# 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++ 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; ` `} ` |

*chevron_right*

*filter_none*

Output:

4

This article is contributed by **Utkarsh Trivedi**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Count distinct points visited on the number line
- Count of obtuse angles in a circle with 'k' equidistant points between 2 given points
- Represent a given set of points by the best possible straight line
- Check whether two points (x1, y1) and (x2, y2) lie on same side of a given line or not
- Find X and Y intercepts of a line passing through the given points
- Find points at a given distance on a line of given slope
- Program to find line passing through 2 Points
- Number of horizontal or vertical line segments to connect 3 points
- Number of ordered points pair satisfying line equation
- Maximum points of intersection n circles
- Maximum number of segments that can contain the given points
- Maximum points of intersection n lines
- Find the maximum possible distance from origin using given points
- Queries on count of points lie inside a circle
- Count Integral points inside a Triangle