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Number of jump required of given length to reach a point of form (d, 0) from origin in 2D plane
  • Difficulty Level : Medium
  • Last Updated : 09 May, 2018

Given three positive integers a, b and d. You are currently at origin (0, 0) on infinite 2D coordinate plane. You are allowed to jump on any point in the 2D plane at euclidean distance either equal to a or b from your current position. The task is to find the minimum number of jump required to reach (d, 0) from (0, 0).

Examples:

Input : a = 2, b = 3, d = 1 
Output : 2
First jump of length a = 2, (0, 0) -> (1/2, √15/2)
Second jump of length a = 2, (1/2, √15/2) -> (1, 0)
Thus, only two jump are required to reach 
(1, 0) from (0, 0).

Input : a = 3, b = 4, d = 11 
Output : 3
(0, 0) -> (4, 0) using length b = 4
(4, 0) -> (8, 0) using length b = 4
(8, 0) -> (11, 0) using length a = 3

First, observe we don’t need to find the intermediate points, we only want the minimum number of jumps required.
So, from (0, 0) we will move in direction of (d, 0) i.e vertically with the jump of length equal to max(a, b) until the Euclidean distance between the current position coordinate and (d, 0) either became less than max(a, b) or equal to 0. This will increase the minimum number of jumps required by 1 at each jump. So this value can be found by floor(d/max(a, b)).

Now, for the rest of the distance, if (d, 0) is a Euclidean distance away, we will make one jump of length a and reach the (d, 0). So, the minimum number of jumps required will be increased by 1 in this case.



Now, let’s solve the case if rest of the distance is not equal to a. Let the rest of distance left be x. Also, observe x will be greater than 0 and less than max(a, b), therefore, 0 < x < 2*max(a, b). We can reach to (d, 0) in 2 steps.
How ?
Lets try to build a triangle ABC such that A is our current position, B is target position i.e (d, 0) and C will be the point such that AC = BC = max(a, b). And this is possible to triangle because sum of two side AC + BC is greater than third side AB. So, one jump is from A to C and another jump from C to B.

Below is the implementation of this approach:

C++




#include <bits/stdc++.h>
using namespace std;
  
// Return the minimum jump of length either a or b
// required to reach (d, 0) from (0, 0).
int minJumps(int a, int b, int d)
{
    // Assigning maximum of a and b to b
    // and assigning minimum of a and b to a.
    int temp = a;
    a = min(a, b);
    b = max(temp, b);
  
    // if d is greater than or equal to b.
    if (d >= b)
        return (d + b - 1) / b;
  
    // if d is 0
    if (d == 0)
        return 0;
  
    // if d is equal to a.
    if (d == a)
        return 1;
  
    // else make triangle, and only 2 
    // steps required.
    return 2;
}
  
int main()
{
    int a = 3, b = 4, d = 11;
    cout << minJumps(a, b, d) << endl;
    return 0;
}


Java




// Java code to find the minimum number
// of jump required to reach 
// (d, 0) from (0, 0).
import java.io.*;
  
class GFG {
  
    // Return the minimum jump of length either a or b
    // required to reach (d, 0) from (0, 0).
    static int minJumps(int a, int b, int d)
    {
        // Assigning maximum of a and b to b
        // and assigning minimum of a and b to a.
        int temp = a;
        a = Math.min(a, b);
        b = Math.max(temp, b);
  
        // if d is greater than or equal to b.
        if (d >= b)
            return (d + b - 1) / b;
  
        // if d is 0
        if (d == 0)
            return 0;
  
        // if d is equal to a.
        if (d == a)
            return 1;
  
        // else make triangle, and only 2
        // steps required.
        return 2;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int a = 3, b = 4, d = 11;
        System.out.println(minJumps(a, b, d));
    }
}
  
// This code is contributed by vt_m


Python3




# Python3 code to find the minimum 
# number of jump required to reach 
# (d, 0) from (0, 0)
  
def minJumps(a, b, d):
      
    temp = a
    a = min(a, b)
    b = max(temp, b)
      
    if (d >= b):
        return (d + b - 1) / b
      
    # if d is 0
    if (d == 0):
        return 0
  
    # if d is equal to a.
    if (d == a):
        return 1
  
    # else make triangle, and 
    # only 2  steps required.
    return 2
          
# Driver Code
a, b, d = 3, 4, 11
print (int(minJumps(a, b, d)))
  
# This code is contributed by _omg


C#




// C# code to find the minimum number
// of jump required to reach 
// (d, 0) from (0, 0).
using System;
  
class GFG {
  
    // Return the minimum jump of length either a or b
    // required to reach (d, 0) from (0, 0).
    static int minJumps(int a, int b, int d)
    {
        // Assigning maximum of a and b to b
        // and assigning minimum of a and b to a.
        int temp = a;
        a = Math.Min(a, b);
        b = Math.Max(temp, b);
  
        // if d is greater than or equal to b.
        if (d >= b)
            return (d + b - 1) / b;
  
        // if d is 0
        if (d == 0)
            return 0;
  
        // if d is equal to a.
        if (d == a)
            return 1;
  
        // else make triangle, and only 2
        // steps required.
        return 2;
    }
      
    // Driver code
    public static void Main()
    {
        int a = 3, b = 4, d = 11;
        Console.WriteLine(minJumps(a, b, d));
    }
}
  
// This code is contributed by vt_m


PHP




<?php
// PHP code to find the minimum
// number of jump required to
// reach (d, 0) from (0, 0).
  
// Return the minimum jump 
// of length either a or b
// required to reach (d, 0)
// from (0, 0).
function minJumps( $a, $b, $d)
{
      
    // Assigning maximum of
    // a and b to b and
    // assigning minimum of 
    // a and b to a.
    $temp = $a;
    $a = min($a, $b);
    $b = max($temp, $b);
  
    // if d is greater than
    // or equal to b.
    if ($d >= $b)
        return ($d + $b - 1) / $b;
  
    // if d is 0
    if ($d == 0)
        return 0;
  
    // if d is equal to a.
    if ($d == $a)
        return 1;
  
    // else make triangle,
    // and only 2 
    // steps required.
    return 2;
}
  
    // Driver Code
    $a = 3;
    $b = 4; 
    $d = 11;
    echo floor(minJumps($a, $b, $d));
      
// This code is contributed by anuj_67.
?>



Output

3

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