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 collisionsint 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 Codeint 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 approachimport 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 Codepublic 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 collisionsdef 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 CodeN = 5# Given set of points with speedD = [[5, 12, 1], [16, 63, 5], \ [-10, 24, 2], [7, 24, 2], \ [-24, 7, 2]]X = 0Y = 0# Function Callprint(solve(D, N, X, Y)) |
C#
// C# program to implement// the above approachusing 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 Codepublic 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 Codelet 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> |
4.0
Time Complexity: O(N log N)
Auxiliary Space: O(N) because using extra space for vector T
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