Problem Description

You are given an array of points on a 2D coordinate plane, where each point is represented as [x, y]. Your task is to find the maximum area of a triangle that can be formed by selecting any three different points from the given array.

The problem asks you to:

1. Consider all possible combinations of three points from the input array
2. Calculate the area of the triangle formed by each combination of three points
3. Return the largest area found

The answer should be accurate within 10^-5 of the actual value, meaning small floating-point differences are acceptable.

The solution uses a brute force approach with three nested loops to check every possible combination of three points. For each triplet of points (x1, y1), (x2, y2), and (x3, y3), it calculates the triangle area using the cross product formula:

• First, it computes two vectors from the first point to the other two points: (u1, v1) = (x2 - x1, y2 - y1) and (u2, v2) = (x3 - x1, y3 - y1)
• Then it calculates the area as |u1 * v2 - u2 * v1| / 2, which is half the absolute value of the cross product
• The maximum area across all possible triangles is tracked and returned

This formula works because the cross product of two vectors gives the area of the parallelogram they form, and a triangle is exactly half of that parallelogram.

Intuition

When we need to find the largest triangle from a set of points, the key insight is that we must check all possible combinations since there's no way to predict which three points will form the largest triangle just by looking at their coordinates. The triangle with maximum area might not necessarily include the points that are farthest apart.

To calculate the area of a triangle given three points, we can use the geometric property that relates triangles to parallelograms. If we place one vertex at the origin and create two vectors to the other vertices, these vectors define a parallelogram. The triangle formed by the three points is exactly half of this parallelogram.

The cross product of two vectors gives us the area of the parallelogram they span. Mathematically, if we have vectors (u1, v1) and (u2, v2), their cross product magnitude is |u1 * v2 - u2 * v1|. This represents the parallelogram area, so dividing by 2 gives us the triangle area.

We translate this concept to our three points by:

1. Treating one point (x1, y1) as a reference (like placing it at the origin conceptually)
2. Creating vectors from this point to the other two points: (x2 - x1, y2 - y1) and (x3 - x1, y3 - y1)
3. Applying the cross product formula and dividing by 2

Since we don't know which three points will give the maximum area, we systematically try every possible combination of three points using three nested loops, calculate each triangle's area, and keep track of the maximum value found.

Learn more about Math patterns.

Solution Approach

The solution implements a brute force approach that examines all possible triangles formed by selecting three points from the given array.

Algorithm Steps:

1. Initialize the maximum area: Start with ans = 0 to track the largest triangle area found.

2. Triple nested iteration: Use three nested loops to generate all possible combinations of three points:

   • First loop selects point (x1, y1)
   • Second loop selects point (x2, y2)
   • Third loop selects point (x3, y3)

3. Calculate vectors: For each combination of three points, create two vectors from the first point to the other two:

   • Vector 1: (u1, v1) = (x2 - x1, y2 - y1)
   • Vector 2: (u2, v2) = (x3 - x1, y3 - y1)

4. Compute triangle area: Apply the cross product formula:

   • Calculate the cross product: u1 * v2 - u2 * v1
   • Take the absolute value to ensure positive area
   • Divide by 2 to get the triangle area: t = abs(u1 * v2 - u2 * v1) / 2

5. Update maximum: Compare the current triangle area with the maximum found so far: ans = max(ans, t)

6. Return result: After checking all combinations, return the maximum area found.

Time Complexity: O(n³) where n is the number of points, since we have three nested loops iterating through all points.

Space Complexity: O(1) as we only use a constant amount of extra space for variables to store coordinates and calculated values.

The beauty of this approach lies in its simplicity - by systematically checking every possible triangle and using the efficient cross product formula for area calculation, we guarantee finding the maximum area without needing complex geometric optimizations.

Example Walkthrough

Let's walk through the solution with a small example to see how it finds the maximum triangle area.

Example Input: points = [[0,0], [0,2], [2,0], [1,1]]

We'll systematically check all possible combinations of three points:

Combination 1: Points [0,0], [0,2], [2,0]

• First point (x1, y1) = (0, 0)
• Second point (x2, y2) = (0, 2)
• Third point (x3, y3) = (2, 0)
• Vector 1: (u1, v1) = (0-0, 2-0) = (0, 2)
• Vector 2: (u2, v2) = (2-0, 0-0) = (2, 0)
• Cross product: u1 * v2 - u2 * v1 = 0 * 0 - 2 * 2 = -4
• Area = |−4| / 2 = 2

Combination 2: Points [0,0], [0,2], [1,1]

