LeetCode #149 — HARD

Max Points on a Line

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given an array of points where points[i] = [x_i, y_i] represents a point on the X-Y plane, return the maximum number of points that lie on the same straight line.

Example 1:

Input: points = [[1,1],[2,2],[3,3]]
Output: 3

Example 2:

Input: points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
Output: 4

Constraints:

1 <= points.length <= 300
points[i].length == 2
-10⁴ <= x_i, y_i <= 10⁴
All the points are unique.

Step 01

Brute Force Baseline

Problem summary: Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane, return the maximum number of points that lie on the same straight line.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math

Example 1

[[1,1],[2,2],[3,3]]

Example 2

[[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #149: Max Points on a Line
class Solution {
    public int maxPoints(int[][] points) {
        int n = points.length;
        int ans = 1;
        for (int i = 0; i < n; ++i) {
            int x1 = points[i][0], y1 = points[i][1];
            for (int j = i + 1; j < n; ++j) {
                int x2 = points[j][0], y2 = points[j][1];
                int cnt = 2;
                for (int k = j + 1; k < n; ++k) {
                    int x3 = points[k][0], y3 = points[k][1];
                    int a = (y2 - y1) * (x3 - x1);
                    int b = (y3 - y1) * (x2 - x1);
                    if (a == b) {
                        ++cnt;
                    }
                }
                ans = Math.max(ans, cnt);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #149: Max Points on a Line
func maxPoints(points [][]int) int {
	n := len(points)
	ans := 1
	for i := 0; i < n; i++ {
		x1, y1 := points[i][0], points[i][1]
		for j := i + 1; j < n; j++ {
			x2, y2 := points[j][0], points[j][1]
			cnt := 2
			for k := j + 1; k < n; k++ {
				x3, y3 := points[k][0], points[k][1]
				a := (y2 - y1) * (x3 - x1)
				b := (y3 - y1) * (x2 - x1)
				if a == b {
					cnt++
				}
			}
			if ans < cnt {
				ans = cnt
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #149: Max Points on a Line
class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        n = len(points)
        ans = 1
        for i in range(n):
            x1, y1 = points[i]
            for j in range(i + 1, n):
                x2, y2 = points[j]
                cnt = 2
                for k in range(j + 1, n):
                    x3, y3 = points[k]
                    a = (y2 - y1) * (x3 - x1)
                    b = (y3 - y1) * (x2 - x1)
                    cnt += a == b
                ans = max(ans, cnt)
        return ans

// Accepted solution for LeetCode #149: Max Points on a Line
use std::collections::HashMap;

impl Solution {
    pub fn max_points(points: Vec<Vec<i32>>) -> i32 {
        let mut res = 1;

        for i in 0..points.len() {
            let point1 = &points[i];
            let mut count = HashMap::new();
            for j in (i + 1)..points.len() {
                let point2 = &points[j];
                let slope;
                if point2[0] == point1[0] {
                    slope = i32::MAX;
                } else {
                    slope = ((point2[1] as f64 - point1[1] as f64)
                        / (point2[0] as f64 - point1[0] as f64)
                        * 100000.0) as i32;
                }

                *count.entry(slope).or_insert(1) += 1;
                res = res.max(*count.get(&slope).unwrap());
            }
        }
        res
    }
}

// Accepted solution for LeetCode #149: Max Points on a Line
function maxPoints(points: number[][]): number {
    let res = 1;

    for (let i = 0; i < points.length; i++) {
        const count = new Map();
        const point1 = points[i];
        for (let j = i + 1; j < points.length; j++) {
            const point2 = points[j];
            let slope;
            if (point2[0] === point1[0]) {
                slope = Number.MAX_SAFE_INTEGER;
            } else {
                slope = (point2[1] - point1[1]) / (point2[0] - point1[0]);
            }
            !count.has(slope)
                ? count.set(slope, 2)
                : count.set(slope, count.get(slope) + 1);

            res = Math.max(res, count.get(slope));
        }
    }
    return res;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

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.

OPTIMIZED

O(n) time

O(1) space

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.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.