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How to check if a given point lies inside or outside a polygon?

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  • Difficulty Level : Hard
  • Last Updated : 14 Oct, 2022
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Given a polygon and a point ‘p‘, find if ‘p‘ lies inside the polygon or not. The points lying on the border are considered inside.


check if a given point lies inside or outside a polygon 1

Approach: The idea to solve this problem is based on How to check if two given line segments intersect, and to be used as follows:

  1. Draw a horizontal line to the right of each point and extend it to infinity
  2. Count the number of times the line intersects with polygon edges.
  3. A point is inside the polygon if either count of intersections is odd or point lies on an edge of polygon.  If none of the conditions is true, then point lies outside.

check if a given point lies inside or outside a polygon 2

How to handle point ‘g’ in the above figure? 

Note that we should return true if the point lies on the line or the same as one of the vertices of the given polygon. To handle this, after checking if the line from ‘p’ to extreme intersects, we check whether ‘p’ is collinear with vertices of the current line of polygon. If it is collinear, then we check if the point ‘p’ lies on current side of polygon, if it lies, we return true, else false.

Following is the implementation of the above approach: 


#include <bits/stdc++.h>
using namespace std;
struct Point {
    int x, y;
struct line {
    Point p1, p2;
bool onLine(line l1, Point p)
    // Check whether p is on the line or not
    if (p.x <= max(l1.p1.x, l1.p2.x)
        && p.x <= min(l1.p1.x, l1.p2.x)
        && (p.y <= max(l1.p1.y, l1.p2.y)
            && p.y <= min(l1.p1.y, l1.p2.y)))
        return true;
    return false;
int direction(Point a, Point b, Point c)
    int val = (b.y - a.y) * (c.x - b.x)
              - (b.x - a.x) * (c.y - b.y);
    if (val == 0)
        // Colinear
        return 0;
    else if (val < 0)
        // Anti-clockwise direction
        return 2;
    // Clockwise direction
    return 1;
bool isIntersect(line l1, line l2)
    // Four direction for two lines and points of other line
    int dir1 = direction(l1.p1, l1.p2, l2.p1);
    int dir2 = direction(l1.p1, l1.p2, l2.p2);
    int dir3 = direction(l2.p1, l2.p2, l1.p1);
    int dir4 = direction(l2.p1, l2.p2, l1.p2);
    // When intersecting
    if (dir1 != dir2 && dir3 != dir4)
        return true;
    // When p2 of line2 are on the line1
    if (dir1 == 0 && onLine(l1, l2.p1))
        return true;
    // When p1 of line2 are on the line1
    if (dir2 == 0 && onLine(l1, l2.p2))
        return true;
    // When p2 of line1 are on the line2
    if (dir3 == 0 && onLine(l2, l1.p1))
        return true;
    // When p1 of line1 are on the line2
    if (dir4 == 0 && onLine(l2, l1.p2))
        return true;
    return false;
bool checkInside(Point poly[], int n, Point p)
    // When polygon has less than 3 edge, it is not polygon
    if (n < 3)
        return false;
    // Create a point at infinity, y is same as point p
    line exline = { p, { 9999, p.y } };
    int count = 0;
    int i = 0;
    do {
        // Forming a line from two consecutive points of
        // poly
        line side = { poly[i], poly[(i + 1) % n] };
        if (isIntersect(side, exline)) {
            // If side is intersects exline
            if (direction(side.p1, p, side.p2) == 0)
                return onLine(side, p);
        i = (i + 1) % n;
    } while (i != 0);
    // When count is odd
    return count & 1;
// Driver code
int main()
    Point polygon[]
        = { { 0, 0 }, { 10, 0 }, { 10, 10 }, { 0, 10 } };
    Point p = { 5, 3 };
    int n = 4;
    // Function call
    if (checkInside(polygon, n, p))
        cout << "Point is inside.";
        cout << "Point is outside.";
    return 0;


Point is inside.

Time Complexity: O(n) where n is the number of vertices in the given polygon.
Auxiliary Space: O(1), since no extra space has been taken.

This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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