Count of collisions at a point (X, Y)

Given a matrix arr[][] of size N * 3 such that each row consists of 3 properties defining an element, i.e. (x-coordinate, y-coordinate, speed), and two integers X and Y, the task is to count the number of possible pairs of collisions at the point (X, Y), if every element in the given array is moving in a straight line towards the given point (X, Y) with their respective speeds.
Examples:

Input: arr[] = [ [5, 12, 1], [16, 63, 5], [-10, 24, 2], [7, 24, 2], [-24, 7, 2] ], X = 0, Y = 0
Output: 4
Explanation:
Possible collisions are between the following pair of elements: (arr[0], arr[1]), (arr[0], arr[2]), (arr[1], arr[2]) and (arr[3], arr[4])
Hence, total collisions are 4.
Input: arr[] = [ [1, 42, 9] ], X = 1, Y = 1
Output: 0
Explanation:
Since a single point is present, there will be no collisions.
Hence, total collisions are 0.

Approach: The idea is to simply find the time at which every point will reach the given point (X, Y). If two points reach the origin at the same time, then they will surely collide. Now, to calculate the time taken by each point do the following:

For a given point (x, y) and speed S
Distance D is given by:
D = = ? ((x2 – x1)² + (y2 – y1)²) = ? ((x)² + (y)²)
Time = Distance/Speed, squaring on both sides to eliminate root from distance,
Time² = Distance²/Speed²

After calculating time² for each point, simply check which of them are same and count the number of collisions.
Below are the steps:

Traverse the array arr[].
Create a new array Time[] and for each point, calculate time and append to this array and sort it.
Traverse Time[] and count the number of elements which arrive the given point at the same time, using the count calculate the number of collisions at that time.
Finally, return the total count of collisions.

Below is the implementation of the above approach:

C++14

// C++14 program to implement
// the above approach
#include <bits/stdc++.h>

using namespace std;
 
// Function to find the count of
// possible pairs of collisions

int solve(vector<vector<int>> &D, int N, 

                           int X, int Y)
{

    // Stores the time at which

    // points reach the origin

    vector<double> T;
 
    // Calculate time for each point

    for(int i = 0; i < N; i++)

    {

        int x = D[i][0];

        int y = D[i][1];
 
        double speed = D[i][2];
 
        double time = ((x * x - X * X) +

                       (y * y - Y * Y)) /

                       (speed * speed);
 
        T.push_back(time);

    }
 
    // Sort the times

    sort(T.begin(), T.end());
 
    int i = 0;

    int total = 0;
 
    // Counting total collisions

    while (i < T.size() - 1)

    {

        // Count of elements arriving at

        // a given point at the same time

        int count = 1;
 
        while (i < T.size() - 1 and 

                T[i] == T[i + 1])

        {

            count += 1;

            i += 1;

        }
 
        total += (count * (count - 1)) / 2;

        i += 1;

    }

    return total;
}
 
// Driver Code

int main()
{

    int N = 5;
 
    // Given set of points with speed

    vector<vector<int>> D = { { 5, 12, 1 },

                              { 16, 63, 5 },

                              { -10, 24, 2 },

                              { 7, 24, 2 },

                              { -24, 7, 2 } };
 
    int X = 0, Y = 0;
 
    // Function call

    cout << (solve(D, N, X, Y));
 
    return 0;
}
 
// This code is contributed by mohit kumar 29

Java

// Java program to implement
// the above approach

import java.util.*;

import java.lang.*;
 
class GFG{
 
// Function to find the count of 
// possible pairs of collisions 

static double solve(int[][] D, int N, 

                        int X, int Y) 
{ 

    // Stores the time at which 

    // points reach the origin 

    ArrayList<Double> T = new ArrayList<>(); 
 
    // Calculate time for each point 

    for(int i = 0; i < N; i++) 

    { 

        int x = D[i][0]; 

        int y = D[i][1]; 
 
        double speed = D[i][2]; 
 
        double time = ((x * x - X * X) + 

                       (y * y - Y * Y)) / 

                       (speed * speed); 
 
        T.add(time); 

    } 
 
    // Sort the times 

    Collections.sort(T); 
 
    int i = 0; 

    int total = 0; 
 
    // Counting total collisions 

    while (i < (T.size() - 1)) 

    { 

        // Count of elements arriving at 

        // a given point at the same time 

        int count = 1; 
 
        while ((i < (T.size() - 1)) && 

               (Double.compare(T.get(i),

                               T.get(i + 1)) == 0))

        { 

            count += 1; 

            i += 1; 

        } 
 
        total += (count * (count - 1)) / 2; 

        i += 1; 

    } 

    return total; 
} 
 
// Driver Code

public static void main (String[] args)
{

    int arr[] = { 1, 2, 3, 4, 5 }; 

    int N = 5; 

