Question

2152. Minimum Number of Lines to Cover Points

中文文档

Description

You are given an array points where points[i] = [xi, yi] represents a point on an X-Y plane.

Straight lines are going to be added to the X-Y plane, such that every point is covered by at least one line.

Return _the minimum number of straight lines needed to cover all the points_.

 

Example 1:

Input: points = [[0,1],[2,3],[4,5],[4,3]]
Output: 2
Explanation: The minimum number of straight lines needed is two. One possible solution is to add:
- One line connecting the point at (0, 1) to the point at (4, 5).
- Another line connecting the point at (2, 3) to the point at (4, 3).

Example 2:

Input: points = [[0,2],[-2,-2],[1,4]]
Output: 1
Explanation: The minimum number of straight lines needed is one. The only solution is to add:
- One line connecting the point at (-2, -2) to the point at (1, 4).

 

Constraints:

• 1 <= points.length <= 10
• points[i].length == 2
• -100 <= xi, yi <= 100
• All the points are unique.

Solutions

Solution 1

class Solution:
  def minimumLines(self, points: List[List[int]]) -> int:
    def check(i, j, k):
      x1, y1 = points[i]
      x2, y2 = points[j]
      x3, y3 = points[k]
      return (x2 - x1) * (y3 - y1) == (x3 - x1) * (y2 - y1)

    @cache
    def dfs(state):
      if state == (1 << n) - 1:
        return 0
      ans = inf
      for i in range(n):
        if not (state >> i & 1):
          for j in range(i + 1, n):
            nxt = state | 1 << i | 1 << j
            for k in range(j + 1, n):
              if not (nxt >> k & 1) and check(i, j, k):
                nxt |= 1 << k
            ans = min(ans, dfs(nxt) + 1)
          if i == n - 1:
            ans = min(ans, dfs(state | 1 << i) + 1)
      return ans

    n = len(points)
    return dfs(0)

class Solution {
  private int[] f;
  private int[][] points;
  private int n;

  public int minimumLines(int[][] points) {
    n = points.length;
    this.points = points;
    f = new int[1 << n];
    return dfs(0);
  }

  private int dfs(int state) {
    if (state == (1 << n) - 1) {
      return 0;
    }
    if (f[state] != 0) {
      return f[state];
    }
    int ans = 1 << 30;
    for (int i = 0; i < n; ++i) {
      if (((state >> i) & 1) == 0) {
        for (int j = i + 1; j < n; ++j) {
          int nxt = state | 1 << i | 1 << j;
          for (int k = j + 1; k < n; ++k) {
            if (((state >> k) & 1) == 0 && check(i, j, k)) {
              nxt |= 1 << k;
            }
          }
          ans = Math.min(ans, dfs(nxt) + 1);
        }
        if (i == n - 1) {
          ans = Math.min(ans, dfs(state | 1 << i) + 1);
        }
      }
    }
    return f[state] = ans;
  }

  private boolean check(int i, int j, int k) {
    int x1 = points[i][0], y1 = points[i][1];
    int x2 = points[j][0], y2 = points[j][1];
    int x3 = points[k][0], y3 = points[k][1];
    return (x2 - x1) * (y3 - y1) == (x3 - x1) * (y2 - y1);
  }
}

class Solution {
public:
  int minimumLines(vector<vector<int>>& points) {
    auto check = [&](int i, int j, int k) {
      int x1 = points[i][0], y1 = points[i][1];
      int x2 = points[j][0], y2 = points[j][1];
      int x3 = points[k][0], y3 = points[k][1];
      return (x2 - x1) * (y3 - y1) == (x3 - x1) * (y2 - y1);
    };
    int n = points.size();
    int f[1 << n];
    memset(f, 0, sizeof f);
    function<int(int)> dfs = [&](int state) -> int {
      if (state == (1 << n) - 1) return 0;
      if (f[state]) return f[state];
      int ans = 1 << 30;
      for (int i = 0; i < n; ++i) {
        if (!(state >> i & 1)) {
          for (int j = i + 1; j < n; ++j) {
            int nxt = state | 1 << i | 1 << j;
            for (int k = j + 1; k < n; ++k) {
              if (!(nxt >> k & 1) && check(i, j, k)) {
                nxt |= 1 << k;
              }
            }
            ans = min(ans, dfs(nxt) + 1);
          }
          if (i == n - 1) {
            ans = min(ans, dfs(state | 1 << i) + 1);
          }
        }
      }
      return f[state] = ans;
    };
    return dfs(0);
  }
};

func minimumLines(points [][]int) int {
  check := func(i, j, k int) bool {
    x1, y1 := points[i][0], points[i][1]
    x2, y2 := points[j][0], points[j][1]
    x3, y3 := points[k][0], points[k][1]
    return (x2-x1)*(y3-y1) == (x3-x1)*(y2-y1)
  }
  n := len(points)
  f := make([]int, 1<<n)
  var dfs func(int) int
  dfs = func(state int) int {
    if state == (1<<n)-1 {
      return 0
    }
    if f[state] > 0 {
      return f[state]
    }
    ans := 1 << 30
    for i := 0; i < n; i++ {
      if (state >> i & 1) == 0 {
        for j := i + 1; j < n; j++ {
          nxt := state | 1<<i | 1<<j
          for k := j + 1; k < n; k++ {
            if (nxt>>k&1) == 0 && check(i, j, k) {
              nxt |= 1 << k
            }
          }
          ans = min(ans, dfs(nxt)+1)
        }
        if i == n-1 {
          ans = min(ans, dfs(state|1<<i)+1)
        }
      }
    }
    f[state] = ans
    return ans
  }
  return dfs(0)
}