Maximum Number of Collinear Points
Source: Asked to me by a friend - who was asked this question in an interview at Facebook
Problem:
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. *
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. *
1) Take any point x and find the slope from x to every other point - O(n)
ReplyDelete2) 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)
a line is defined by the equation ax+b = y;
ReplyDeleteInitialize 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)
Your complexity analysis assumes that HashMap could be updated in O(1). So, overall complexity should be expected O(n*n).
Deletea line is defined by its equation ax+b = y;
ReplyDeleteInitialize 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)
Can we use something like this ?
ReplyDeleteSort the points based on x and y coordinates.
Then basically use DP approach to get the longest increasing sequence with the same slope