# Points on a Line- Amazon Top Interview Questions

### Problem Statement :

```You are given a two-dimensional list of integers coordinates. Each list contains two integers [x, y] representing a point on the Cartesian plane.

Return the maximum number of points that lie on some line.

Constraints

n ≤ 1,000 where n is the length of coordinates

Example 1

Input

coordinates = [
[5, 1],
[7, 2],
[9, 3],
[0, 0],
[1, 1],
[5, 5],
[6, 6]
]

Output

4

Explanation

The points [[0, 0], [1, 1], [5, 5], [6, 6]] all fall in a line.```

### Solution :

```                        ```Solution in C++ :

int solve(vector<vector<int>>& p) {
int global_max = 0, n = p.size();
unordered_map<double, int> slope_map;
for (int i = 0; i < n; i++) {
slope_map.clear();
int x1 = p[i], y1 = p[i];
int isSamePoint = 0, local_max = 0;
for (int j = 0; j < n; j++) {
int x2 = p[j], y2 = p[j];
if (x1 == x2 and y1 == y2)
isSamePoint++;
else {
double slope = (1.0 * (y1 - y2)) / (x2 - x1);
slope_map[slope]++;
}
}
for (auto& [x, y] : slope_map) local_max = max(local_max, y);
global_max = max(global_max, local_max + isSamePoint);
}
return global_max;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[][] points) {
int n = points.length;
if (n == 0)
return 0;
int ans = 1;
for (int i = 0; i < n; i++) {
var map = new HashMap<Double, Integer>();
for (int j = 0; j < n; j++) {
if (j == i)
continue;
double slope =
((double) points[j] - points[i]) / (points[j] - points[i]);
int count = map.getOrDefault(slope, 0);
map.put(slope, count + 1);
ans = Math.max(ans, count + 2);
}
}
return ans;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, coordinates):
if not coordinates:
return 0
if len(coordinates) <= 2:
return len(coordinates)

max_line = 0
seen = set()
for i, p1 in enumerate(coordinates):
max_count = float("-inf")
slope_counter = collections.Counter()
for j, p2 in enumerate(coordinates):
if i == j:
continue
x1, y1 = p1
x2, y2 = p2
try:
slope = (y2 - y1) / (x2 - x1)
except:
slope = float("inf")
slope_counter[slope] += 1
if slope_counter[slope] > max_count:
max_count = slope_counter[slope]
max_line = max(max_line, max_count + 1)
return max_line```
```

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