What is the time complexity of Detect Cycles in 2D Grid?

The optimal solution has a time complexity of O(m*n).

What is the space complexity of Detect Cycles in 2D Grid?

The space complexity is O(m*n).

What data structures are used to solve Detect Cycles in 2D Grid?

This problem uses Array, Depth-First Search, Breadth-First Search, Union Find, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Detect Cycles in 2D Grid?

Insight 1: The problem can be effectively solved using the Union-Find (Disjoint Set Union) data structure. This allows us to efficiently track connected components of cells with the same value. Insight 2: The essence of detecting a cycle lies in identifying if two adjacent cells with the same value already belong to the same connected component. If they do, merging them would create a cycle.

Which companies ask LeetCode 1559. Detect Cycles in 2D Grid in interviews?

This problem is frequently asked at Nutanix.

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Detect Cycles in 2D Grid - LeetCode 1559 Solution

Detect Cycles in 2D Grid - Complete Solution Guide

Detect Cycles in 2D Grid is LeetCode problem 1559, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a 2D array of characters grid of size m x n , you need to find if there exists any cycle consisting of the same value in grid . A cycle is a path of length 4 or more in the grid that starts and ends at the same cell. From a given cell, you can move to one of the cells adjacent to it - in one of the four directions (up, down, left, or right), if it has the same value of the current cell. Also, you cannot move to the cell that you visited in your last move. For example, the cycle (1, 1) -> (

Detailed Explanation

The problem asks us to determine if there exists a cycle in a 2D grid of characters, where a cycle is defined as a path of length 4 or more that starts and ends at the same cell, and all cells in the cycle must have the same value. Movement between cells is restricted to the four cardinal directions (up, down, left, right), and you cannot immediately return to the cell you just came from. The grid contains only lowercase English letters.

Solution Approach

The provided solution uses the Union-Find algorithm to detect cycles in the grid. Each cell in the grid is treated as a node in a graph, and edges are formed between adjacent cells with the same value. The Union-Find data structure is used to maintain the connected components of this graph. The algorithm iterates through each cell in the grid, and for each cell, it checks its right and down neighbors. If a neighbor has the same value as the current cell, the algorithm attempts to union the corresponding sets of the two cells. If the two cells already belong to the same set, then adding an edge between them would create a cycle, and the algorithm returns true. If the algorithm iterates through the entire grid without finding a cycle, it returns false.

Step-by-Step Algorithm

Step 1: Initialize a Union-Find data structure with `m * n` nodes, where `m` is the number of rows and `n` is the number of columns in the grid. Each cell in the grid is mapped to a unique integer in the range [0, m*n-1]. Initially, each cell is in its own set.
Step 2: Iterate through each cell (r, c) in the grid.
Step 3: For each cell (r, c), check its right neighbor (r, c+1) and its down neighbor (r+1, c).
Step 4: If a neighbor exists and has the same value as the current cell, attempt to 'union' the current cell and the neighbor using the Union-Find data structure.
Step 5: The 'union' operation first finds the root of the sets containing both cells. If the roots are the same, then the cells are already in the same set, and adding an edge between them would create a cycle. In this case, return `true`.
Step 6: If the roots are different, then merge the two sets by setting the parent of one root to the other root. This indicates that these two cells are now part of the same connected component.
Step 7: If after iterating through all the cells, no cycle is detected, return `false`.

Key Insights

Insight 1: The problem can be effectively solved using the Union-Find (Disjoint Set Union) data structure. This allows us to efficiently track connected components of cells with the same value.
Insight 2: The essence of detecting a cycle lies in identifying if two adjacent cells with the same value already belong to the same connected component. If they do, merging them would create a cycle.
Insight 3: The grid indices need to be converted into a 1D representation for the Union-Find data structure, as Union-Find operates on a set of integers.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Depth-First Search, Breadth-First Search, Union Find, Matrix.

Companies

Asked at: Nutanix.

Detect Cycles in 2D Grid | LeetCode 1559

Medium

Array Depth-First Search Breadth-First Search Union Find Matrix

Time: O(m*n)Space: O(m*n)

Solution Code

Available in:

class Solution {
    public boolean containsCycle(char[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[] parent = new int[m * n];
        for (int i = 0; i < m * n; i++) {
            parent[i] = i;
        }

        function find(int i) {
            if (parent[i] == i) {
                return i;
            }
            parent[i] = find(parent[i]);
            return parent[i];
        }

        function union(int i, int j) {
            int root_i = find(i);
            int root_j = find(j);
            if (root_i == root_j) {
                return false;
            }
            parent[root_j] = root_i;
            return true;
        }

        function find(int i) {
            if (parent[i] == i) {
                return i;
            }
            parent[i] = find(parent[i]);
            return parent[i];
        }

        function union(int i, int j) {
            int root_i = find(i);
            int root_j = find(j);
            if (root_i == root_j) {
                return false;
            }
            parent[root_j] = root_i;
            return true;
        }
        
