Line Segment Intersection - Problem

Determine if two line segments intersect. Given two line segments defined by their endpoints, return true if they intersect, false otherwise.

Each line segment is defined by two points: [(x1, y1), (x2, y2)] for the first segment and [(x3, y3), (x4, y4)] for the second segment.

Special cases to handle:

Segments are collinear (lie on the same line)
Segments touch only at endpoints
One or both segments are actually points (zero length)
Segments overlap partially or completely

Return true for any intersection, including touching at endpoints.

Input & Output

Example 1 — Basic Intersection

$ Input: segment1 = [[0,0],[4,3]], segment2 = [[2,0],[2,4]]

› Output: true

💡 Note: The first segment goes from (0,0) to (4,3), the second goes from (2,0) to (2,4). They intersect at point (2, 1.5).

Example 2 — No Intersection

$ Input: segment1 = [[0,0],[2,2]], segment2 = [[3,3],[5,5]]

› Output: false

💡 Note: Both segments lie on the same line y = x but don't overlap - there's a gap between (2,2) and (3,3).

Example 3 — Endpoint Touch

$ Input: segment1 = [[0,0],[2,2]], segment2 = [[2,2],[4,0]]

› Output: true

💡 Note: The segments meet exactly at point (2,2) where the first ends and second begins.

Constraints

-10⁴ ≤ coordinates ≤ 10⁴
Segments can have zero length (point)
Handle floating point precision carefully

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 15 f Meta 12

The key insight is using orientation tests with cross products to determine if two line segments straddle each other. Best approach is the Orientation-Based Method which avoids floating point division and handles all edge cases cleanly. Time: O(1), Space: O(1)

Common Approaches

✓ Orientation-Based Method

⏱️ Time: O(1) Space: O(1)

Calculate orientations using cross products to determine which side of a line each point lies on. Two segments intersect if they straddle each other.

Parametric Intersection Check

⏱️ Time: O(1) Space: O(1)

Express each line segment parametrically and solve the system of equations to find intersection parameters. Check if parameters are within valid range [0,1] for both segments.

Orientation-Based Method — Algorithm Steps

Calculate orientation of triplet (p, q, r) using cross product
Check if segment1 straddles segment2 by testing orientations
Check if segment2 straddles segment1 by testing orientations
Handle special collinear cases with overlap checking

Visualization

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Step-by-Step Walkthrough

Calculate Orientations

Find orientation of each endpoint relative to other segment

Check Straddle

Segments intersect if they straddle each other

Handle Collinear

Special cases when points lie on same line

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int orientation(int p[2], int q[2], int r[2]) {
    long long val = (long long)(q[1] - p[1]) * (r[0] - p[0]) - (long long)(q[0] - p[0]) * (r[1] - p[1]);
    if (val == 0) return 0;
    return val > 0 ? 1 : 2;
}

int onSegment(int p[2], int q[2], int r[2]) {
    int maxX = p[0] > r[0] ? p[0] : r[0];
    int minX = p[0] < r[0] ? p[0] : r[0];
    int maxY = p[1] > r[1] ? p[1] : r[1];
    int minY = p[1] < r[1] ? p[1] : r[1];
    
    return q[0] <= maxX && q[0] >= minX && q[1] <= maxY && q[1] >= minY;
}

int solution(int segment1[2][2], int segment2[2][2]) {
    int o1 = orientation(segment1[0], segment1[1], segment2[0]);
    int o2 = orientation(segment1[0], segment1[1], segment2[1]);
    int o3 = orientation(segment2[0], segment2[1], segment1[0]);
    int o4 = orientation(segment2[0], segment2[1], segment1[1]);
    
    if (o1 != o2 && o3 != o4) {
        return 1;
    }
    
    if ((o1 == 0 && onSegment(segment1[0], segment2[0], segment1[1])) ||
        (o2 == 0 && onSegment(segment1[0], segment2[1], segment1[1])) ||
        (o3 == 0 && onSegment(segment2[0], segment1[0], segment2[1])) ||
        (o4 == 0 && onSegment(segment2[0], segment1[1], segment2[1]))) {
        return 1;
    }
    
    return 0;
}

int main() {
    int segment1[2][2], segment2[2][2];
    scanf("[[%d,%d],[%d,%d]]", &segment1[0][0], &segment1[0][1], &segment1[1][0], &segment1[1][1]);
    scanf("[[%d,%d],[%d,%d]]", &segment2[0][0], &segment2[0][1], &segment2[1][0], &segment2[1][1]);
    
    int result = solution(segment1, segment2);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Fixed number of orientation calculations and comparisons

✓ Linear Growth

Space Complexity

O(1)

Only stores orientation values and temporary variables

✓ Linear Space

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Line Segment Intersection - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Orientation-Based Method — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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