Maximum Number of Collinear Points

Source: Asked to me by a friend - who was asked this question in an interview at Facebook

Problem:

Given n points on a 2D plane, find the equation of the line with maximum number of collinear points. What is the time complexity of your algorithm?








Solution:

Highlight the part between the * symbols for the answer.
* O(n^2) algorithm: the optimal line passes through at least two of the points (otherwise rotate/translate it until it does without losing collinear points). So for each point p, compute the slope from p to every other point and tally frequencies in a hash map (use exact rational slopes dy/dx reduced by gcd, plus a sentinel for vertical lines - avoid floating point). The most frequent slope from p, plus p itself, gives the best line through p. Take the maximum over all p.

Complexity: n points x O(n) slope computations and hash updates = O(n^2) expected time, O(n) space per round. (Deterministic O(n^2 log n) via sorting slopes instead of hashing.) This is worst-case optimal up to log factors - the problem is 3SUM-hard.

Solution by piyushmahesh (per-point slope tally) and an Anonymous commenter (per-pair line hashing) from the comments. *

Comments

  1. piyushmahesh11:47 PM, December 29, 2014

    1) Take any point x and find the slope from x to every other point - O(n)
    2) Find the number of points for each slope by hashing or building a binary search tree - O(nlogn)
    3) Find the max number of point for any slope
    4) Repeat this for every point and find the max among them

    Total time complexity - O(n*n*logn)

    ReplyDelete
  2. Unknown4:18 AM, October 31, 2015

    a line is defined by the equation ax+b = y;

    Initialize a hashmap with key (a,b) and an integer value;

    for each edge of two points :
    compute a,b and increment the integer value associated with (a,b) in the hashmap
    O(n*n)

    compute max of the values in the hashmap and return key associated
    O(n*n)

    Total time complexity O(n*n)

    ReplyDelete
    Replies
    1. Unknown10:09 PM, December 06, 2018

      Your complexity analysis assumes that HashMap could be updated in O(1). So, overall complexity should be expected O(n*n).

      Delete
  3. Unknown4:22 AM, October 31, 2015

    a line is defined by its equation ax+b = y;

    Initialize a hashmap with key (a, b) and integer value

    for each edge of two points, compute a and b,
    increment the value associated with key (a,b)
    O(n*n)

    compute the max of values in the hashmap and return key associated
    O(n*n)

    Total time complexity : O(n*n)

    ReplyDelete
  4. Unknown1:52 AM, December 07, 2015

    Can we use something like this ?

    Sort the points based on x and y coordinates.

    Then basically use DP approach to get the longest increasing sequence with the same slope

    ReplyDelete
    Replies
    1. Pratik Poddar3:10 PM, November 06, 2026

      Replying to Unknown's comment, Can we use something like this ?Sort the points based on x and y coordinates. - Sorting does not linearize the problem: equal CONSECUTIVE slopes after sorting by x do certify collinearity, but a DP must track the slope against every previous point - the state space is O(n) slopes per element and you end up at O(n^2) states with expensive transitions. The slope-hash per anchor point is simpler and optimal at expected O(n^2). (Reply created using AI)

      Delete

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