Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
You are given a 2D integer array points, where points[i] = [xi, yi]. You are also given an integer w. Your task is to cover all the given points with rectangles.
Each rectangle has its lower end at some point (x1, 0) and its upper end at some point (x2, y2), where x1 <= x2, y2 >= 0, and the condition x2 - x1 <= w must be satisfied for each rectangle.
A point is considered covered by a rectangle if it lies within or on the boundary of the rectangle.
Return an integer denoting the minimum number of rectangles needed so that each point is covered by at least one rectangle.
Note: A point may be covered by more than one rectangle.
Example 1:
Input: points = [[2,1],[1,0],[1,4],[1,8],[3,5],[4,6]], w = 1
Output: 2
Explanation:
The image above shows one possible placement of rectangles to cover the points:
(1, 0) and its upper end at (2, 8)(3, 0) and its upper end at (4, 8)Example 2:
Input: points = [[0,0],[1,1],[2,2],[3,3],[4,4],[5,5],[6,6]], w = 2
Output: 3
Explanation:
The image above shows one possible placement of rectangles to cover the points:
(0, 0) and its upper end at (2, 2)(3, 0) and its upper end at (5, 5)(6, 0) and its upper end at (6, 6)Example 3:
Input: points = [[2,3],[1,2]], w = 0
Output: 2
Explanation:
The image above shows one possible placement of rectangles to cover the points:
(1, 0) and its upper end at (1, 2)(2, 0) and its upper end at (2, 3)Constraints:
1 <= points.length <= 105points[i].length == 20 <= xi == points[i][0] <= 1090 <= yi == points[i][1] <= 1090 <= w <= 109(xi, yi) are distinct.Problem summary: You are given a 2D integer array points, where points[i] = [xi, yi]. You are also given an integer w. Your task is to cover all the given points with rectangles. Each rectangle has its lower end at some point (x1, 0) and its upper end at some point (x2, y2), where x1 <= x2, y2 >= 0, and the condition x2 - x1 <= w must be satisfied for each rectangle. A point is considered covered by a rectangle if it lies within or on the boundary of the rectangle. Return an integer denoting the minimum number of rectangles needed so that each point is covered by at least one rectangle. Note: A point may be covered by more than one rectangle.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Greedy
[[2,1],[1,0],[1,4],[1,8],[3,5],[4,6]] 1
[[0,0],[1,1],[2,2],[3,3],[4,4],[5,5],[6,6]] 2
[[2,3],[1,2]] 0
minimum-area-rectangle)k-closest-points-to-origin)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3111: Minimum Rectangles to Cover Points
class Solution {
public int minRectanglesToCoverPoints(int[][] points, int w) {
Arrays.sort(points, (a, b) -> a[0] - b[0]);
int ans = 0, x1 = -1;
for (int[] p : points) {
int x = p[0];
if (x > x1) {
++ans;
x1 = x + w;
}
}
return ans;
}
}
// Accepted solution for LeetCode #3111: Minimum Rectangles to Cover Points
func minRectanglesToCoverPoints(points [][]int, w int) (ans int) {
sort.Slice(points, func(i, j int) bool { return points[i][0] < points[j][0] })
x1 := -1
for _, p := range points {
if x := p[0]; x > x1 {
ans++
x1 = x + w
}
}
return
}
# Accepted solution for LeetCode #3111: Minimum Rectangles to Cover Points
class Solution:
def minRectanglesToCoverPoints(self, points: List[List[int]], w: int) -> int:
points.sort()
ans, x1 = 0, -1
for x, _ in points:
if x > x1:
ans += 1
x1 = x + w
return ans
// Accepted solution for LeetCode #3111: Minimum Rectangles to Cover Points
impl Solution {
pub fn min_rectangles_to_cover_points(mut points: Vec<Vec<i32>>, w: i32) -> i32 {
points.sort_by(|a, b| a[0].cmp(&b[0]));
let mut ans = 0;
let mut x1 = -1;
for p in points {
let x = p[0];
if x > x1 {
ans += 1;
x1 = x + w;
}
}
ans
}
}
// Accepted solution for LeetCode #3111: Minimum Rectangles to Cover Points
function minRectanglesToCoverPoints(points: number[][], w: number): number {
points.sort((a, b) => a[0] - b[0]);
let [ans, x1] = [0, -1];
for (const [x, _] of points) {
if (x > x1) {
++ans;
x1 = x + w;
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.
Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.