#3030

Medium

LeetCode Problem Workspace

Find the Grid of Region Average

Find the Grid of Region Average requires identifying regions in a matrix where pixel differences are below a threshold, then averaging them.

array matrix

Problem Statement

You are given a grayscale image as an m x n grid, where each element represents a pixel intensity between 0 and 255. Additionally, a non-negative integer threshold is provided, determining the maximum allowed difference in intensity between adjacent pixels for them to belong to the same region.

Regions are defined as 3x3 subgrids in the image where the intensity difference between any two adjacent pixels does not exceed the given threshold. After identifying regions, compute the average intensity for each region and replace all pixel values in the grid by the corresponding average intensity value of their respective regions.

Examples

Example 1

Input: image = [[5,6,7,10],[8,9,10,10],[11,12,13,10]], threshold = 3

Output: [[9,9,9,9],[9,9,9,9],[9,9,9,9]]

There are two regions as illustrated above. The average intensity of the first region is 9, while the average intensity of the second region is 9.67 which is rounded down to 9. The average intensity of both of the regions is (9 + 9) / 2 = 9. As all the pixels belong to either region 1, region 2, or both of them, the intensity of every pixel in the result is 9. Please note that the rounded-down values are used when calculating the average of multiple regions, hence the calculation is done using 9 as the average intensity of region 2, not 9.67.

Example 2

Input: image = [[10,20,30],[15,25,35],[20,30,40],[25,35,45]], threshold = 12

Output: [[25,25,25],[27,27,27],[27,27,27],[30,30,30]]

There are two regions as illustrated above. The average intensity of the first region is 25, while the average intensity of the second region is 30. The average intensity of both of the regions is (25 + 30) / 2 = 27.5 which is rounded down to 27. All the pixels in row 0 of the image belong to region 1, hence all the pixels in row 0 in the result are 25. Similarly, all the pixels in row 3 in the result are 30. The pixels in rows 1 and 2 of the image belong to region 1 and region 2, hence their assigned value is 27 in the result.

Example 3

Input: image = [[5,6,7],[8,9,10],[11,12,13]], threshold = 1

Output: [[5,6,7],[8,9,10],[11,12,13]]

There is only one 3 x 3 subgrid, while it does not have the condition on difference of adjacent pixels, for example, the difference between image[0][0] and image[1][0] is |5 - 8| = 3 > threshold = 1 . None of them belong to any valid regions, so the result should be the same as image .

Constraints

3 <= n, m <= 500
0 <= image[i][j] <= 255
0 <= threshold <= 255

Solution Approach

Brute Force Search for Regions

In this approach, iterate over each 3x3 subgrid in the image and check the absolute difference between all adjacent pixels. If the difference is less than or equal to the threshold, the pixels belong to the same region, and we compute their average intensity.

Region Classification with DFS

Another approach is to use Depth-First Search (DFS) to find connected regions. Start from an unvisited pixel, explore all connected pixels within the threshold, and classify them as part of a region. After identifying all regions, calculate the average intensity for each region.

Optimized Region Search with Union-Find

For efficiency, use the Union-Find algorithm to group adjacent pixels that belong to the same region. After processing all pixels, merge them into regions and calculate the average intensity for each region before updating the grid.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity varies depending on the approach. A brute-force approach may take O(m * n) time to evaluate each 3x3 subgrid. The DFS or Union-Find approaches can reduce redundant checks and help in better time complexity, potentially improving to O(m * n) for region identification and averaging.

What Interviewers Usually Probe

Tests the candidate's ability to handle matrix manipulations and region-based problems.
Validates understanding of dynamic programming or search algorithms like DFS and Union-Find.
Assesses efficiency in handling large grids with potentially expensive computations.

Common Pitfalls or Variants

Common pitfalls

Not handling edge cases like small grid sizes (less than 3x3) properly.
Forgetting to round the calculated average intensity when required by the problem statement.
Using an inefficient algorithm that leads to timeout or excessive space complexity, particularly with large grids.

Follow-up variants

Extend the problem to larger regions (e.g., 4x4 instead of 3x3).
Modify the threshold condition to allow for more or less lenient pixel differences.
Add constraints on the number of regions that can exist in the image.

FAQ

How do I find regions in the 'Find the Grid of Region Average' problem?

Start by checking each 3x3 subgrid for adjacent pixels whose intensity difference is within the threshold. This defines the region.

What is the time complexity of solving the 'Find the Grid of Region Average' problem?

The time complexity can vary based on the approach. Brute force takes O(m * n), while optimized techniques like DFS or Union-Find may improve it.

How do I calculate the average intensity of a region?

Sum the pixel intensities of all the pixels in a region and divide by the number of pixels in the region, rounding down if necessary.

What is the main challenge in solving the 'Find the Grid of Region Average' problem?

The main challenge lies in efficiently identifying regions that meet the threshold condition and then calculating their average intensities.

Can I use dynamic programming to solve this problem?

While dynamic programming is not the most suitable for this problem, techniques like DFS or Union-Find are often more effective for handling regions and connectivity.

terminal

Solution

Solution 1

#### Python3

class Solution:
    def resultGrid(self, image: List[List[int]], threshold: int) -> List[List[int]]:
        n, m = len(image), len(image[0])
        ans = [[0] * m for _ in range(n)]
        ct = [[0] * m for _ in range(n)]
        for i in range(n - 2):
            for j in range(m - 2):
                region = True
                for k in range(3):
                    for l in range(2):
                        region &= (
                            abs(image[i + k][j + l] - image[i + k][j + l + 1])
                            <= threshold
                        )
                for k in range(2):
                    for l in range(3):
                        region &= (
                            abs(image[i + k][j + l] - image[i + k + 1][j + l])
                            <= threshold
                        )

                if region:
                    tot = 0
                    for k in range(3):
                        for l in range(3):
                            tot += image[i + k][j + l]
                    for k in range(3):
                        for l in range(3):
                            ct[i + k][j + l] += 1
                            ans[i + k][j + l] += tot // 9

        for i in range(n):
            for j in range(m):
                if ct[i][j] == 0:
                    ans[i][j] = image[i][j]
                else:
                    ans[i][j] //= ct[i][j]

        return ans

Continue Practicing

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