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

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

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

<script>

// Javascript code to find number
// of days after which
// tank will become empty

// Utility method to get
// sum of first n numbers
function getCumulateSum(n)
{
    return parseInt((n * (n + 1)) / 2, 10);
}

// Method returns minimum
// number of days after
// which tank will become empty
function 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
    let lo = 0;
    let hi = 1e4;
    let mid;

    // Loop until low is
    // less than high
    while (lo < hi)
    {
        mid = parseInt((lo + hi) / 2, 10);

        // 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
let C = 5;
let l = 2;

document.write(minDaysToEmpty(C, l));

// This code is contributed by rameshtravel07
</script>

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 answer 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;
}

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

<?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
?>

Javascript

<script>

// Javascript 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;

    let eq_root = (Math.sqrt(1 + 8 *
                            (C - l)) - 1) / 2;
    return (Math.ceil(eq_root) + l);
}

// Driver code
document.write(minDaysToEmpty(5, 2) + "</br>");
document.write(minDaysToEmpty(6514683, 4965));

// This code is contributed by suresh07

</script>

Output :

4
8573

Thanks to Andrey Khayrutdinov for suggesting this solution.
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