Leetcode 200 Number of Islands (Java) With Video Explanation

// SOLVING THIS WITH AN AI ASSISTANT (2026)

If you are working through this problem with an AI coding assistant — Claude, ChatGPT, Cursor chat, Gemini, GitHub Copilot, Aider, or any agent — the goal isn’t to ask for the answer. It is to use the tool to understand the pattern. The prompt sequence I’d run:

Spec it back to me first. “In your own words, what is this problem actually testing? What’s the smallest example that fails the naive approach?”
Brute-force first, optimize after. “Write the simplest correct solution, even if it’s O(n²). Don’t optimize. Just make it correct, with comments explaining each step.”
Ask for the upgrade. “Now show me the optimal solution. What insight makes it possible? What pattern is this an instance of?”
Stress-test it. “Generate 10 edge cases — empty input, single element, duplicates, max size, sorted, reverse-sorted. Run my solution against each.”

The pattern matters more than the answer. If the agent just hands you optimized code, you’ve trained yourself to lose interviews.

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example :

Input:
11000
11000
00100
00011

Output: 3

Solution:
We will solve this problem using DFS(Depth First Search) Algorithm tracing all the 1’s until we find no 1’s on 4 sides recursively then we will increase the count.

class Solution {
    public int numIslands(char[][] grid) {
        if(grid.length==0) return 0;
        int rows = grid.length;
        int columns = grid[0].length;
        int noOfIslands = 0;
        for(int i=0;i<rows;i++){
            for(int j=0;j<columns;j++){
                if(grid[i][j]=='1'){
                    dfs(grid, i, j);
                    noOfIslands++;
                }
            }
        }
        
        return noOfIslands;
    }
    
    private void dfs(char[][] grid, int i, int j){
        if(i<0 || j<0 ||i>=grid.length|| j>=grid[0].length|| grid[i][j]!='1') return;
        grid[i][j] = '2';
        dfs(grid, i+1, j);
        dfs(grid, i, j+1);
        dfs(grid, i-1, j);
        dfs(grid, i, j-1);
        
    }
}

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rajendra