LeetCode #3394 — MEDIUM

Check if Grid can be Cut into Sections

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

Problem Statement

You are given an integer n representing the dimensions of an n x n grid, with the origin at the bottom-left corner of the grid. You are also given a 2D array of coordinates rectangles, where rectangles[i] is in the form [startx, starty, endx, endy], representing a rectangle on the grid. Each rectangle is defined as follows:

  • (startx, starty): The bottom-left corner of the rectangle.
  • (endx, endy): The top-right corner of the rectangle.

Note that the rectangles do not overlap. Your task is to determine if it is possible to make either two horizontal or two vertical cuts on the grid such that:

  • Each of the three resulting sections formed by the cuts contains at least one rectangle.
  • Every rectangle belongs to exactly one section.

Return true if such cuts can be made; otherwise, return false.

Example 1:

Input: n = 5, rectangles = [[1,0,5,2],[0,2,2,4],[3,2,5,3],[0,4,4,5]]

Output: true

Explanation:

The grid is shown in the diagram. We can make horizontal cuts at y = 2 and y = 4. Hence, output is true.

Example 2:

Input: n = 4, rectangles = [[0,0,1,1],[2,0,3,4],[0,2,2,3],[3,0,4,3]]

Output: true

Explanation:

We can make vertical cuts at x = 2 and x = 3. Hence, output is true.

Example 3:

Input: n = 4, rectangles = [[0,2,2,4],[1,0,3,2],[2,2,3,4],[3,0,4,2],[3,2,4,4]]

Output: false

Explanation:

We cannot make two horizontal or two vertical cuts that satisfy the conditions. Hence, output is false.

Constraints:

  • 3 <= n <= 109
  • 3 <= rectangles.length <= 105
  • 0 <= rectangles[i][0] < rectangles[i][2] <= n
  • 0 <= rectangles[i][1] < rectangles[i][3] <= n
  • No two rectangles overlap.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an integer n representing the dimensions of an n x n grid, with the origin at the bottom-left corner of the grid. You are also given a 2D array of coordinates rectangles, where rectangles[i] is in the form [startx, starty, endx, endy], representing a rectangle on the grid. Each rectangle is defined as follows: (startx, starty): The bottom-left corner of the rectangle. (endx, endy): The top-right corner of the rectangle. Note that the rectangles do not overlap. Your task is to determine if it is possible to make either two horizontal or two vertical cuts on the grid such that: Each of the three resulting sections formed by the cuts contains at least one rectangle. Every rectangle belongs to exactly one section. Return true if such cuts can be made; otherwise, return false.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

5
[[1,0,5,2],[0,2,2,4],[3,2,5,3],[0,4,4,5]]

Example 2

4
[[0,0,1,1],[2,0,3,4],[0,2,2,3],[3,0,4,3]]

Example 3

4
[[0,2,2,4],[1,0,3,2],[2,2,3,4],[3,0,4,2],[3,2,4,4]]
Step 02

Core Insight

What unlocks the optimal approach

  • For each rectangle, consider ranges <code>[start_x, end_x]</code> and <code>[start_y, end_y]</code> separately.
  • For x and y directions, check whether we can split it into 3 parts.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3394: Check if Grid can be Cut into Sections
class Solution {
    // Helper class to mimic C++ pair<int, int>
    static class Pair {
        int value;
        int type;

        Pair(int value, int type) {
            this.value = value;
            this.type = type;
        }
    }

    private boolean countLineIntersections(List<Pair> coordinates) {
        int lines = 0;
        int overlap = 0;

        for (Pair coord : coordinates) {
            if (coord.type == 0) {
                overlap--;
            } else {
                overlap++;
            }

            if (overlap == 0) {
                lines++;
            }
        }

        return lines >= 3;
    }

    public boolean checkValidCuts(int n, int[][] rectangles) {
        List<Pair> yCoordinates = new ArrayList<>();
        List<Pair> xCoordinates = new ArrayList<>();

        for (int[] rectangle : rectangles) {
            // rectangle = [x1, y1, x2, y2]
            yCoordinates.add(new Pair(rectangle[1], 1)); // y1, start
            yCoordinates.add(new Pair(rectangle[3], 0)); // y2, end

            xCoordinates.add(new Pair(rectangle[0], 1)); // x1, start
            xCoordinates.add(new Pair(rectangle[2], 0)); // x2, end
        }

        Comparator<Pair> comparator = (a, b) -> {
            if (a.value != b.value) return Integer.compare(a.value, b.value);
            return Integer.compare(a.type, b.type); // End (0) before Start (1)
        };

        Collections.sort(yCoordinates, comparator);
        Collections.sort(xCoordinates, comparator);

        return countLineIntersections(yCoordinates) || countLineIntersections(xCoordinates);
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

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