• Difficulty Level : Medium
• Last Updated : 29 Apr, 2021

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 using namespace std; // Utility method to get sum of first n numbersint getCumulateSum(int n){    return (n * (n + 1)) / 2;} // Method returns minimum number of days after// which tank will become emptyint 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 methodsint 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 emptypublic 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 numbersdef getCumulateSum(n):     return int((n * (n + 1)) / 2)  # Method returns minimum number of days# after  which tank will become emptydef 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 codeC, l = 5, 2print(minDaysToEmpty(C, l)) # This code is contributed by Smitha Dinesh Semwal.

## C#

 // C# code to find number// of days after which// tank will become emptyusing 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: 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 emptyint 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 methodsint 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 emptyimport java.lang.*;class GFG {     // Method returns minimum number of days// after which tank will become emptystatic 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 codepublic 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 emptyimport math # Method returns minimum number of days # after which tank will become emptydef 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 codeprint(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 emptyusing System; class GFG{     // Method returns minimum// number of days after// which tank will become emptystatic 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 codestatic 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 codeecho 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.
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