# 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 colinear 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 colinear 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; ` `} ` |

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## Python3

`def` `maxPoints(points): ` ` ` `n` `=` `len` `(points) ` ` ` ` ` `# upto two points all points will be part of the line ` ` ` `if` `n<` `3` `: ` ` ` `return` `n ` ` ` ` ` `max_val` `=` `0` ` ` ` ` `# looping for each point ` ` ` `for` `i ` `in` `points: ` ` ` ` ` `# Creating a dictionary for every new ` ` ` `# point to save memory ` ` ` `d ` `=` `{} ` ` ` `dups ` `=` `0` ` ` `cur_max ` `=` `0` ` ` ` ` `# pairing with all other points ` ` ` `for` `j ` `in` `points: ` ` ` `if` `i!` `=` `j: ` ` ` `if` `j[` `0` `]` `=` `=` `i[` `0` `]: ` `#vertical line ` ` ` `slope` `=` `'inf'` ` ` `else` `: ` ` ` `slope` `=` `float` `(j[` `1` `]` `-` `i[` `1` `])` `/` `float` `(j[` `0` `]` `-` `i[` `0` `]) ` ` ` ` ` `# Increasing the frequency of slope and ` ` ` `# updating cur_max for current point(i) ` ` ` `d[slope] ` `=` `d.get(slope,` `0` `)` `+` `1` ` ` `cur_max` `=` `max` `(cur_max, d[slope]) ` ` ` ` ` `# if both points are equal same increase ` ` ` `# duplicates count. ` ` ` `# Please note that this will also increment ` ` ` `# when we map it with itself. ` ` ` `# we still do it because we will not have to ` ` ` `# add the extra one at the end. ` ` ` `else` `: ` ` ` `dups` `+` `=` `1` ` ` ` ` `max_val` `=` `max` `(max_val, cur_max` `+` `dups) ` ` ` ` ` `return` `max_val ` ` ` `# Driver code ` `points ` `=` `[(` `-` `1` `, ` `1` `), (` `0` `, ` `0` `), (` `1` `, ` `1` `), (` `2` `, ` `2` `), (` `3` `, ` `3` `), (` `3` `, ` `4` `)] ` `print` `(maxPoints(points)) ` |

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