Max Points on a Line LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Max Points on a Line in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemMax Points on a Line– LeetCode Problem

Max Points on a Line– LeetCode Problem

Problem:

Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane, return the maximum number of points that lie on the same straight line.

Example 1:

plane1
Input: points = [[1,1],[2,2],[3,3]]
Output: 3

Example 2:

plane2
Input: points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
Output: 4

Constraints:

  • 1 <= points.length <= 300
  • points[i].length == 2
  • -104 <= xi, yi <= 104
  • All the points are unique.
Max Points on a Line– LeetCode Solutions
Max Points on a Line Solution in C++:
class Solution {
 public:
  int maxPoints(vector<vector<int>>& points) {
    int ans = 0;

    for (int i = 0; i < points.size(); ++i) {
      unordered_map<pair<int, int>, int, pairHash> slopeCount;
      const vector<int> p1{points[i]};
      int samePoints = 1;
      int maxPoints = 0;  // maximum number of points with the same slope
      for (int j = i + 1; j < points.size(); ++j) {
        const vector<int> p2{points[j]};
        if (p1 == p2)
          ++samePoints;
        else
          maxPoints = max(maxPoints, ++slopeCount[getSlope(p1, p2)]);
      }
      ans = max(ans, samePoints + maxPoints);
    }

    return ans;
  }

 private:
  pair<int, int> getSlope(const vector<int>& p1, const vector<int>& p2) {
    const int dx = p2[0] - p1[0];
    const int dy = p2[1] - p1[1];
    if (dx == 0)
      return {0, p1[0]};
    if (dy == 0)
      return {p1[1], 0};
    const int d = __gcd(dx, dy);
    return {dx / d, dy / d};
  }

  struct pairHash {
    size_t operator()(const pair<int, int>& p) const {
      return p.first ^ p.second;
    }
  };
};
Max Points on a Line Solution in Java:
class Solution {
  public int maxPoints(int[][] points) {
    int ans = 0;

    for (int i = 0; i < points.length; ++i) {
      Map<Pair<Integer, Integer>, Integer> slopeCount = new HashMap<>();
      int[] p1 = points[i];
      int samePoints = 1;
      int maxPoints = 0; // maximum number of points with the same slope
      for (int j = i + 1; j < points.length; ++j) {
        int[] p2 = points[j];
        if (p1[0] == p2[0] && p1[1] == p2[1])
          ++samePoints;
        else {
          Pair<Integer, Integer> slope = getSlope(p1, p2);
          slopeCount.merge(slope, 1, Integer::sum);
          maxPoints = Math.max(maxPoints, slopeCount.get(slope));
        }
      }
      ans = Math.max(ans, samePoints + maxPoints);
    }

    return ans;
  }

  private Pair<Integer, Integer> getSlope(int[] p1, int[] p2) {
    final int dx = p2[0] - p1[0];
    final int dy = p2[1] - p1[1];
    if (dx == 0)
      return new Pair<>(0, p1[0]);
    if (dy == 0)
      return new Pair<>(p1[1], 0);
    final int d = gcd(dx, dy);
    return new Pair<>(dx / d, dy / d);
  }

  private int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
  }
}
Max Points on a Line Solution in Python:
class Solution:
  def maxPoints(self, points: List[List[int]]) -> int:
    ans = 0

    def gcd(a: int, b: int) -> int:
      return a if b == 0 else gcd(b, a % b)

    def getSlope(p1: List[int], p2: List[int]) -> Tuple[int, int]:
      dx = p2[0] - p1[0]
      dy = p2[1] - p1[1]
      if dx == 0:
        return (0, p1[0])
      if dy == 0:
        return (p1[1], 0)
      d = gcd(dx, dy)
      return (dx // d, dy // d)

    for i, p1 in enumerate(points):
      slopeCount = defaultdict(int)
      samePoints = 1
      maxPoints = 0
      for j in range(i + 1, len(points)):
        p2 = points[j]
        if p1 == p2:
          samePoints += 1
        else:
          slope = getSlope(p1, p2)
          slopeCount[slope] += 1
          maxPoints = max(maxPoints, slopeCount[slope])
      ans = max(ans, samePoints + maxPoints)

    return ans

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