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Count of collisions at a point (X, Y)
  • Last Updated : 14 Sep, 2020

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:
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:
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

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// 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

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Java

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// 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

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Python3

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# 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))

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C#

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// 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

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Output: 

4.0

Time Complexity: O(N log N) 
Auxiliary Space: O(1)
 

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