Leetcode #2152: Minimum Number of Lines to Cover Points
In this guide, we solve Leetcode #2152 Minimum Number of Lines to Cover Points in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Bit Manipulation, Geometry, Array, Hash Table, Math, Dynamic Programming, Backtracking, Bitmask
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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).
Python Solution
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)
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)
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.