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                int idx = r * n + c;

                if (r + 1 < m && grid[r][c] == grid[r + 1][c]) {
                    int neighbor_idx = (r + 1) * n + c;
                    if (!union(idx, neighbor_idx, parent)) {
                        return true;
                    }
                }

                if (c + 1 < n && grid[r][c] == grid[r][c + 1]) {
                    int neighbor_idx = r * n + (c + 1);
                    if (!union(idx, neighbor_idx, parent)) {
                        return true;
                    }
                }
            }
        }

        return false;
    }

    private int find(int i, int[] parent) {
        if (parent[i] == i) {
            return i;
        }
        parent[i] = find(parent[i], parent);
        return parent[i];
    }

    private boolean union(int i, int j, int[] parent) {
        int root_i = find(i, parent);
        int root_j = find(j, parent);
        if (root_i == root_j) {
            return false;
        }
        parent[root_j] = root_i;
        return true;
    }
}

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Problem Info

Description

Problem 1559: Detect Cycles in 2D Grid - Detailed explanation and solution approaches.

Topics

Array Depth-First Search Breadth-First Search Union Find Matrix

Companies

Nutanix

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1559

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize a Union-Find data structure with `m * n` nodes, where `m` is the number of rows and `n` is the number of columns in the grid. Each cell in the grid is mapped to a unique integer in the range [0, m*n-1]. Initially, each cell is in its own set.
2
Step 2: Iterate through each cell (r, c) in the grid.
3
Step 3: For each cell (r, c), check its right neighbor (r, c+1) and its down neighbor (r+1, c).
4
Step 4: If a neighbor exists and has the same value as the current cell, attempt to 'union' the current cell and the neighbor using the Union-Find data structure.
5
Step 5: The 'union' operation first finds the root of the sets containing both cells. If the roots are the same, then the cells are already in the same set, and adding an edge between them would create a cycle. In this case, return `true`.
6
Step 6: If the roots are different, then merge the two sets by setting the parent of one root to the other root. This indicates that these two cells are now part of the same connected component.
7
Step 7: If after iterating through all the cells, no cycle is detected, return `false`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(m*n)

📝

Why?

The algorithm iterates through the entire grid, contributing O(m*n). The Union-Find operations (find and union) are nearly constant time on average due to path compression and union by rank (though union by rank isn't implemented here, path compression alone leads to nearly constant time). The parent array in the Union-Find data structure requires space proportional to the number of cells in the grid.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Depth-First Search (DFS)

DFS can also be used to detect cycles. Start a DFS from each cell, and keep track of the cells visited in the current path. If you encounter a cell that is already in the current path (and is not the immediate parent), then you have found a cycle.

Time: O(m*n), where m is the number of rows and n is the number of columns.Space: O(m*n) in the worst case, due to the recursion stack and the visited set.

Trade-offs: DFS might be more intuitive for some people, but Union-Find is generally faster in practice. DFS needs to maintain the visited set for each DFS call, and has higher overhead due to recursion. Union-Find only maintains parent pointers and performs relatively simple operations. However, the recursive nature of DFS may hit stack overflow issues for very large grids, a problem that union-find doesn't suffer from.

✨

Key Insights

1Insight 1: The problem can be effectively solved using the Union-Find (Disjoint Set Union) data structure. This allows us to efficiently track connected components of cells with the same value.
2Insight 2: The essence of detecting a cycle lies in identifying if two adjacent cells with the same value already belong to the same connected component. If they do, merging them would create a cycle.
3Insight 3: The grid indices need to be converted into a 1D representation for the Union-Find data structure, as Union-Find operates on a set of integers.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: A grid with only one row or one column. The algorithm correctly handles this case because it checks array boundaries before accessing neighboring cells.
Edge case 2: A grid where all the characters are different. In this case, the algorithm will not find any cycles and will correctly return `false` because no union operations will lead to connecting already connected nodes.
Edge case 3: A grid where there's a cycle of length exactly 4. The solution will detect this cycle as it is of length >= 4.
Edge case 4: A grid with cycles of different characters. The union-find operations are only performed between adjacent cells with the same character. Therefore the cycles of different characters are processed independently from each other, and do not interfere.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check array boundaries before accessing neighboring cells, which can lead to out-of-bounds errors.

✗Common mistake 2: Not correctly converting the 2D grid coordinates into a 1D index for the Union-Find data structure.

✗Common mistake 3: Confusing the conditions for creating a cycle. A cycle is detected when attempting to union two cells that are already in the same set (i.e., have the same root).

✗Common mistake 4: Implementing find and union incorrectly, which would impact the cycle detection logic negatively.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network connectivity analysis. Determining if there are redundant paths in a network, which can improve reliability.

🏦Real-world scenario 2: Image processing. Identifying connected regions of pixels with similar colors or features, which can be used for object recognition or segmentation.

🎮Real-world scenario 3: Social network analysis. Finding communities or groups of people who are closely connected to each other.

📱Real-world scenario 4: Circuit board design. Checking for short circuits or unwanted connections between components.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 28, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Detect Cycles in 2D Grid"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?