Consider two people moving in opposite direction with speeds U meters/second and V meters/second respectively. The task is to find how long will take to make the distance between them X meters.
Examples:
Input: U = 3, V = 3, X = 3
Output: 0.5
After 0.5 seconds, policeman A will be at distance 1.5 meters
and policeman B will be at distance 1.5 meters in the opposite direction
The distance between the two policemen is 1.5 + 1.5 = 3
Input: U = 5, V = 2, X = 4
Output: 0.571429
Approach: It can be solved using distance = speed * time. Here, distance would be equal to the given range i.e. distance = X and speed would be the sum of the two speeds because they are moving in the opposite direction i.e. speed = U + V.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the time for which// the two policemen can communicatedouble getTime(int u, int v, int x){ double speed = u + v; // time = distance / speed double time = x / speed; return time;}// Driver codeint main(){ double u = 3, v = 3, x = 3; cout << getTime(u, v, x); return 0;} |
Java
// Java implementation of the approachclass GFG { // Function to return the time for which // the two policemen can communicate static double getTime(int u, int v, int x) { double speed = u + v; // time = distance / speed double time = x / speed; return time; } // Driver code public static void main(String[] args) { int u = 3, v = 3, x = 3; System.out.println(getTime(u, v, x)); }}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 implementation of the approach # Function to return the time # for which the two policemen # can communicate def getTime(u, v, x): speed = u + v # time = distance / speed time = x / speed return time # Driver code if __name__ == "__main__": u, v, x = 3, 3, 3 print(getTime(u, v, x)) # This code is contributed# by Rituraj Jain |
C#
// C# implementation of the approachusing System;class GFG { // Function to return the time for which // the two policemen can communicate static double getTime(int u, int v, int x) { double speed = u + v; // time = distance / speed double time = x / speed; return time; } // Driver code public static void Main() { int u = 3, v = 3, x = 3; Console.WriteLine(getTime(u, v, x)); }}// This code is contributed// by Akanksha Rai |
PHP
<?php// PHP implementation of the approach// Function to return the time for which// the two policemen can communicatefunction getTime($u, $v, $x){ $speed = $u + $v; // time = distance / speed $time = $x / $speed; return $time;}// Driver code$u = 3; $v = 3; $x = 3;echo getTime($u, $v, $x);// This code is contributed // by Akanksha Rai?> |
Javascript
<script>// JavaScript implementation of the approach// Function to return the time for which// the two policemen can communicatefunction getTime(u, v, x){ let speed = u + v; // time = distance / speed let time = x / speed; return time;}// Driver code let u = 3, v = 3, x = 3; document.write(getTime(u, v, x));// This code is contributed by Surbhi Tyagi.</script> |
0.5
Time Complexity: O(1)
Auxiliary Space: O(1)
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