problemmediumalgorithmsleetcode-2152leetcode 2152leetcode2152

Minimum Number of Lines to Cover Points

MediumUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

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 theminimum number of straight lines needed to cover all the points.

Examples

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.

Solution

Method 1 – DP with Bitmask (Backtracking)

Intuition

For each subset of points, we can cover them with a single line if they are collinear. Use DP with bitmask: for each mask, try all possible lines, and recursively cover the rest. The answer is the minimum number of lines needed to cover all points.

Approach

  1. For each pair of points, find all points collinear with them.
  2. Use DP: dp[mask] = min number of lines to cover points in mask.
  3. For each mask, try all possible lines, update dp.
  4. Return dp[full_mask].

Code

C++
#include <vector>
using namespace std;
class Solution {
public:
    int minLines(vector<vector<int>>& points) {
        int n = points.size();
        vector<vector<int>> col(n, vector<int>(n, 0));
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                int mask = 0;
                for (int k = 0; k < n; ++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];
                    if ((x2-x1)*(y3-y1) == (y2-y1)*(x3-x1)) mask |= 1<<k;
                }
                col[i][j] = mask;
            }
        }
        vector<int> dp(1<<n, n);
        dp[0] = 0;
        for (int mask = 0; mask < (1<<n); ++mask) {
            if (dp[mask] == n) continue;
            int first = 0;
            while (first < n && !(mask & (1<<first))) ++first;
            if (first == n) continue;
            for (int j = 0; j < n; ++j) {
                int new_mask = mask | col[first][j];
                dp[new_mask] = min(dp[new_mask], dp[mask] + 1);
            }
        }
        return dp[(1<<n)-1];
    }
};
Go
func minLines(points [][]int) int {
    n := len(points)
    col := make([][]int, n)
    for i := range col { col[i] = make([]int, n) }
    for i := 0; i < n; i++ {
        for j := 0; j < n; j++ {
            mask := 0
            for k := 0; k < n; k++ {
                x1, y1 := points[i][0], points[i][1]
                x2, y2 := points[j][0], points[j][1]
                x3, y3 := points[k][0], points[k][1]
                if (x2-x1)*(y3-y1) == (y2-y1)*(x3-x1) {
                    mask |= 1 << k
                }
            }
            col[i][j] = mask
        }
    }
    dp := make([]int, 1<<n)
    for i := range dp { dp[i] = n }
    dp[0] = 0
    for mask := 0; mask < 1<<n; mask++ {
        if dp[mask] == n { continue }
        first := 0
        for first < n && (mask&(1<<first)) == 0 { first++ }
        if first == n { continue }
        for j := 0; j < n; j++ {
            newMask := mask | col[first][j]
            if dp[newMask] > dp[mask]+1 {
                dp[newMask] = dp[mask]+1
            }
        }
    }
    return dp[(1<<n)-1]
}
Java
class Solution {
    public int minLines(int[][] points) {
        int n = points.length;
        int[][] col = new int[n][n];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                int mask = 0;
                for (int k = 0; k < n; ++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];
                    if ((x2-x1)*(y3-y1) == (y2-y1)*(x3-x1)) mask |= 1<<k;
                }
                col[i][j] = mask;
            }
        }
        int[] dp = new int[1<<n];
        java.util.Arrays.fill(dp, n);
        dp[0] = 0;
        for (int mask = 0; mask < (1<<n); ++mask) {
            if (dp[mask] == n) continue;
            int first = 0;
            while (first < n && (mask & (1<<first)) == 0) ++first;
            if (first == n) continue;
            for (int j = 0; j < n; ++j) {
                int new_mask = mask | col[first][j];
                dp[new_mask] = Math.min(dp[new_mask], dp[mask] + 1);
            }
        }
        return dp[(1<<n)-1];
    }
}
Kotlin
class Solution {
    fun minLines(points: Array<IntArray>): Int {
        val n = points.size
        val col = Array(n) { IntArray(n) }
