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 are given a 2D array coords of size n x 2, representing the coordinates of n points in an infinite Cartesian plane.

Find twice the maximum area of a triangle with its corners at any three elements from coords, such that at least one side of this triangle is parallel to the x-axis or y-axis.

Formally, if the maximum area of such a triangle is A, return 2 * A. If no such triangle exists, return -1.

Note that a triangle cannot have zero area.

Input & Output

Example 1 — Basic Triangle

$ Input: coords = [[1,3],[5,3],[3,5]]

› Output: 8

💡 Note: Points (1,3) and (5,3) form a horizontal side of length 4. The third point (3,5) is at distance 2 from this horizontal line. Area = 4 × 2 / 2 = 4. Return 2 × 4 = 8.

Example 2 — Vertical Side

$ Input: coords = [[2,1],[2,4],[6,2]]

› Output: 12

💡 Note: Points (2,1) and (2,4) form a vertical side of length 3. The third point (6,2) is at distance 4 from this vertical line. Area = 3 × 4 / 2 = 6. Return 2 × 6 = 12.

Example 3 — No Valid Triangle

$ Input: coords = [[1,1],[2,3],[4,5]]

› Output: -1

💡 Note: No two points share the same x or y coordinate, so no triangle can have an axis-parallel side. Return -1.

Constraints

3 ≤ coords.length ≤ 1000
-10⁹ ≤ coords[i][0], coords[i][1] ≤ 10⁹
All coordinate pairs are unique

Visualization

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Understanding the Visualization

Input Points

Given coordinate points in 2D plane

Find Axis-Parallel Sides

Identify pairs sharing x or y coordinates

Calculate Areas

Compute triangle areas and return twice the maximum

Key Takeaway

🎯 Key Insight: For axis-parallel sides, two points must share identical x or y coordinates

Asked in

G Google 25 a Amazon 18 M Microsoft 15

The key insight is that if a triangle has at least one axis-parallel side, then two points must share the same x-coordinate (vertical side) or y-coordinate (horizontal side). Best approach uses hash maps to group points efficiently. Time: O(n³), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check All Triplets	O(n³)	O(1)	Check every possible combination of three points and calculate area when one side is axis-parallel
Hash Map Optimization	O(n³)	O(n)	Group points by shared coordinates and check combinations more efficiently

Brute Force - Check All Triplets — Algorithm Steps

Iterate through all possible triplets of points
For each triplet, check if any pair shares x or y coordinates
Calculate area using cross product formula
Track maximum area found

Visualization

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Step-by-Step Walkthrough

Select Triplet

Pick any 3 points from the coordinate array

Check Axis-Parallel

Verify if any two points share x or y coordinates

Calculate Area

Use cross product to find twice the triangle area

Code -

solution.c — C

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

int abs_val(int x) {
    return x < 0 ? -x : x;
}

int max_val(int a, int b) {
    return a > b ? a : b;
}

int solution(int coords[][2], int n) {
    if (n < 3) return -1;
    
    int maxArea = -1;
    
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            for (int k = j + 1; k < n; k++) {
                int x1 = coords[i][0], y1 = coords[i][1];
                int x2 = coords[j][0], y2 = coords[j][1];
                int x3 = coords[k][0], y3 = coords[k][1];
                
                // Check if any side is parallel to axis
                int hasAxisParallel = 0;
                if (x1 == x2 || x1 == x3 || x2 == x3) {
                    hasAxisParallel = 1;
                } else if (y1 == y2 || y1 == y3 || y2 == y3) {
                    hasAxisParallel = 1;
                }
                
                if (hasAxisParallel) {
                    // Calculate twice the area using cross product
                    int areaTwice = abs_val((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1));
                    if (areaTwice > 0) {
                        maxArea = max_val(maxArea, areaTwice);
                    }
                }
            }
        }
    }
    
    return maxArea;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for 2D array
    int coords[1000][2];
    int n = 0;
    char *ptr = line + 1; // skip first '['
    
    while (*ptr && *ptr != ']') {
        if (*ptr == '[') {
            ptr++;
            coords[n][0] = atoi(ptr);
            while (*ptr && *ptr != ',') ptr++;
            if (*ptr == ',') ptr++;
            coords[n][1] = atoi(ptr);
            while (*ptr && *ptr != ']') ptr++;
            if (*ptr == ']') ptr++;
            n++;
        } else {
            ptr++;
        }
    }
    
    int result = solution(coords, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Triple nested loop to check all combinations of three points

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to track maximum area

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Triplets — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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