Number of Islands

The problem

Given a 2D grid map of '1's (land) and '0's (water), count the number of islands. An island is formed by connecting adjacent lands horizontally or vertically, and is surrounded by water on all sides.

Every unvisited land cell is an undiscovered piece of territory. When a sweep finds one, you have discovered an entirely new island, so increment the counter and immediately flood the entire component into visited state before resuming the scan. The grid itself can serve as its own visited log by sinking land to water.

Full solution

In-place grid sinking BFS: flood each newly found island to water as it is counted, so no external visited set is needed and, unlike recursive DFS, the queue never risks blowing the call stack on a large grid.

from collections import deque


def num_islands(grid: list[list[str]]) -> int:
    rows, cols = len(grid), len(grid[0])
    count = 0
    for r in range(rows):
        for c in range(cols):
            if grid[r][c] != "1":
                continue
            count += 1
            # BFS sinks the whole component to "0" so it is never counted again,
            # and a queue avoids the recursion-depth blowup a DFS risks on a big grid.
            queue = deque([(r, c)])
            grid[r][c] = "0"
            while queue:
                cr, cc = queue.popleft()
                for nr, nc in ((cr - 1, cc), (cr + 1, cc), (cr, cc - 1), (cr, cc + 1)):
                    if 0 <= nr < rows and 0 <= nc < cols and grid[nr][nc] == "1":
                        grid[nr][nc] = "0"
                        queue.append((nr, nc))
    return count

function numIslands(grid) {
  const rows = grid.length;
  const cols = grid[0].length;
  let count = 0;
  for (let r = 0; r < rows; r++) {
    for (let c = 0; c < cols; c++) {
      if (grid[r][c] !== "1") continue;
      count++;
      const queue = [[r, c]];
      grid[r][c] = "0";
      while (queue.length) {
        const [cr, cc] = queue.shift();
        for (const [nr, nc] of [[cr - 1, cc], [cr + 1, cc], [cr, cc - 1], [cr, cc + 1]]) {
          if (nr >= 0 && nr < rows && nc >= 0 && nc < cols && grid[nr][nc] === "1") {
            grid[nr][nc] = "0";
            queue.push([nr, nc]);
          }
        }
      }
    }
  }
  return count;
}

The theory behind it

Graph — the ground this problem stands on. All Graph problems

What Graph is

A graph is a network of individual points, called vertices or nodes, connected by lines called edges. Think of a subway transit map, an electrical circuit, or a web of social friends. Unlike a tree, a graph has no designated top node and no parent-child hierarchy. Connections can run one-way or both ways, and paths can loop back on themselves to form closed cycles.

When to reach for it

Reach for graph algorithms when inputs describe relationships, networks, flights between cities, course prerequisites, or clone networks. Signals include finding the shortest route across unweighted connections, ordering tasks that depend on earlier tasks, counting isolated clusters, or checking whether a path contains an infinite loop. Whenever problems present pairs of related entities and ask for reachability, distances, or dependencies, graph representations apply.

How the pattern works

First convert edge lists into an adjacency list, mapping each node to an array of its neighbors. Choose your exploration strategy based on the goal: use a queue and breadth-first search to find the shortest path in unweighted networks, or use recursion and depth-first search to explore full paths and detect cycles. Because graphs can have loops, always track visited nodes in a set or boolean array. Add nodes to the visited set at the moment they enter the queue so they are never visited twice.

What each operation costs

Which roles need this problem

Graph is a core topic for these 11 roles — if you're targeting one of them, this problem is early in your path, not optional.

Secondary for 6 more roles, including Performance Engineer, Search Engineer, Information Retrieval Engineer.