Find if the glass will be empty or not when the rate of drinking is given

Given a cylinder glass with diameter equals D centimeters. The initial level of water in the glass equals H centimeters from the bottom. You drink the water with a speed of M milliliters per second. But if you do not drink the water from the glass, the level of water increases by N centimeters per second. The task is to find the time required to make the glass empty or find if it is possible to make the glass empty or not.

Examples:

Input: D = 1, H = 1, M = 1, N = 1
Output: 3.65979

Input: D = 1, H = 2, M = 3, N = 100
Output: -1

Approach: This is a Geometry question. It is known that the area of the glass is pie * r2 where r represents the radius i.e. (D / 2). So to find the rate at which the water is being consumed per second, divide the volume given (it is known that 1 milliliter equals 1 cubic centimeter) with the area.
If the value is less than the rate at which the water is being poured in the glass, if it is not being drunk then the answer will be No else the glass can be empty.
To find the time, divide h / (v / (pie * r2) – e) and that is the time when the glass will become empty.



Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
double pie = 3.1415926535897;
  
// Function to return the time when
// the glass will be empty
double findsolution(double d, double h,
                    double m, double n)
{
    double k = (4 * m) / (pie * d * d);
  
    // Check the condition when the
    // glass will never be empty
    if (n > k)
        return -1;
  
    // Find the time
    double ans = (h / (k - n));
    return ans;
}
  
// Driver code
int main()
{
    double d = 1, h = 1, m = 1, n = 1;
  
    cout << findsolution(d, h, m, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
  
class GFG
{
  
static double pie = 3.1415926535897;
  
// Function to return the time when
// the glass will be empty
static double findsolution(double d, double h,
                    double m, double n)
{
    double k = (4 * m) / (pie * d * d);
  
    // Check the condition when the
    // glass will never be empty
    if (n > k)
        return -1;
  
    // Find the time
    double ans = (h / (k - n));
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    double d = 1, h = 1, m = 1, n = 1;
  
    System.out.printf("%.5f",findsolution(d, h, m, n));
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 implementation of the approach
pie = 3.1415926535897
  
# Function to return the time when
# the glass will be empty
def findsolution(d, h, m, n):
  
    k = (4 * m) / (pie * d * d)
  
    # Check the condition when the
    # glass will never be empty
    if (n > k):
        return -1
  
    # Find the time
    ans = (h / (k - n))
    return round(ans, 5)
  
# Driver code
d = 1
h = 1
m = 1
n = 1
  
print(findsolution(d, h, m, n))
  
# This code is contributed by Mohit Kumar

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
static double pie = 3.1415926535897;
  
// Function to return the time when
// the glass will be empty
static double findsolution(double d, double h,
                           double m, double n)
{
    double k = (4 * m) / (pie * d * d);
  
    // Check the condition when the
    // glass will never be empty
    if (n > k)
        return -1;
  
    // Find the time
    double ans = (h / (k - n));
    return ans;
}
  
// Driver code
public static void Main(String[] args)
{
    double d = 1, h = 1, m = 1, n = 1;
  
    Console.Write("{0:F5}"
            findsolution(d, h, m, n));
}
}
  
// This code is contributed by 29AjayKumar

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Output:

3.65979

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