What is the time complexity of Find the Grid of Region Average?

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

What data structures are used to solve Find the Grid of Region Average?

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

What are the key insights for solving Find the Grid of Region Average?

Insight 1: The core task is to identify valid 3x3 regions that meet the threshold criteria for adjacent pixel intensity differences. Insight 2: Each pixel can belong to multiple 3x3 regions, and we need to track both the sum of the region averages and the number of regions a pixel belongs to.

Which companies ask LeetCode 3030. Find the Grid of Region Average in interviews?

This problem is frequently asked at jio.

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Find the Grid of Region Average - LeetCode 3030 Solution

Q: What is the space complexity of Find the Grid of Region Average?

The space complexity is O(m*n).

Find the Grid of Region Average - Complete Solution Guide

Find the Grid of Region Average is LeetCode problem 3030, 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 m x n grid image which represents a grayscale image, where image[i][j] represents a pixel with intensity in the range [0..255] . You are also given a non-negative integer threshold . Two pixels are adjacent if they share an edge. A region is a 3 x 3 subgrid where the absolute difference in intensity between any two adjacent pixels is less than or equal to threshold . All pixels in a region belong to that region, note that a pixel can belong to multiple regions. You need to calculat

Detailed Explanation

The problem asks us to process a grayscale image represented as a 2D grid (matrix) and smooth it by averaging the intensity of pixels within certain regions. A region is defined as a 3x3 subgrid where the absolute difference between any two adjacent pixels is less than or equal to a given threshold. The goal is to create a new grid, 'result', where each pixel's value is the average intensity of all regions it belongs to, rounded down. If a pixel doesn't belong to any region, its value remains unchanged from the original image. The averaging process requires rounding down the individual region averages before taking the final average if a pixel belongs to multiple regions. The inputs are the image grid, the threshold value, and the outputs are the resulting processed grid.

Solution Approach

The solution iterates through all possible 3x3 subgrids within the input image. For each subgrid, it checks if it's a valid region by verifying that the absolute difference between adjacent pixels is no more than the given threshold. If it's a valid region, the solution calculates the average intensity of that region (using integer division for rounding down) and updates two auxiliary matrices: `region_sums` and `region_counts`. `region_sums` stores the sum of region averages for each pixel, and `region_counts` stores the number of regions each pixel belongs to. Finally, the solution iterates through the image again, calculates the average intensity for each pixel based on the values in `region_sums` and `region_counts`, and stores the result in a new matrix, `result`. If a pixel doesn't belong to any region, its original intensity value is preserved.

Step-by-Step Algorithm

Step 1: Initialize two auxiliary matrices, `region_sums` and `region_counts`, with the same dimensions as the input image. Fill them with zeros.
Step 2: Iterate through the image with a sliding window of size 3x3. Specifically, iterate from row 0 to m-3 and column 0 to n-3, where m and n are the dimensions of the image.
Step 3: For each 3x3 subgrid, check if it forms a valid region. This involves checking the absolute difference between horizontally and vertically adjacent pixels against the threshold. If any difference exceeds the threshold, the subgrid is invalid; move on to the next subgrid.
Step 4: If the subgrid is a valid region, calculate the average intensity by summing the intensities of all nine pixels and dividing by 9 (integer division).
Step 5: Update the `region_sums` and `region_counts` matrices. For each pixel within the current 3x3 region, add the calculated average intensity to the corresponding cell in `region_sums`, and increment the corresponding cell in `region_counts`.
Step 6: After processing all possible 3x3 subgrids, create a result matrix with the same dimensions as the input image.
Step 7: Iterate through the image again. For each pixel, if the corresponding cell in `region_counts` is greater than 0 (meaning the pixel belongs to at least one region), calculate the average intensity by dividing the value in `region_sums` by the value in `region_counts` (again, integer division). Store this average in the corresponding cell of the result matrix. Otherwise, if the value in `region_counts` is 0, copy the original pixel value from the input image into the result matrix.
Step 8: Return the result matrix.

Key Insights

Insight 1: The core task is to identify valid 3x3 regions that meet the threshold criteria for adjacent pixel intensity differences.
Insight 2: Each pixel can belong to multiple 3x3 regions, and we need to track both the sum of the region averages and the number of regions a pixel belongs to.
Insight 3: Integer division is crucial for both calculating initial region averages and final results, effectively rounding down at each step as the problem requires.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Matrix.

Companies

Asked at: jio.

Find the Grid of Region Average | LeetCode 3030

Medium

Array Matrix

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

Solution Code

Available in:

class Solution {
    public int[][] resultGrid(int[][] image, int threshold) {
        int m = image.length;
        int n = image[0].length;

        int[][] regionSums = new int[m][n];
        int[][] regionCounts = new int[m][n];

        for (int r = 0; r < m - 2; r++) {
            for (int c = 0; c < n - 2; c++) {

                boolean isValidRegion = true;

                for (int i = r; i < r + 3; i++) {
                    if (Math.abs(image[i][c] - image[i][c + 1]) > threshold ||
                            Math.abs(image[i][c + 1] - image[i][c + 2]) > threshold) {
                        isValidRegion = false;
                        break;
                    }
                }
                if (!isValidRegion) {
                    continue;
                }

