Check if the given point lies inside given N points of a Convex Polygon

Given coordinates of the N points of a Convex Polygon. The task is to check if the given point (X, Y) lies inside the polygon.

Examples:

Input: N = 7, Points: {(1, 1), (2, 1), (3, 1), (4, 1), (4, 2), (4, 3), (4, 4)}, Query: X = 3, Y = 2
Below is the image of plotting of the given points:

Output: YES

Input: N = 7, Points: {(1, 1), (2, 1), (3, 1), (4, 1), (4, 2), (4, 3), (4, 4)}, Query: X = 3, Y = 9
Output: NO

Approach: The idea is to use Graham Scan Algorithm to find if the given point lies inside the Convex Polygon or not. Below are some of the observations:

• Suppose the point (X, Y) is a point in the set of points of the convex polygon. If the Graham Scan Algorithm is used on this set of points, another set of points would be obtained, which makes up the Convex Hull.
• If the point (X, Y) lies inside the polygon, it won’t lie on the Convex Hull and hence won’t be present in the newly generated set of points of the Convex Hull.
• If the point (X, Y) lies outside the polygon, it will then lie on the Convex Hull formed and hence would be present in the newly generated set of points of the Convex Hull.

Below are the steps to solve the problem:

1. Sort the given points along with the query point in the increasing order of their abscissa values. If the abscissa values(x-coordinates) of any two points are the same, then sort them on the basis of their ordinate value.
2. Set the bottom-left point as the start point and top-right point as the end point of the convex hull.
3. Iterate over all the points and find out the points, forming the convex polygon, that lies in between the start and endpoints in the clockwise direction. Store these points in a vector.
4. Iterate over all the points and find out the points, forming the convex polygon, that lies in between the start and endpoints in the counter-clockwise direction. Store these points in the vector.
5. Check if the query point exists in the vector then the point lies outside the convex hull. So return “No”.
6. If the point doesn’t exist in the vector, then the point lies inside the convex hull print “Yes”.

Below is the implementation based on the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Sorting Function to sort points
bool cmp(pair<int, int>& a,
         pair<int, int>& b)
{

    if (a.first == b.first)
        return a.second < b.second;
    return a.first < b.first;
}

// Function To Check Clockwise
// Orientation
int cw(pair<int, int>& a,
       pair<int, int>& b,
       pair<int, int>& c)
{

    int p = a.first * (b.second - c.second)
            + b.first * (c.second - a.second)
            + c.first * (a.second - b.second);

    return p < 0ll;
}

// Function To Check Counter
// Clockwise Orientation
int ccw(pair<int, int>& a,
        pair<int, int>& b,
        pair<int, int>& c)
{

    int p = a.first * (b.second - c.second)
            + b.first * (c.second - a.second)
            + c.first * (a.second - b.second);

    return p > 0ll;
}

// Graham Scan algorithm to find Convex
// Hull from given points
vector<pair<int, int> > convexHull(
    vector<pair<int, int> >& v)
{
    // Sort the points
    sort(v.begin(),
         v.end(), cmp);

    int n = v.size();
    if (n <= 3)
        return v;

    // Set starting and ending points as
    // left bottom and top right
    pair<int, int> p1 = v[0];
    pair<int, int> p2 = v[n - 1];

    // Vector to store points in
    // upper half and lower half
    vector<pair<int, int> > up, down;

    // Insert StartingEnding Points
    up.push_back(p1);
    down.push_back(p1);

    // Iterate over points
    for (int i = 1; i < n; i++) {

        if (i == n - 1 || !ccw(p1, v[i], p2)) {

            while (up.size() > 1
                   && ccw(up[up.size() - 2],
                          up[up.size() - 1],
                          v[i])) {

                // Exclude this point
                // if we can form better

                up.pop_back();
            }

            up.push_back(v[i]);
        }

        if (i == n - 1 || !cw(p1, v[i], p2)) {

            while (down.size() > 1
                   && cw(down[down.size() - 2],
                         down[down.size() - 1],
                         v[i])) {

                // Exclude this point
                // if we can form better
                down.pop_back();
            }
            down.push_back(v[i]);
        }
    }

    // Combine upper and  lower half
    for (int i = down.size() - 2;
         i > 0; i--)
        up.push_back(down[i]);

    // Remove duplicate points
    up.resize(unique(up.begin(),
                     up.end())
              - up.begin());

    // Return the points on Convex Hull
    return up;
}

// Function to find if point lies inside
// a convex polygon
bool isInside(vector<pair<int, int> > points,
              pair<int, int> query)
{
    // Include the query point in the
    // polygon points
    points.push_back(query);

    // Form a convex hull from the points
    points = convexHull(points);

    // Iterate over the points
    // of convex hull
    for (auto x : points) {

        // If the query point lies
        // on the convex hull
        // then it wasn't inside
        if (x == query)
            return false;
    }

    // Otherwise it was Inside
    return true;
}

// Driver Code
int main()
{

    // Points of the polygon
    // given in any order
    int n = 7;
    vector<pair<int, int> > points;

    points = { { 1, 1 }, { 2, 1 }, { 3, 1 },
               { 4, 1 }, { 4, 2 }, { 4, 3 },
               { 4, 4 } };

    // Query Points
    pair<int, int> query = { 3, 2 };

    // Check if its inside
    if (isInside(points, query)) {
        cout << "YES" << endl;
    }
    else {
        cout << "NO" << endl;
    }

    return 0;
}

Output:

YES

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