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++
// 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 emptydouble 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 codeint main(){ double d = 1, h = 1, m = 1, n = 1; cout << findsolution(d, h, m, n); return 0;} |
Java
// Java implementation of the approachclass GFG{static double pie = 3.1415926535897;// Function to return the time when// the glass will be emptystatic 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 codepublic 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 |
Python3
# Python3 implementation of the approachpie = 3.1415926535897# Function to return the time when# the glass will be emptydef 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 coded = 1h = 1m = 1n = 1print(findsolution(d, h, m, n))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing System;class GFG{static double pie = 3.1415926535897;// Function to return the time when// the glass will be emptystatic 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 codepublic 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 |
Javascript
<script>// JavaScript implementation of the approach const pie = 3.1415926535897; // Function to return the time when // the glass will be empty function findsolution(d , h , m , n) { var k = (4 * m) / (pie * d * d); // Check the condition when the // glass will never be empty if (n > k) return -1; // Find the time var ans = (h / (k - n)); return ans; } // Driver code var d = 1, h = 1, m = 1, n = 1; document.write(findsolution(d, h, m, n).toFixed(5));// This code contributed by Rajput-Ji </script> |
3.65979
Time Complexity: O(1)
Auxiliary Space: O(1)
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