0950HardArrayHash TableMathGeometryEuclidean AlgorithmGreatest Common Divisor

Most Pins on One Straight Line

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Pin positions are given as points, each [x, y], and all of them are different.

Return the greatest number of pins that lie on a single straight line.

Examples

Example 1

Input: points = [[2, 7], [4, 11], [6, 15], [9, 2], [12, 2]]
Output: 3

The pins at 2 and 7, 4 and 11, and 6 and 15 all lie on one line, climbing two across and four up each step, so three pins share it and no line holds more.

Example 2

Input: points = [[0, 0], [0, 1], [0, 2], [0, 3]]
Output: 4

All four pins sit in one column, which is a straight line.

Example 3

Input: points = [[0, 0]]
Output: 1

A single pin lies on a line of its own.

Constraints

1 <= points.length <= 300
points[i].length == 2
-10000 <= points[i][0] <= 10000
-10000 <= points[i][1] <= 10000
All the pin positions are different

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def max_points(points: list[list[int]]) -> int:

Java

public int maxPoints(int[][] points)

The intended approach

Time O(n^2 log V) for n pins with values up to V, the logarithm being the divisor step.Space O(n) for the groups around one pin.

Any line holding two or more pins passes through some pin, so fix one pin and group the others by the direction from it: pins in the same direction, or the exact opposite, sit on one line through it. The largest group plus the fixed pin itself is the best line through that pin, and taking the best over all the fixed pins gives the answer. Naming a direction has to be exact, so avoid dividing. Take the step in `x` and `y`, divide both by their greatest common divisor, and settle the sign the same way every time, say by flipping both when the `x` step is negative, or when it is zero and the `y` step is negative. That gives one name per line direction with no rounding anywhere. A single pin on its own is a line holding one. Settling the sign is a safe habit rather than a necessity here: every line has an outermost pin, and from that pin the rest of the line lies one way, so the count is found there even if a direction and its opposite were counted apart. Doing it anyway means the answer does not depend on that argument.

Pythonentry point: max_points

def max_points(points):
    from math import gcd
    n = len(points)
    if n < 3:
        return n
    best = 1
    for i in range(n):
        groups = {}
        for j in range(n):
            if i == j:
                continue
            dx = points[j][0] - points[i][0]
            dy = points[j][1] - points[i][1]
            g = gcd(abs(dx), abs(dy))
            if g:
                dx //= g
                dy //= g
            if dx < 0 or (dx == 0 and dy < 0):
                dx = -dx
                dy = -dy
            key = (dx, dy)
            groups[key] = groups.get(key, 0) + 1
        for key in groups:
            if groups[key] + 1 > best:
                best = groups[key] + 1
    return best

The obvious version

Take every pair of pins as a line and count how many pins lie on it, testing each with a cross product so nothing is divided. Cubic in the pin count and it groups nothing.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_max_points(points):
    n = len(points)
    if n < 3:
        return n
    best = 1
    for i in range(n):
        for j in range(i + 1, n):
            ax = points[j][0] - points[i][0]
            ay = points[j][1] - points[i][1]
            count = 0
            for k in range(n):
                bx = points[k][0] - points[i][0]
                by = points[k][1] - points[i][1]
                if ax * by - ay * bx == 0:
                    count += 1
            if count > best:
                best = count
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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