• First point (x1, y1) = (0, 0)
• Second point (x2, y2) = (0, 2)
• Third point (x3, y3) = (1, 1)
• Vector 1: (u1, v1) = (0-0, 2-0) = (0, 2)
• Vector 2: (u2, v2) = (1-0, 1-0) = (1, 1)
• Cross product: u1 * v2 - u2 * v1 = 0 * 1 - 1 * 2 = -2
• Area = |−2| / 2 = 1

Combination 3: Points [0,0], [2,0], [1,1]

• First point (x1, y1) = (0, 0)
• Second point (x2, y2) = (2, 0)
• Third point (x3, y3) = (1, 1)
• Vector 1: (u1, v1) = (2-0, 0-0) = (2, 0)
• Vector 2: (u2, v2) = (1-0, 1-0) = (1, 1)
• Cross product: u1 * v2 - u2 * v1 = 2 * 1 - 1 * 0 = 2
• Area = |2| / 2 = 1

Combination 4: Points [0,2], [2,0], [1,1]

• First point (x1, y1) = (0, 2)
• Second point (x2, y2) = (2, 0)
• Third point (x3, y3) = (1, 1)
• Vector 1: (u1, v1) = (2-0, 0-2) = (2, -2)
• Vector 2: (u2, v2) = (1-0, 1-2) = (1, -1)
• Cross product: u1 * v2 - u2 * v1 = 2 * (-1) - 1 * (-2) = -2 + 2 = 0
• Area = |0| / 2 = 0

The maximum area found is 2, which comes from the triangle formed by points [0,0], [0,2], and [2,0]. This makes intuitive sense as these three points form a right triangle with legs of length 2 each, giving an area of (2 * 2) / 2 = 2.

Notice how the algorithm doesn't miss any combination - it checks all 4 possible triangles from our 4 points. The fourth combination turned out to have zero area because those three points are collinear (they all lie on the same line y = 2 - x).

Solution Implementation

Time and Space Complexity

Time Complexity: O(n³) where n is the number of points in the input list.

The code uses three nested loops, each iterating through all points in the list. For each combination of three points, it performs constant-time operations to calculate the area using the cross product formula. Since we're examining all possible combinations of three points from n points, the total number of iterations is n × n × n = n³.

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space. It stores a few variables (ans, x1, y1, x2, y2, x3, y3, u1, v1, u2, v2, t) regardless of the input size. No additional data structures are created that grow with the input size.

The area calculation uses the formula: Area = |u1 * v2 - u2 * v1| / 2, where (u1, v1) and (u2, v2) are vectors from the first point to the second and third points respectively. This is derived from the cross product formula for triangle area.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Selecting the Same Point Multiple Times

The current implementation doesn't prevent selecting the same point multiple times. If we pick the same point twice or three times, we get a degenerate triangle with zero area. While this doesn't affect correctness (area = 0 won't be maximum), it wastes computation time.

Solution: Add checks to ensure we're selecting three distinct points:

for i in range(len(points)):
    for j in range(i + 1, len(points)):
        for k in range(j + 1, len(points)):
            # Now guaranteed to have three different points

2. Integer Overflow in Cross Product Calculation

When dealing with large coordinate values, the cross product calculation vector1_x * vector2_y - vector2_x * vector1_y might overflow if using integer arithmetic in other languages. While Python handles large integers automatically, this is a concern in languages like C++ or Java.

Solution: Use appropriate data types (double/float) for intermediate calculations or check for potential overflow conditions.

3. Floating Point Precision Issues

Division by 2.0 and floating-point arithmetic can introduce small precision errors. While the problem allows for 10^-5 tolerance, accumulated errors in extreme cases might be problematic.

Solution: Consider using integer arithmetic until the final division:

# Keep as integer until the end
cross_product = vector1_x * vector2_y - vector2_x * vector1_y
area = abs(cross_product) / 2.0  # Only divide once at the end

4. Inefficient Triple Loop Without Early Termination

The naive triple loop checks many redundant combinations (e.g., points (A,B,C) and (B,A,C) form the same triangle).

Solution: Use index-based loops to avoid redundant combinations:

n = len(points)
for i in range(n - 2):
    for j in range(i + 1, n - 1):
        for k in range(j + 1, n):
            x1, y1 = points[i]
            x2, y2 = points[j]
            x3, y3 = points[k]
            # Calculate area...

This reduces the number of iterations from n³ to C(n,3) = n(n-1)(n-2)/6.

5. Not Handling Edge Cases

The code doesn't explicitly check for edge cases like fewer than 3 points in the input array.

Solution: Add validation at the beginning:

if len(points) < 3:
    return 0.0