Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
Given an array of points on the X-Y plane points where points[i] = [xi, yi], return the area of the largest triangle that can be formed by any three different points. Answers within 10-5 of the actual answer will be accepted.
Example 1:
Input: points = [[0,0],[0,1],[1,0],[0,2],[2,0]] Output: 2.00000 Explanation: The five points are shown in the above figure. The red triangle is the largest.
Example 2:
Input: points = [[1,0],[0,0],[0,1]] Output: 0.50000
Constraints:
3 <= points.length <= 50-50 <= xi, yi <= 50Problem summary: Given an array of points on the X-Y plane points where points[i] = [xi, yi], return the area of the largest triangle that can be formed by any three different points. Answers within 10-5 of the actual answer will be accepted.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math
[[0,0],[0,1],[1,0],[0,2],[2,0]]
[[1,0],[0,0],[0,1]]
largest-perimeter-triangle)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #812: Largest Triangle Area
class Solution {
public double largestTriangleArea(int[][] points) {
double ans = 0;
for (int[] p1 : points) {
int x1 = p1[0], y1 = p1[1];
for (int[] p2 : points) {
int x2 = p2[0], y2 = p2[1];
for (int[] p3 : points) {
int x3 = p3[0], y3 = p3[1];
int u1 = x2 - x1, v1 = y2 - y1;
int u2 = x3 - x1, v2 = y3 - y1;
double t = Math.abs(u1 * v2 - u2 * v1) / 2.0;
ans = Math.max(ans, t);
}
}
}
return ans;
}
}
// Accepted solution for LeetCode #812: Largest Triangle Area
func largestTriangleArea(points [][]int) float64 {
ans := 0.0
for _, p1 := range points {
x1, y1 := p1[0], p1[1]
for _, p2 := range points {
x2, y2 := p2[0], p2[1]
for _, p3 := range points {
x3, y3 := p3[0], p3[1]
u1, v1 := x2-x1, y2-y1
u2, v2 := x3-x1, y3-y1
t := float64(abs(u1*v2-u2*v1)) / 2.0
ans = math.Max(ans, t)
}
}
}
return ans
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
# Accepted solution for LeetCode #812: Largest Triangle Area
class Solution:
def largestTriangleArea(self, points: List[List[int]]) -> float:
ans = 0
for x1, y1 in points:
for x2, y2 in points:
for x3, y3 in points:
u1, v1 = x2 - x1, y2 - y1
u2, v2 = x3 - x1, y3 - y1
t = abs(u1 * v2 - u2 * v1) / 2
ans = max(ans, t)
return ans
// Accepted solution for LeetCode #812: Largest Triangle Area
impl Solution {
pub fn largest_triangle_area(points: Vec<Vec<i32>>) -> f64 {
let mut ans: f64 = 0.0;
for point1 in &points {
let (x1, y1) = (point1[0], point1[1]);
for point2 in &points {
let (x2, y2) = (point2[0], point2[1]);
for point3 in &points {
let (x3, y3) = (point3[0], point3[1]);
let u1 = x2 - x1;
let v1 = y2 - y1;
let u2 = x3 - x1;
let v2 = y3 - y1;
let t = ((u1 * v2 - u2 * v1) as f64).abs() / 2.0;
ans = ans.max(t);
}
}
}
ans
}
}
// Accepted solution for LeetCode #812: Largest Triangle Area
function largestTriangleArea(points: number[][]): number {
let ans = 0;
for (const [x1, y1] of points) {
for (const [x2, y2] of points) {
for (const [x3, y3] of points) {
const u1 = x2 - x1,
v1 = y2 - y1;
const u2 = x3 - x1,
v2 = y3 - y1;
const t = Math.abs(u1 * v2 - u2 * v1) / 2;
ans = Math.max(ans, t);
}
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.