                for (int j = c; j < c + 3; j++) {
                    if (Math.abs(image[r][j] - image[r + 1][j]) > threshold ||
                            Math.abs(image[r + 1][j] - image[r + 2][j]) > threshold) {
                        isValidRegion = false;
                        break;
                    }
                }
                if (!isValidRegion) {
                    continue;
                }

                int currentSum = 0;
                for (int i = r; i < r + 3; i++) {
                    for (int j = c; j < c + 3; j++) {
                        currentSum += image[i][j];
                    }
                }

                int regionAvg = currentSum / 9;

                for (int i = r; i < r + 3; i++) {
                    for (int j = c; j < c + 3; j++) {
                        regionSums[i][j] += regionAvg;
                        regionCounts[i][j] += 1;
                    }
                }
            }
        }

        int[][] result = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (regionCounts[i][j] == 0) {
                    result[i][j] = image[i][j];
                } else {
                    result[i][j] = regionSums[i][j] / regionCounts[i][j];
                }
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 3030: Find the Grid of Region Average - Detailed explanation and solution approaches.

Topics

Array Matrix

Companies

jio

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3030

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

8 steps

1
Step 1: Initialize two auxiliary matrices, `region_sums` and `region_counts`, with the same dimensions as the input image. Fill them with zeros.
2
Step 2: Iterate through the image with a sliding window of size 3x3. Specifically, iterate from row 0 to m-3 and column 0 to n-3, where m and n are the dimensions of the image.
3
Step 3: For each 3x3 subgrid, check if it forms a valid region. This involves checking the absolute difference between horizontally and vertically adjacent pixels against the threshold. If any difference exceeds the threshold, the subgrid is invalid; move on to the next subgrid.
4
Step 4: If the subgrid is a valid region, calculate the average intensity by summing the intensities of all nine pixels and dividing by 9 (integer division).
5
Step 5: Update the `region_sums` and `region_counts` matrices. For each pixel within the current 3x3 region, add the calculated average intensity to the corresponding cell in `region_sums`, and increment the corresponding cell in `region_counts`.
6
Step 6: After processing all possible 3x3 subgrids, create a result matrix with the same dimensions as the input image.
7
Step 7: Iterate through the image again. For each pixel, if the corresponding cell in `region_counts` is greater than 0 (meaning the pixel belongs to at least one region), calculate the average intensity by dividing the value in `region_sums` by the value in `region_counts` (again, integer division). Store this average in the corresponding cell of the result matrix. Otherwise, if the value in `region_counts` is 0, copy the original pixel value from the input image into the result matrix.
8
Step 8: Return the result matrix.

💡

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 time complexity is O(m*n) because the algorithm iterates through the image twice. The first iteration involves checking all possible 3x3 subgrids, which takes O((m-2)*(n-2)) time, simplified to O(m*n). The second iteration involves constructing the result matrix, which also takes O(m*n) time. The space complexity is O(m*n) because the algorithm uses two auxiliary matrices (`region_sums` and `region_counts`) and the `result` matrix, all of which have the same dimensions as the input image.

📖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

Precomputed Prefix Sum

Instead of repeatedly calculating the sum of each 3x3 subgrid, we can precompute a prefix sum matrix. This allows calculating the sum of any rectangular region in O(1) time.

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

Trade-offs: The time complexity remains the same as the core algorithm is the sliding window for checking the threshold conditions. The space complexity is the same. This might offer marginal performance improvement in specific implementations due to constant factor optimization but no impact on big O.

✨

Key Insights

1Insight 1: The core task is to identify valid 3x3 regions that meet the threshold criteria for adjacent pixel intensity differences.
2Insight 2: Each pixel can belong to multiple 3x3 regions, and we need to track both the sum of the region averages and the number of regions a pixel belongs to.
3Insight 3: Integer division is crucial for both calculating initial region averages and final results, effectively rounding down at each step as the problem requires.

📚

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: The input image does not contain any valid 3x3 regions. In this case, the `region_counts` matrix will be all zeros, and the result matrix will simply be a copy of the input image. The code handles this correctly by copying the original pixel value when `region_counts[i][j]` is 0.
Edge case 2: The threshold is very small (e.g., 0 or 1), leading to very few valid regions. The code correctly handles this scenario by only considering the valid regions and calculating averages based on those regions only.
Edge case 3: The threshold is very large, such as 255, making every 3x3 subgrid a valid region. The code will still function correctly, calculating the average of all valid regions for each pixel.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to round down the average intensity of each region and/or the final average. Integer division is the key to meeting the problem's requirement.

✗Common mistake 2: Incorrectly checking the threshold condition, such as missing horizontal or vertical adjacency checks, or using the wrong comparison operator.

✗Common mistake 3: Not handling the case where a pixel does not belong to any region, resulting in incorrect values in the output grid. The code needs to copy the original pixel value in this scenario.

✅

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 smoothing or blurring in image processing. This problem simulates a simple form of spatial filtering, where pixel values are averaged with their neighbors to reduce noise or enhance certain features.

🏦Real-world scenario 2: Data analysis where you want to smooth out noisy data points by averaging values within certain local regions, as long as the changes between adjacent values are within a defined tolerance. For example, smoothing temperature readings across a geographical area where the temperature difference between adjacent locations is expected to be within a specific range.

🎮Real-world scenario 3: Fault detection in sensor networks. Identifying regions where sensor readings are inconsistent by comparing adjacent values against a threshold.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: October 29, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

❓

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.

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 "Find the Grid of Region Average"?

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

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