Flood Fill
Given a 2D pixel grid, a starting position, and a new color, paint the connected region of the same color with the new color and return the modified grid.
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Problem
An image is represented by an m x n integer grid image where image[i][j] represents the pixel value of the image. You are also given three integers sr, sc, and color. Perform a flood fill starting from the pixel image[sr][sc].
Input
A 2D integer array `image` (m×n), starting row `sr` and column `sc`, and an integer `color`.
Output
The modified 2D grid after filling.
Examples
Input: image = [[1,1,1],[1,1,0],[1,0,1]], sr = 1, sc = 1, color = 2
Output: [[2,2,2],[2,2,0],[2,0,1]]
Starting at (1,1), the connected region of 1s gets painted to 2. The isolated 1 at (2,2) is not reachable, so it stays.
Input: image = [[0,0,0],[0,0,0]], sr = 0, sc = 0, color = 0
Output: [[0,0,0],[0,0,0]]
The starting pixel is already color 0, return the image unchanged.
The brute-force approach
For each candidate cell, trace every possible path from the start position and check if the cell is reachable through same-color neighbors.
# No meaningful brute force: connectivity can only be determined
# by actually exploring outward from the starting pixel.
# The DFS approach IS the direct solution, there's no simpler version.Checking connectivity naively for each cell would mean re-scanning the grid repeatedly. O((m×n)²). The only efficient way to find the connected region is to explore outward from the start in one pass.
Spotting the pattern
This is a Graph Traversal problem. The key question to ask yourself:
How do I visit every cell in the connected region exactly once, without revisiting?
Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.
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