Count maximum points on same line
Given N point on a 2D plane as pair of (x, y) co-ordinates, we need to find maximum number of point which lie on the same line.
Examples:
Input : points[] = {-1, 1}, {0, 0}, {1, 1}, {2, 2}, {3, 3}, {3, 4} Output : 4 Then maximum number of point which lie on same line are 4, those point are {0, 0}, {1, 1}, {2, 2}, {3, 3}
We can solve above problem by following approach – For each point p, calculate its slope with other points and use a map to record how many points have same slope, by which we can find out how many points are on same line with p as their one point. For each point keep doing the same thing and update the maximum number of point count found so far.
Some things to note in implementation are:
1) if two point are (x1, y1) and (x2, y2) then their slope will be (y2 – y1) / (x2 – x1) which can be a double value and can cause precision problems. To get rid of the precision problems, we treat slope as pair ((y2 – y1), (x2 – x1)) instead of ratio and reduce pair by their gcd before inserting into map. In below code points which are vertical or repeated are treated separately.
2) If we use unordered_map in c++ or HashMap in Java for storing the slope pair, then total time complexity of solution will be O(n^2)
C++
/* C/C++ program to find maximum number of point which lie on same line */ #include <bits/stdc++.h> #include <boost/functional/hash.hpp> using namespace std; // method to find maximum collinear point int maxPointOnSameLine(vector< pair< int , int > > points) { int N = points.size(); if (N < 2) return N; int maxPoint = 0; int curMax, overlapPoints, verticalPoints; // here since we are using unordered_map // which is based on hash function //But by default we don't have hash function for pairs //so we'll use hash function defined in Boost library unordered_map<pair< int , int >, int ,boost:: hash<pair< int , int > > > slopeMap; // looping for each point for ( int i = 0; i < N; i++) { curMax = overlapPoints = verticalPoints = 0; // looping from i + 1 to ignore same pair again for ( int j = i + 1; j < N; j++) { // If both point are equal then just // increase overlapPoint count if (points[i] == points[j]) overlapPoints++; // If x co-ordinate is same, then both // point are vertical to each other else if (points[i].first == points[j].first) verticalPoints++; else { int yDif = points[j].second - points[i].second; int xDif = points[j].first - points[i].first; int g = __gcd(xDif, yDif); // reducing the difference by their gcd yDif /= g; xDif /= g; // increasing the frequency of current slope // in map slopeMap[make_pair(yDif, xDif)]++; curMax = max(curMax, slopeMap[make_pair(yDif, xDif)]); } curMax = max(curMax, verticalPoints); } // updating global maximum by current point's maximum maxPoint = max(maxPoint, curMax + overlapPoints + 1); // printf("maximum collinear point // which contains current point // are : %d\n", curMax + overlapPoints + 1); slopeMap.clear(); } return maxPoint; } // Driver code int main() { const int N = 6; int arr[N][2] = {{-1, 1}, {0, 0}, {1, 1}, {2, 2}, {3, 3}, {3, 4}}; vector< pair< int , int > > points; for ( int i = 0; i < N; i++) points.push_back(make_pair(arr[i][0], arr[i][1])); cout << maxPointOnSameLine(points) << endl; return 0; } |
Python3
# python3 program to find maximum number of 2D points that lie on the same line. from collections import defaultdict from math import gcd from typing import DefaultDict, List , Tuple IntPair = Tuple [ int , int ] def normalized_slope(a: IntPair, b: IntPair) - > IntPair: """ Returns normalized (rise, run) tuple. We won't return the actual rise/run result in order to avoid floating point math, which leads to faulty comparisons. See """ run = b[ 0 ] - a[ 0 ] # normalize undefined slopes to (1, 0) if run = = 0 : return ( 1 , 0 ) # normalize to left-to-right if run < 0 : a, b = b, a run = b[ 0 ] - a[ 0 ] rise = b[ 1 ] - a[ 1 ] # Normalize by greatest common divisor. # math.gcd only works on positive numbers. gcd_ = gcd( abs (rise), run) return ( rise / / gcd_, run / / gcd_, ) def maximum_points_on_same_line(points: List [ List [ int ]]) - > int : # You need at least 3 points to potentially have non-collinear points. # For [0, 2] points, all points are on the same line. if len (points) < 3 : return len (points) # Note that every line we find will have at least 2 points. # There will be at least one line because len(points) >= 3. # Therefore, it's safe to initialize to 0. max_val = 0 for a_index in range ( 0 , len (points) - 1 ): # All lines in this iteration go through point a. # Note that lines a-b and a-c cannot be parallel. # Therefore, if lines a-b and a-c have the same slope, they're the same # line. a = tuple (points[a_index]) # Fresh lines already have a, so default=1 slope_counts: DefaultDict[IntPair, int ] = defaultdict( lambda : 1 ) for b_index in range (a_index + 1 , len (points)): b = tuple (points[b_index]) slope_counts[normalized_slope(a, b)] + = 1 max_val = max ( max_val, max (slope_counts.values()), ) return max_val print (maximum_points_on_same_line([ [ - 1 , 1 ], [ 0 , 0 ], [ 1 , 1 ], [ 2 , 2 ], [ 3 , 3 ], [ 3 , 4 ], ])) # This code is contributed by Jose Alvarado Torre |
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