LeetCode #959 — MEDIUM

Regions Cut By Slashes

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

Solve on LeetCode
The Problem

Problem Statement

An n x n grid is composed of 1 x 1 squares where each 1 x 1 square consists of a '/', '\', or blank space ' '. These characters divide the square into contiguous regions.

Given the grid grid represented as a string array, return the number of regions.

Note that backslash characters are escaped, so a '\' is represented as '\\'.

Example 1:

Input: grid = [" /","/ "]
Output: 2

Example 2:

Input: grid = [" /","  "]
Output: 1

Example 3:

Input: grid = ["/\\","\\/"]
Output: 5
Explanation: Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.

Constraints:

  • n == grid.length == grid[i].length
  • 1 <= n <= 30
  • grid[i][j] is either '/', '\', or ' '.
Patterns Used
🔗Union-Find→

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: An n x n grid is composed of 1 x 1 squares where each 1 x 1 square consists of a '/', '\', or blank space ' '. These characters divide the square into contiguous regions. Given the grid grid represented as a string array, return the number of regions. Note that backslash characters are escaped, so a '\' is represented as '\\'.

Baseline thinking

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

Pattern signal: Array · Hash Map · Union-Find

Example 1

[" /","/ "]

Example 2

[" /","  "]

Example 3

["/\\","\\/"]
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
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 #959: Regions Cut By Slashes
class Solution {
    private int[] p;
    private int size;

    public int regionsBySlashes(String[] grid) {
        int n = grid.length;
        size = n * n * 4;
        p = new int[size];
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                int k = i * n + j;
                if (i < n - 1) {
                    union(4 * k + 2, (k + n) * 4);
                }
                if (j < n - 1) {
                    union(4 * k + 1, (k + 1) * 4 + 3);
                }
                char v = grid[i].charAt(j);
                if (v == '/') {
                    union(4 * k, 4 * k + 3);
                    union(4 * k + 1, 4 * k + 2);
                } else if (v == '\\') {
                    union(4 * k, 4 * k + 1);
                    union(4 * k + 2, 4 * k + 3);
                } else {
                    union(4 * k, 4 * k + 1);
                    union(4 * k + 1, 4 * k + 2);
                    union(4 * k + 2, 4 * k + 3);
                }
            }
        }
        return size;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    private void union(int a, int b) {
        int pa = find(a);
        int pb = find(b);
        if (pa == pb) {
            return;
        }
        p[pa] = pb;
        --size;
    }
}
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(α(n))
Space
O(n)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND
O(α(n)) time
O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”
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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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