        for (i in 0 until n) {
            for (j in 0 until n) {
                var mask = 0
                for (k in 0 until n) {
                    val (x1, y1) = points[i]
                    val (x2, y2) = points[j]
                    val (x3, y3) = points[k]
                    if ((x2-x1)*(y3-y1) == (y2-y1)*(x3-x1)) mask = mask or (1 shl k)
                }
                col[i][j] = mask
            }
        }
        val dp = IntArray(1 shl n) { n }
        dp[0] = 0
        for (mask in 0 until (1 shl n)) {
            if (dp[mask] == n) continue
            var first = 0
            while (first < n && (mask and (1 shl first)) == 0) first++
            if (first == n) continue
            for (j in 0 until n) {
                val newMask = mask or col[first][j]
                dp[newMask] = minOf(dp[newMask], dp[mask] + 1)
            }
        }
        return dp[(1 shl n) - 1]
    }
}
Python
class Solution:
    def minLines(self, points: list[list[int]]) -> int:
        n = len(points)
        col = [[0]*n for _ in range(n)]
        for i in range(n):
            for j in range(n):
                mask = 0
                for k in range(n):
                    x1, y1 = points[i]
                    x2, y2 = points[j]
                    x3, y3 = points[k]
                    if (x2-x1)*(y3-y1) == (y2-y1)*(x3-x1):
                        mask |= 1<<k
                col[i][j] = mask
        dp = [n]*(1<<n)
        dp[0] = 0
        for mask in range(1<<n):
            if dp[mask] == n: continue
            first = 0
            while first < n and not (mask & (1<<first)): first += 1
            if first == n: continue
            for j in range(n):
                new_mask = mask | col[first][j]
                dp[new_mask] = min(dp[new_mask], dp[mask]+1)
        return dp[(1<<n)-1]
Rust
impl Solution {
    pub fn min_lines(points: Vec<Vec<i32>>) -> i32 {
        let n = points.len();
        let mut col = vec![vec![0; n]; n];
        for i in 0..n {
            for j in 0..n {
                let mut mask = 0;
                for k in 0..n {
                    let (x1, y1) = (points[i][0], points[i][1]);
                    let (x2, y2) = (points[j][0], points[j][1]);
                    let (x3, y3) = (points[k][0], points[k][1]);
                    if (x2-x1)*(y3-y1) == (y2-y1)*(x3-x1) { mask |= 1<<k; }
                }
                col[i][j] = mask;
            }
        }
        let mut dp = vec![n; 1<<n];
        dp[0] = 0;
        for mask in 0..(1<<n) {
            if dp[mask] == n { continue; }
            let mut first = 0;
            while first < n && (mask & (1<<first)) == 0 { first += 1; }
            if first == n { continue; }
            for j in 0..n {
                let new_mask = mask | col[first][j];
                dp[new_mask] = dp[new_mask].min(dp[mask]+1);
            }
        }
        dp[(1<<n)-1]
    }
}
TypeScript
class Solution {
    minLines(points: number[][]): number {
        const n = points.length;
        const col = Array.from({length: n}, () => Array(n).fill(0));
        for (let i = 0; i < n; ++i) {
            for (let j = 0; j < n; ++j) {
                let mask = 0;
                for (let k = 0; k < n; ++k) {
                    const [x1, y1] = points[i], [x2, y2] = points[j], [x3, y3] = points[k];
                    if ((x2-x1)*(y3-y1) === (y2-y1)*(x3-x1)) mask |= 1<<k;
                }
                col[i][j] = mask;
            }
        }
        const dp = Array(1<<n).fill(n);
        dp[0] = 0;
        for (let mask = 0; mask < (1<<n); ++mask) {
            if (dp[mask] === n) continue;
            let first = 0;
            while (first < n && !(mask & (1<<first))) ++first;
            if (first === n) continue;
            for (let j = 0; j < n; ++j) {
                const new_mask = mask | col[first][j];
                dp[new_mask] = Math.min(dp[new_mask], dp[mask]+1);
            }
        }
        return dp[(1<<n)-1];
    }
}

Complexity

  • ⏰ Time complexity: O(n^3 * 2^n) — n = number of points.
  • 🧺 Space complexity: O(2^n) — for DP array.

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