dfs-1 solutions

[dbkuppagiri]

https://leetcode.com/problems/01-matrix/

https://leetcode.com/problems/flood-fill/

[super30admin]

Your floodFill solution is correct and efficient. The code is clean and well-organized. Here are some points to consider:

Strengths:

• You correctly handle the base case where the original color is the same as the new color by returning early (though in your code, the DFS function would not be called in that case because the base condition in DFS checks for image[row][col] === color). Actually, in your code, if the original color is the same as the new color, the DFS function is called, but it immediately returns because the starting pixel already has the new color. This is correct but slightly inefficient because you call DFS unnecessarily. You can add a check at the beginning to return immediately if originalColor === color, which would avoid the DFS call entirely. This is done in the reference solution.

Areas for improvement:

• Consider adding a check at the beginning of floodFill to return the image if originalColor === color. This would avoid the DFS call and make the code slightly more efficient in that edge case.
• Although the constraints are small, be aware that DFS with recursion might lead to stack overflow for very large images. In such cases, BFS (using a queue) is preferred because it uses iterative loops and avoids deep recursion. However, for this problem, it is acceptable.

Example of the improvement:

var floodFill = function (image, sr, sc, color) {
    let originalColor = image[sr][sc];
    if (originalColor === color) return image;  // Add this line to avoid unnecessary DFS
    // ... rest of the code
};

Overall, your solution is good.