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2152. Minimum Number of Lines to Cover Points

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

  • 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);
        }
    };
    
  • 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)
    
    
  • 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)
    }
    

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