# Accepted java O(n^2) solution using HashMap and slope

• the idea is to basically calculate the slope between each point for a fixed point, and do this for all points.
hash table helps count same slopes.
things to note:

• Beware of duplicate points
• In java, 0 has two values for type Double, 0.0 and -0.0, be careful with these
``````    public class Solution {
public int maxPoints(Point[] points) {
if (points.length <= 2) return points.length;

// map slope to count
HashMap<Double, Integer> hm = new HashMap<Double, Integer>();
int maxCount = 1;

for (int i = 0; i < points.length; i++) {
// hash each slope and count duplicates
hm = new HashMap<Double, Integer>();
int duplicate = 0;
int maxR = 0;

for (int j = i + 1; j < points.length; j++) {
double slope;
if (isDupl(points[i], points[j])) {
duplicate++;
continue;
}
else slope = slopeFor(points[i], points[j]);

if (hm.containsKey(slope)) {
int prevCount = hm.get(slope);
hm.put(slope, prevCount + 1);
maxR = Math.max(maxR, prevCount + 1);
}
else {
hm.put(slope, 1);
maxR = Math.max(maxR, 1);
}
}
maxCount = Math.max(maxCount, maxR + duplicate);
}
return maxCount + 1;
}

private boolean isDupl(Point a, Point b) {
return a.x == b.x && a.y == b.y;
}

private double slopeFor(Point a, Point b) {
double dx = a.x - b.x;
double dy = a.y - b.y;
if (dx == 0) return Double.POSITIVE_INFINITY;
else if (dy == 0) return 0;
return dy / dx;
}
}``````

• what if you have two lines with a tiny difference in slope?
Say this difference is so tiny that double type will not catch it

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