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Check if any pair of semicircles intersect or not

Given an array arr[] consisting of N integers such that each pair of points (x, y) represents the endpoints of the semicircle as (x, 0) and (y, 0), the task is to check if any pair of semicircles intersect or not. If found to be true, then print “Yes”. Otherwise, print “No”.

Examples:

Input: arr[] = {0, 15, 5, 10}
Output: No

Input: arr[] = {0, 10, 5, 15}
Output: Yes

Approach: The given problem can be solved by checking if any pairs of semicircles intersect or not and then print the elements accordingly. Follow the steps below to solve the problem:

  • Store all the possible semicircles in a vector with their corresponding x coordinates and y coordinates.
  • Now, generate all possible pairs of the above vectors and perform the following steps:
    • Let the pair of coordinates be X and Y and check if the circles intersect or not using the following conditions:
      • If the value of (X[0] < Y[0], X[1] < Y[1] and Y[0] < X[2]) or (Y[0] < X[0], Y[1] < X[1], and X[0] < Y[1]), then the semicircle intersects. Otherwise, it doesn’t intersect.
      • If the above two circles intersect, then print “Yes” and break out of the loop.
  • After completing the above steps, if any pair of circles doesn’t intersect, then print “No”.

C++

// C++ program for the above approach
 
#include <iostream>
#include <vector>
using namespace std;
 
// Function to check if any pairs of
// the semicircles intersects or not
bool checkIntersection(int arr[], int N)
{
 
    // Stores the coordinates of all
    // the semicircles
    vector<pair<int, int> > vec;
 
    for (int i = 0; i < N - 1; i++) {
 
        // x and y are coordinates
        int x = arr[i], y = arr[i + 1];
 
        // Store the minimum and maximum
        // value of the pair
        int minn = min(x, y);
        int maxx = max(x, y);
 
        // Push the pair in vector
        vec.push_back({ minn, maxx });
    }
 
    // Compare one pair with other pairs
    for (int i = 0; i < vec.size(); i++) {
 
        pair<int, int> x = vec[i];
 
        // Generating the second pair
        for (int j = 0;
             j < vec.size(); j++) {
 
            // Extract all the second
            // pairs one by one
            pair<int, int> y = vec[j];
 
            // 1st condition
            bool cond1
                = (x.first < y.first
                   and x.second < y.second
                   and y.first < x.second);
 
            // 2nd condition
            bool cond2
                = (y.first < x.first
                   and y.second < x.second
                   and x.first < y.second);
 
            // If any one condition
            // is true
            if (cond1 or cond2) {
                return true;
            }
        }
    }
 
    // If any pair of semicircles
    // doesn't exists
    return false;
}
 
// Driver Code
int main()
{
    int arr[] = { 0, 15, 5, 10 };
    int N = sizeof(arr) / sizeof(int);
 
    if (checkIntersection(arr, N))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
static class pair
{
    int first, second;
     
    public pair(int first, int second) 
    {
        this.first = first;
        this.second = second;
    }   
}
 
// Function to check if any pairs of
// the semicircles intersects or not
static boolean checkIntersection(int arr[], int N)
{
 
    // Stores the coordinates of all
    // the semicircles
    Vector<pair > vec = new Vector<>();
 
    for(int i = 0; i < N - 1; i++)
    {
         
        // x and y are coordinates
        int x = arr[i], y = arr[i + 1];
 
        // Store the minimum and maximum
        // value of the pair
        int minn = Math.min(x, y);
        int maxx = Math.max(x, y);
 
        // Push the pair in vector
        vec.add(new pair( minn, maxx ));
    }
 
    // Compare one pair with other pairs
    for(int i = 0; i < vec.size(); i++)
    {
        pair x = vec.elementAt(i);
 
        // Generating the second pair
        for(int j = 0;
                j < vec.size(); j++)
        {
 
            // Extract all the second
            // pairs one by one
            pair y = vec.elementAt(j);
 
            // 1st condition
            boolean cond1 = (x.first < y.first &&
                            x.second < y.second &&
                             y.first < x.second);
 
            // 2nd condition
            boolean cond2 = (y.first < x.first &&
                            y.second < x.second &&
                             x.first < y.second);
 
            // If any one condition
            // is true
            if (cond1 || cond2)
            {
                return true;
            }
        }
    }
 
    // If any pair of semicircles
    // doesn't exists
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 0, 15, 5, 10 };
    int N = arr.length;
 
    if (checkIntersection(arr, N))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Function to check if any pairs of
# the semicircles intersects or not
def checkIntersection(arr, N):
     
