LeetCode 3030: Find the Grid of Region Average Solution

Master LeetCode problem 3030 (Find the Grid of Region Average), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3030. Find the Grid of Region Average

Medium

Array Matrix

Problem Explanation

Explanation

To solve this problem, we can iterate over each pixel in the image and for each pixel, we can find all the adjacent pixels and check if they form a valid region based on the given threshold. We can use depth-first search (DFS) to find all connected pixels within a region. Once we have identified all pixels in a region, we can calculate the average intensity of that region and update the result grid accordingly. Finally, we return the result grid.

Algorithm:

Create a result grid of the same size as the input image.
Iterate over each pixel in the image.
For each pixel, perform a DFS to find all connected pixels within a region based on the threshold.
Calculate the average intensity of the region and update the result grid with the rounded-down average intensity.
Continue this process for all pixels in the image.
Return the result grid.

Time Complexity: The time complexity of this algorithm is O(m * n) where m is the number of rows and n is the number of columns in the input grid.

Space Complexity: The space complexity of this algorithm is O(m * n) for the result grid and additional space for the DFS stack.

Solution Code

class Solution {
    public int[][] findAverageGrid(int[][] image, int threshold) {
        int m = image.length;
        int n = image[0].length;
        int[][] result = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                boolean[][] visited = new boolean[m][n];
                result[i][j] = dfs(image, threshold, i, j, visited);
            }
        }
        
        return result;
    }
    
    private int dfs(int[][] image, int threshold, int i, int j, boolean[][] visited) {
        if (i < 0 || i >= image.length || j < 0 || j >= image[0].length || visited[i][j]) {
            return 0;
        }
        
        visited[i][j] = true;
        int sum = image[i][j];
        int count = 1;
        
        int[] dx = {0, 0, 1, -1};
        int[] dy = {1, -1, 0, 0};
        
        for (int k = 0; k < 4; k++) {
            int ni = i + dx[k];
            int nj = j + dy[k];
            if (ni >= 0 && ni < image.length && nj >= 0 && nj < image[0].length &&
                    Math.abs(image[i][j] - image[ni][nj]) <= threshold) {
                int val = dfs(image, threshold, ni, nj, visited);
                sum += val;
                count++;
            }
        }
        
        return sum / count;
    }
}

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