    // Given set of points with speed 

    int[][] D = { { 5, 12, 1 }, 

                  { 16, 63, 5 }, 

                  { -10, 24, 2 }, 

                  { 7, 24, 2 }, 

                  { -24, 7, 2 } }; 

    int X = 0, Y = 0; 

    // Function call 

    System.out.println(solve(D, N, X, Y)); 
}
}
 
// This code is contributed by offbeat

Python3

# Python3 Program to implement
# the above approach
 
# Function to find the count of
# possible pairs of collisions

def solve(D, N, X, Y):
 
    # Stores the time at which 

    # points reach the origin

    T = [] 

    # Calculate time for each point

    for i in range(N):
 
        x = D[i][0] 

        y = D[i][1]
 
        speed = D[i][2]
 
        time = ((x * x - X * X) +

                (y * y - Y * Y)) / (speed * speed) 

        T.append(time)

    # Sort the times

    T.sort()

    i = 0

    total = 0

    # Counting total collisions

    while i<len(T)-1:

        # Count of elements arriving at

        # a given point at the same time

        count = 1
 
        while i<len(T)-1 and T[i] == T[i + 1]:

            count += 1

            i+= 1

        total+= (count*(count-1))/2

        i+= 1
 
    return total
 
# Driver Code
 
N = 5
 
# Given set of points with speed

D = [[5, 12, 1], [16, 63, 5], \

    [-10, 24, 2], [7, 24, 2], \

    [-24, 7, 2]]
 
X = 0

Y = 0
 
# Function Call

print(solve(D, N, X, Y))

// C# program to implement
// the above approach

using System;

using System.Collections;
 
class GFG{
 
// Function to find the count of 
// possible pairs of collisions 

static double solve(int[, ] D, int N, 

                        int X, int Y) 
{ 

    // Stores the time at which 

    // points reach the origin 

    ArrayList T = new ArrayList(); 
 
    // Calculate time for each point 

    for(int i = 0; i < N; i++) 

    { 

        int x = D[i, 0]; 

        int y = D[i, 1]; 
 
        double speed = D[i, 2]; 
 
        double time = ((x * x - X * X) + 

                       (y * y - Y * Y)) / 

                       (speed * speed); 
 
        T.Add(time); 

    } 
 
    // Sort the times 

    T.Sort(); 
 
    int j = 0; 

    int total = 0; 
 
    // Counting total collisions 

    while (j < (T.Count - 1)) 

    { 

        // Count of elements arriving at 

        // a given point at the same time 

        int count = 1; 

        while ((j < (T.Count - 1)) && 

              (Convert.ToDouble(T[j]).CompareTo(

               Convert.ToDouble(T[j + 1])) == 0))

        { 

            count += 1; 

            j += 1; 

        } 

        total += (count * (count - 1)) / 2; 

        j += 1; 

    } 

    return total; 
} 
 
// Driver Code

public static void Main (String[] args)
{

    int N = 5; 

    // Given set of points with speed 

    int [,] D = new int [,] { { 5, 12, 1 }, 

                              { 16, 63, 5 }, 

                              { -10, 24, 2 }, 

                              { 7, 24, 2 }, 

                              { -24, 7, 2 } }; 

    int X = 0, Y = 0; 

    // Function call 

    Console.WriteLine(solve(D, N, X, Y)); 
}
}
 
// This code is contributed by jana_sayantan

Javascript

<script>
 
// JavaScript program to implement
// the above approach
 
// Function to find the count of 
// possible pairs of collisions 

function solve(D,N,X,Y)
{

    // Stores the time at which 

    // points reach the origin 

    let T = [];

    // Calculate time for each point 

    for(let i = 0; i < N; i++) 

    { 

        let x = D[i][0]; 

        let y = D[i][1]; 

        let speed = D[i][2]; 

        let time = ((x * x - X * X) + 

                       (y * y - Y * Y)) / 

                       (speed * speed); 

        T.push(time); 

    } 

    // Sort the times 

    T.sort(function(a,b){return a-b;});

    let i = 0; 

    let total = 0; 

    // Counting total collisions 

    while (i < (T.length - 1)) 

    { 

        // Count of elements arriving at 

        // a given point at the same time 

        let count = 1; 

        while ((i < (T.length - 1)) && 

               (T[i]==T[i+1]))

        { 

            count += 1; 

            i += 1; 

        } 

        total += (count * (count - 1)) / 2; 

        i += 1; 

    } 

    return total; 
}
 
// Driver Code
let arr=[1, 2, 3, 4, 5];
let N = 5; 

 // Given set of points with speed 
let D = [[5, 12, 1], [16, 63, 5],

    [-10, 24, 2], [7, 24, 2],

    [-24, 7, 2]];
 
let X = 0, Y = 0; 
// Function call 
document.write(solve(D, N, X, Y).toFixed(1)); 
 
// This code is contributed by patel2127
 
</script>