    # Stores the coordinates of all
    # the semicircles
    vec = []
 
    for i in range(N - 1):
         
        # x and y are coordinates
        x = arr[i]
        y = arr[i + 1]
 
        # Store the minimum and maximum
        # value of the pair
        minn = min(x, y)
        maxx = max(x, y)
 
        # Push the pair in vector
        vec.append([minn, maxx])
 
    # Compare one pair with other pairs
    for i in range(len(vec)):
        x = vec[i]
 
        # Generating the second pair
        for j in range(len(vec)):
             
            # Extract all the second
            # pairs one by one
            y = vec[j]
 
            # 1st condition
            cond1 = (x[0] < y[0] and
                     x[1] < y[1] and
                     y[0] < x[1])
 
            # 2nd condition
            cond2 = (y[0] < x[0] and
                     y[1] < x[1] and
                     x[0] < y[1])
 
            # If any one condition
            # is true
            if (cond1 or cond2):
                return True
 
    # If any pair of semicircles
    # doesn't exists
    return False
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 0, 15, 5, 10 ]
    N = len(arr)
 
    if (checkIntersection(arr, N)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by ipg2016107

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
class pair
{
    public int first, second;
     
    public pair(int first, int second) 
    {
        this.first = first;
        this.second = second;
    }   
}
 
// Function to check if any pairs of
// the semicircles intersects or not
static bool checkIntersection(int []arr, int N)
{
 
    // Stores the coordinates of all
    // the semicircles
    List<pair > vec = new List<pair>();
 
    for(int i = 0; i < N - 1; i++)
    {
         
        // x and y are coordinates
        int x = arr[i], y = arr[i + 1];
 
        // Store the minimum and maximum
        // value of the pair
        int minn = Math.Min(x, y);
        int maxx = Math.Max(x, y);
 
        // Push the pair in vector
        vec.Add(new pair( minn, maxx ));
    }
 
    // Compare one pair with other pairs
    for(int i = 0; i < vec.Count; i++)
    {
        pair x = vec[i];
 
        // Generating the second pair
        for(int j = 0;
                j < vec.Count; j++)
        {
             
            // Extract all the second
            // pairs one by one
            pair y = vec[j];
 
            // 1st condition
            bool cond1 = (x.first < y.first &&
                         x.second < y.second &&
                          y.first < x.second);
 
            // 2nd condition
            bool cond2 = (y.first < x.first &&
                         y.second < x.second &&
                          x.first < y.second);
 
            // If any one condition
            // is true
            if (cond1 || cond2)
            {
                return true;
            }
        }
    }
 
    // If any pair of semicircles
    // doesn't exists
    return false;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 0, 15, 5, 10 };
    int N = arr.Length;
 
    if (checkIntersection(arr, N))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
 
// JavaScript program for the above approach
 
 
// Function to check if any pairs of
// the semicircles intersects or not
function checkIntersection(arr, N)
{
 
    // Stores the coordinates of all
    // the semicircles
    let vec = [];
 
    for (let i = 0; i < N - 1; i++) {
 
        // x and y are coordinates
        let x = arr[i], y = arr[i + 1];
 
        // Store the minimum and maximum
        // value of the pair
        let minn = Math.min(x, y);
        let maxx = Math.max(x, y);
 
        // Push the pair in vector
        vec.push([ minn, maxx ]);
    }
 
    // Compare one pair with other pairs
    for (let i = 0; i < vec.length; i++) {
 
        let x = vec[i];
 
        // Generating the second pair
        for (let j = 0;j < vec.length; j++) {
 
            // Extract all the second
            // pairs one by one
            let y = vec[j];
 
            // 1st condition
            let cond1 = (x[0] < y[0]
                && x[1] < y[1]
                && y[0] < x[1]);
 
            // 2nd condition
            let cond2 = (y[0] < x[0]
                && y[1] < x[1]
                && x[0] < y[1]);
 
            // If any one condition
            // is true
            if (cond1 || cond2) {
                return true;
            }
        }
    }
 
    // If any pair of semicircles
    // doesn't exists
    return false;
}
 
// Driver Code
 
let arr = [ 0, 15, 5, 10 ];
let N = arr.length;
 
if (checkIntersection(arr, N))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by shinjanpatra.
</script>

Output: 

No

 

Time Complexity: O(N2)
Auxiliary Space: O(N)


 


Reffered: https://www.geeksforgeeks.org

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Category:
 Coding
Sub Category:
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