What is the time complexity of Largest Plus Sign?

The optimal solution has a time complexity of O(n^2).

What is the space complexity of Largest Plus Sign?

The space complexity is O(n^2).

What data structures are used to solve Largest Plus Sign?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Largest Plus Sign?

Insight 1: Dynamic programming is suitable here because we can calculate the length of consecutive 1s in each direction (up, down, left, right) from each cell and reuse this information to determine the maximum plus sign order. Insight 2: Preprocessing the grid by calculating the lengths of consecutive 1s simplifies the problem significantly compared to brute-force checking all possible plus signs.

Which companies ask LeetCode 764. Largest Plus Sign in interviews?

This problem is frequently asked at Intuit.

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Largest Plus Sign - LeetCode 764 Solution

Largest Plus Sign - Complete Solution Guide

Largest Plus Sign is LeetCode problem 764, 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

You are given an integer n . You have an n x n binary grid grid with all values initially 1 's except for some indices given in the array mines . The i th element of the array mines is defined as mines[i] = [x i , y i ] where grid[x i ][y i ] == 0 . Return the order of the largest axis-aligned plus sign of 1 's contained in grid . If there is none, return 0 . An axis-aligned plus sign of 1 's of order k has some center grid[r][c] == 1 along with four arms of length k - 1 going up, down, left, an

Detailed Explanation

The problem asks us to find the largest 'plus sign' of 1s within an n x n grid. Initially, all cells in the grid are 1s. However, the input `mines` array specifies certain cell coordinates that should be set to 0. The 'order' of a plus sign is determined by the length of its arms (up, down, left, right) extending from a central cell, all of which must consist of 1s. A plus sign of order `k` has arms of length `k-1`. The goal is to find the maximum `k` such that a plus sign of order `k` exists entirely within the grid, composed only of 1s after applying the `mines`. If no such plus sign exists, return 0.

Solution Approach

The solution uses dynamic programming to efficiently determine the order of the largest plus sign. It creates a DP table of the same size as the grid. Each cell in the DP table will store the maximum possible order of a plus sign centered at that cell. The algorithm iterates through the grid four times: from left to right, right to left, top to bottom, and bottom to top. In each iteration, it calculates the length of consecutive 1s in the corresponding direction. Finally, the algorithm iterates through the grid again and takes the minimum of the four directional counts for each cell. This minimum represents the maximum possible order of a plus sign centered at that cell. The largest of these values is the final result.

Step-by-Step Algorithm

Step 1: Initialize a set `banned` to store the coordinates of the mined cells. This allows for efficient checking whether a cell is a mine (0) or not (1).
Step 2: Create a DP table `dp` of size n x n, initialized with 0s. This table will store the maximum order of a plus sign that can be centered at each cell.
Step 3: Iterate through each row from left to right. For each cell, if it is not a mine, increment the `count`. Otherwise, reset the `count` to 0. Store the `count` in the DP table.
Step 4: Iterate through each row from right to left. For each cell, if it is not a mine, increment the `count`. Otherwise, reset the `count` to 0. Update the DP table with the minimum of the current value and the `count`.
Step 5: Iterate through each column from top to bottom. For each cell, if it is not a mine, increment the `count`. Otherwise, reset the `count` to 0. Update the DP table with the minimum of the current value and the `count`.
Step 6: Iterate through each column from bottom to top. For each cell, if it is not a mine, increment the `count`. Otherwise, reset the `count` to 0. Update the DP table with the minimum of the current value and the `count`.
Step 7: Iterate through the DP table and find the maximum value. This value represents the order of the largest plus sign.

Key Insights

Insight 1: Dynamic programming is suitable here because we can calculate the length of consecutive 1s in each direction (up, down, left, right) from each cell and reuse this information to determine the maximum plus sign order.
Insight 2: Preprocessing the grid by calculating the lengths of consecutive 1s simplifies the problem significantly compared to brute-force checking all possible plus signs.
Insight 3: Storing the banned cells (mines) in a hash set allows for quick lookups when determining whether a cell contains a 1 or a 0.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Intuit.

Largest Plus Sign | LeetCode 764

Medium

Array Dynamic Programming

Time: O(n^2)Space: O(n^2)

Solution Code

Available in:

import java.util.HashSet;
import java.util.Set;

class Solution {
    public int orderOfLargestPlusSign(int n, int[][] mines) {
        Set<String> banned = new HashSet<>();
        for (int[] mine : mines) {
            banned.add(mine[0] + "," + mine[1]);
        }
        int[][] dp = new int[n][n];

        // Horizontal passes
        for (int r = 0; r < n; r++) {
            // Left to Right
            int count = 0;
            for (int c = 0; c < n; c++) {
                count = banned.contains(r + "," + c) ? 0 : count + 1;
                dp[r][c] = count;
            }

