Given N points in 2-dimensional space, we need to print the count of the minimum number of lines which traverse through all these N points and which go through a specific (xO, yO) point also.

Examples:

If given points are (-1, 3), (4, 3), (2, 1), (-1, -2), (3, -3) and (xO, yO) point is (1, 0) i.e. every line must go through this point. Then we have to draw at least two lines to cover all these points going through (xO, yO) as shown in below diagram.

We can solve this problem by considering the slope of all points with (xO, yO). If two distinct points have the same slope with (xO, yO) then they can be covered with same line only so we can track slope of each point and whenever we get a new slope we will increase our line count by one.

In below code slope is stored as a pair of integer to get rid of the precision problem and a set is used to keep track of occurred slopes.

Please see below code for better understanding.

// C++ program to get minimum lines to cover // all the points #include <bits/stdc++.h> using namespace std; // Utility method to get gcd of a and b int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } // method returns reduced form of dy/dx as a pair pair<int, int> getReducedForm(int dy, int dx) { int g = gcd(abs(dy), abs(dx)); // get sign of result bool sign = (dy < 0) ^ (dx < 0); if (sign) return make_pair(-abs(dy) / g, abs(dx) / g); else return make_pair(abs(dy) / g, abs(dx) / g); } /* method returns minimum number of lines to cover all points where all lines goes through (xO, yO) */ int minLinesToCoverPoints(int points[][2], int N, int xO, int yO) { // set to store slope as a pair set< pair<int, int> > st; pair<int, int> temp; int minLines = 0; // loop over all points once for (int i = 0; i < N; i++) { // get x and y co-ordinate of current point int curX = points[i][0]; int curY = points[i][1]; temp = getReducedForm(curY - yO, curX - xO); // if this slope is not there in set, // increase ans by 1 and insert in set if (st.find(temp) == st.end()) { st.insert(temp); minLines++; } } return minLines; } // Driver code to test above methods int main() { int xO, yO; xO = 1; yO = 0; int points[][2] = { {-1, 3}, {4, 3}, {2, 1}, {-1, -2}, {3, -3} }; int N = sizeof(points) / sizeof(points[0]); cout << minLinesToCoverPoints(points, N, xO, yO); return 0; }

Output:

2

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