LeetCode #1391 — MEDIUM

Check if There is a Valid Path in a Grid

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

Solve on LeetCode
The Problem

Problem Statement

You are given an m x n grid. Each cell of grid represents a street. The street of grid[i][j] can be:

  • 1 which means a street connecting the left cell and the right cell.
  • 2 which means a street connecting the upper cell and the lower cell.
  • 3 which means a street connecting the left cell and the lower cell.
  • 4 which means a street connecting the right cell and the lower cell.
  • 5 which means a street connecting the left cell and the upper cell.
  • 6 which means a street connecting the right cell and the upper cell.

You will initially start at the street of the upper-left cell (0, 0). A valid path in the grid is a path that starts from the upper left cell (0, 0) and ends at the bottom-right cell (m - 1, n - 1). The path should only follow the streets.

Notice that you are not allowed to change any street.

Return true if there is a valid path in the grid or false otherwise.

Example 1:

Input: grid = [[2,4,3],[6,5,2]]
Output: true
Explanation: As shown you can start at cell (0, 0) and visit all the cells of the grid to reach (m - 1, n - 1).

Example 2:

Input: grid = [[1,2,1],[1,2,1]]
Output: false
Explanation: As shown you the street at cell (0, 0) is not connected with any street of any other cell and you will get stuck at cell (0, 0)

Example 3:

Input: grid = [[1,1,2]]
Output: false
Explanation: You will get stuck at cell (0, 1) and you cannot reach cell (0, 2).

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 300
  • 1 <= grid[i][j] <= 6
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: You are given an m x n grid. Each cell of grid represents a street. The street of grid[i][j] can be: 1 which means a street connecting the left cell and the right cell. 2 which means a street connecting the upper cell and the lower cell. 3 which means a street connecting the left cell and the lower cell. 4 which means a street connecting the right cell and the lower cell. 5 which means a street connecting the left cell and the upper cell. 6 which means a street connecting the right cell and the upper cell. You will initially start at the street of the upper-left cell (0, 0). A valid path in the grid is a path that starts from the upper left cell (0, 0) and ends at the bottom-right cell (m - 1, n - 1). The path should only follow the streets. Notice that you are not allowed to change any street. Return true if there is a valid path in the grid or false otherwise.

Baseline thinking

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

Pattern signal: Array · Union-Find

Example 1

[[2,4,3],[6,5,2]]

Example 2

[[1,2,1],[1,2,1]]

Example 3

[[1,1,2]]

Related Problems

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Step 02

Core Insight

What unlocks the optimal approach

  • Start DFS from the node (0, 0) and follow the path till you stop.
  • When you reach a cell and cannot move anymore check that this cell is (m - 1, n - 1) or not.
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 #1391: Check if There is a Valid Path in a Grid
class Solution {
    private int[] p;
    private int[][] grid;
    private int m;
    private int n;

    public boolean hasValidPath(int[][] grid) {
        this.grid = grid;
        m = grid.length;
        n = grid[0].length;
        p = new int[m * n];
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int e = grid[i][j];
                if (e == 1) {
                    left(i, j);
                    right(i, j);
                } else if (e == 2) {
                    up(i, j);
                    down(i, j);
                } else if (e == 3) {
                    left(i, j);
                    down(i, j);
                } else if (e == 4) {
                    right(i, j);
                    down(i, j);
                } else if (e == 5) {
                    left(i, j);
                    up(i, j);
                } else {
                    right(i, j);
                    up(i, j);
                }
            }
        }
        return find(0) == find(m * n - 1);
    }

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

    private void left(int i, int j) {
        if (j > 0 && (grid[i][j - 1] == 1 || grid[i][j - 1] == 4 || grid[i][j - 1] == 6)) {
            p[find(i * n + j)] = find(i * n + j - 1);
        }
    }

    private void right(int i, int j) {
        if (j < n - 1 && (grid[i][j + 1] == 1 || grid[i][j + 1] == 3 || grid[i][j + 1] == 5)) {
            p[find(i * n + j)] = find(i * n + j + 1);
        }
    }

    private void up(int i, int j) {
        if (i > 0 && (grid[i - 1][j] == 2 || grid[i - 1][j] == 3 || grid[i - 1][j] == 4)) {
            p[find(i * n + j)] = find((i - 1) * n + j);
        }
    }

    private void down(int i, int j) {
        if (i < m - 1 && (grid[i + 1][j] == 2 || grid[i + 1][j] == 5 || grid[i + 1][j] == 6)) {
            p[find(i * n + j)] = find((i + 1) * n + j);
        }
    }
}
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.

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