            // Right to Left
            count = 0;
            for (int c = n - 1; c >= 0; c--) {
                count = banned.contains(r + "," + c) ? 0 : count + 1;
                dp[r][c] = Math.min(dp[r][c], count);
            }
        }

        int max_order = 0;
        // Vertical passes
        for (int c = 0; c < n; c++) {
            // Top to Bottom
            int count = 0;
            for (int r = 0; r < n; r++) {
                count = banned.contains(r + "," + c) ? 0 : count + 1;
                dp[r][c] = Math.min(dp[r][c], count);
            }

            // Bottom to Top
            count = 0;
            for (int r = n - 1; r >= 0; r--) {
                count = banned.contains(r + "," + c) ? 0 : count + 1;
                dp[r][c] = Math.min(dp[r][c], count);
                max_order = Math.max(max_order, dp[r][c]);
            }
        }

        return max_order;
    }
}

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Problem Info

Description

Problem 764: Largest Plus Sign - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Intuit

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:764

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 set `banned` to store the coordinates of the mined cells. This allows for efficient checking whether a cell is a mine (0) or not (1).
2
Step 2: Create a DP table `dp` of size n x n, initialized with 0s. This table will store the maximum order of a plus sign that can be centered at each cell.
3
Step 3: Iterate through each row from left to right. For each cell, if it is not a mine, increment the `count`. Otherwise, reset the `count` to 0. Store the `count` in the DP table.
4
Step 4: Iterate through each row from right to left. For each cell, if it is not a mine, increment the `count`. Otherwise, reset the `count` to 0. Update the DP table with the minimum of the current value and the `count`.
5
Step 5: Iterate through each column from top to bottom. For each cell, if it is not a mine, increment the `count`. Otherwise, reset the `count` to 0. Update the DP table with the minimum of the current value and the `count`.
6
Step 6: Iterate through each column from bottom to top. For each cell, if it is not a mine, increment the `count`. Otherwise, reset the `count` to 0. Update the DP table with the minimum of the current value and the `count`.
7
Step 7: Iterate through the DP table and find the maximum value. This value represents the order of the largest plus sign.

💡

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(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The code iterates through the n x n grid a constant number of times (four times for the dynamic programming passes and once for finding the maximum value). The set operations for checking banned cells take O(1) on average. Therefore, the overall time complexity is O(n^2). The DP table `dp` requires O(n^2) space, and the banned set could potentially store all cells, also taking O(n^2) space in the worst case.

📖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

Brute-Force Approach

Iterate through all possible centers of the plus sign and for each center, check all possible orders to see if a plus sign of that order exists.

Time: O(n^3 * k), where k is the maximum order possibleSpace: O(1)

Trade-offs: This approach is simple to understand but very inefficient due to the multiple nested loops. It would likely exceed the time limit for larger input sizes. The DP approach is significantly better.

✨

Key Insights

1Insight 1: Dynamic programming is suitable here because we can calculate the length of consecutive 1s in each direction (up, down, left, right) from each cell and reuse this information to determine the maximum plus sign order.
2Insight 2: Preprocessing the grid by calculating the lengths of consecutive 1s simplifies the problem significantly compared to brute-force checking all possible plus signs.
3Insight 3: Storing the banned cells (mines) in a hash set allows for quick lookups when determining whether a cell contains a 1 or a 0.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `n = 1, mines = [[0, 0]]`. The grid is a single cell, and it's a mine, so the largest plus sign has order 0. The solution correctly handles this.
Edge case 2: `n = 5, mines = []`. There are no mines. The largest plus sign has order `(n+1)//2` = 3, located in the center. The solution will correctly determine this.
Edge case 3: `n = 5, mines = [[2,2]]`. A single mine is in the center. The largest possible plus sign has order 2.
Edge case 4: `mines` contains all cells. The result is 0.
Edge case 5: `mines` is empty. The result is (n+1)//2.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider edge cases, such as when all cells are mines, or when there are no mines.

✗Common mistake 2: Not understanding the definition of the plus sign order, which is the length of each arm plus one. Confusing it with just the arm length.

✗Common mistake 3: Using a brute-force approach without optimization, resulting in a time limit exceeded error.

✗Common mistake 4: Inefficient data structure for storing mined cells resulting in higher time complexity for checking if a cell is mined or not.

✅

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: Image processing - identifying patterns or shapes composed of pixels with a specific value. For instance, finding 'cross' shapes in a medical image to identify certain features.

🏦Real-world scenario 2: Resource allocation - determining the optimal placement of resources or facilities such that they can reach a certain number of surrounding areas, even if some areas are blocked. Imagine placing a network of Wi-Fi hotspots to cover a city area. Some buildings might block signals.

💡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: July 15, 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

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Largest Plus Sign"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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