Find Maximum Area of a Triangle - Practice Coding Problems

Find Maximum Area of a Triangle - Problem

Array Hash Table Math Greedy Geometry Enumeration Medium

You're tasked with finding the maximum area of a triangle from a collection of points on a 2D plane, but with a special constraint: at least one side of the triangle must be parallel to either the x-axis or y-axis.

Given a 2D array coords of size n × 2 representing coordinates of n points, you need to:

Find three points that form a triangle with maximum area
Ensure at least one side is horizontal (parallel to x-axis) or vertical (parallel to y-axis)
Return twice the maximum area (this avoids floating-point precision issues)
Return -1 if no valid triangle exists

Key insight: A triangle has a horizontal side when two points share the same y-coordinate, and a vertical side when two points share the same x-coordinate. The area is maximized by finding the point farthest from the shared coordinate line.

Input & Output

example_1.py — Basic Triangle

$ Input: coords = [[0,0],[0,1],[1,0]]

› Output: 1

💡 Note: The triangle formed by points (0,0), (0,1), and (1,0) has a vertical side from (0,0) to (0,1) and a horizontal side from (0,0) to (1,0). The area is 0.5, so twice the area is 1.

example_2.py — Larger Triangle

$ Input: coords = [[0,0],[2,0],[1,3]]

› Output: 6

💡 Note: The triangle has a horizontal base from (0,0) to (2,0) with length 2, and height 3 to point (1,3). Area = 0.5 × 2 × 3 = 3, so twice the area is 6.

example_3.py — No Valid Triangle

$ Input: coords = [[1,1],[2,2],[3,3]]

› Output: -1

💡 Note: All points lie on the same diagonal line, so no triangle can be formed with horizontal or vertical sides. Return -1 as no valid triangle exists.

Visualization

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Understanding the Visualization

Identify Aligned Points

Group points that share x-coordinates (vertical alignment) or y-coordinates (horizontal alignment)

Find Maximum Bases

For each aligned group, find the pair of points with maximum distance to form the triangle's base

Optimize Third Vertex

For each base, find the third point that maximizes the perpendicular distance (height)

Calculate Maximum Area

Use the formula Area = ½ × base × height, return twice the maximum area found

Key Takeaway

🎯 Key Insight: By leveraging the constraint that at least one side must be axis-aligned, we can reduce the problem from checking all O(n³) triangles to efficiently computing optimal areas using coordinate grouping in O(n²) time.

Time & Space Complexity

Time Complexity

⏱️

O(n²)

O(n) to build hash maps, then O(n²) to find optimal triangles for each coordinate group

⚠ Quadratic Growth

Space Complexity

O(n)

Hash maps to store coordinate groupings

⚡ Linearithmic Space

Constraints

3 ≤ coords.length ≤ 1000
-10⁴ ≤ coords[i][0], coords[i][1] ≤ 10⁴
At least one side of the triangle must be parallel to x-axis or y-axis
Triangle must have non-zero area

Asked in

G Google 42 a Amazon 38 f Meta 26 ⊞ Microsoft 19

The optimal solution uses coordinate grouping with hash maps to efficiently find triangles with maximum area. By grouping points that share the same x or y coordinates, we can quickly identify potential horizontal and vertical sides, then find the third point that maximizes the triangle's area using the formula area = base × height.

Common Approaches

Approach	Time	Space	Notes
✓ Hash Map Grouping (Optimal)	O(n²)	O(n)	Group points by x and y coordinates, then find maximum distances for optimal triangles
Brute Force (Check All Triangles)	O(n³)	O(1)	Check every possible combination of three points and calculate triangle areas

Hash Map Grouping (Optimal) — Algorithm Steps

Create hash maps to group points by x-coordinates and y-coordinates
For points sharing the same x-coordinate, find the pair with maximum vertical distance
For points sharing the same y-coordinate, find the pair with maximum horizontal distance
For each maximum base pair, find the third point with maximum perpendicular distance
Calculate area as base × height and track the maximum

Visualization

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Step-by-Step Walkthrough

Group Coordinates

Create hash maps for x and y coordinate groupings

Find Maximum Bases

For each coordinate group, find the maximum distance pair

Optimize Third Point

Find third point with maximum perpendicular distance

Calculate Areas

Compute area as base × height for each triangle

Code -

solution.c — C

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

typedef struct {
    int value;
    int index;
} Point;

int compare(const void* a, const void* b) {
    Point* pa = (Point*)a;
    Point* pb = (Point*)b;
    return pa->value - pb->value;
}

int largestTriangleArea(int coords[][2], int n) {
    if (n < 3) return -1;
    
    int maxArea = 0;
    
    // Check all possible horizontal and vertical bases
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int x1 = coords[i][0], y1 = coords[i][1];
            int x2 = coords[j][0], y2 = coords[j][1];
            
            // Horizontal base (same y)
            if (y1 == y2 && x1 != x2) {
                int baseWidth = abs(x2 - x1);
                for (int k = 0; k < n; k++) {
                    if (k != i && k != j && coords[k][1] != y1) {
                        int height = abs(coords[k][1] - y1);
                        int area = baseWidth * height;
                        if (area > maxArea) {
                            maxArea = area;
                        }
                    }
                }
            }
            
            // Vertical base (same x)
            if (x1 == x2 && y1 != y2) {
                int baseHeight = abs(y2 - y1);
                for (int k = 0; k < n; k++) {
                    if (k != i && k != j && coords[k][0] != x1) {
                        int width = abs(coords[k][0] - x1);
                        int area = baseHeight * width;
                        if (area > maxArea) {
                            maxArea = area;
                        }
                    }
                }
            }
        }
    }
    
    return maxArea > 0 ? maxArea : -1;
}

int main() {
    int coords[][2] = {{0, 0}, {0, 1}, {1, 0}};
    printf("%d\n", largestTriangleArea(coords, 3));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

O(n) to build hash maps, then O(n²) to find optimal triangles for each coordinate group

⚠ Quadratic Growth

Space Complexity

O(n)

Hash maps to store coordinate groupings

⚡ Linearithmic Space

Constraints

3 ≤ coords.length ≤ 1000
-10⁴ ≤ coords[i][0], coords[i][1] ≤ 10⁴
At least one side of the triangle must be parallel to x-axis or y-axis
Triangle must have non-zero area

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Common Approaches

Hash Map Grouping (Optimal) — Algorithm Steps

Visualization

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