Example 1:

Input: coordinates = [[1,2],[2,3],[3,4],[4,5],[5,6],[6,7]]
Output: true

Example 2:

Input: coordinates = [[1,1],[2,2],[3,4],[4,5],[5,6],[7,7]]
Output: false

Constraints:

• 2 <= coordinates.length <= 1000
• coordinates[i].length == 2
• -10^4 <= coordinates[i][0], coordinates[i][1] <= 10^4
• coordinates contains no duplicate point.

Solution: Slope and Hashset

This is not a easy problem, a few corner cases:

• dx == 0
• dy == 0
• dx < 0

Basically we are counting (dx / gcd(dx, dy), dy / gcd(dx, dy)). We will have only ONE entry if all the points are on the same line.

Time complexity: O(n)
Space complexity: O(1) w/ early exit.

// Author: Huahua
class Solution {
public:
  bool checkStraightLine(vector<vector<int>>& coordinates) {
    set<pair<int, int>> s;
    for (int i = 1; i < coordinates.size(); ++i) {      
      int dx = coordinates[i][0] - coordinates[0][0];
      int dy = coordinates[i][1] - coordinates[0][1];      
      if (dy == 0)
        s.emplace(1, 0);
      else if (dx == 0)
        s.emplace(0, 1);
      else {
        if (dx < 0) dx *= -1, dy *= -1;
        const int d = gcd(dx, dy);
        s.emplace(dx / d, dy / d);
      }
      if (s.size() > 1) return false;
    }
    return true;
  }
};

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