CPMagical CP Library

Problem solution · Java

Minimum Number of Lines to Cover Points

Minimum Number of Lines to Cover Points: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
55 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Minimum Number of Lines to Cover Points, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 55 lines of Java from the credited upstream file 2152.java.
  • The implementation visibly relies on sequence storage.
  • 3 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Number of Lines to Cover Points · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int minimumLines(int[][] points) {    final int allCovered = (1 << points.length) - 1;    int[] mem = new int[allCovered];    Arrays.fill(mem, -1);    return dfs(points, 0, allCovered, mem);  }   private int dfs(int[][] points, int covered, int allCovered, int[] mem) {    if (covered == allCovered)      return 0;    if (mem[covered] != -1)      return mem[covered];     final int n = points.length;    int ans = n / 2 + ((n & 1) == 1 ? 1 : 0);     for (int i = 0; i < n; ++i) {      if ((covered >> i & 1) == 1)        continue;      for (int j = 0; j < n; ++j) {        if (i == j)          continue;        // Connect the points[i] with the points[j].        int newCovered = covered | 1 << i | 1 << j;        // Mark the points covered by this line.        Pair<Integer, Integer> slope = getSlope(points[i], points[j]);        for (int k = 0; k < n; ++k)          if (getSlope(points[i], points[k]).equals(slope))            newCovered |= 1 << k;        ans = Math.min(ans, 1 + dfs(points, newCovered, allCovered, mem));      }    }     return mem[covered] = ans;  }   private Pair<Integer, Integer> getSlope(int[] p, int[] q) {    final int dx = p[0] - q[0];    final int dy = p[1] - q[1];    if (dx == 0)      return new Pair<>(0, p[0]);    if (dy == 0)      return new Pair<>(p[1], 0);    final int d = gcd(dx, dy);    final int x = dx / d;    final int y = dy / d;    return x > 0 ? new Pair<>(x, y) : new Pair<>(-x, -y);  }   private int gcd(int a, int b) {    return b == 0 ? a : gcd(b, a % b);  }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