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Graphs and Network Algorithms · Graphs

Adjacency List

An adjacency list stores each vertex with its neighbors, usually saving space when most vertices connect to only a few others.

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Student Focus

This representation is used heavily in coding practice because it fits most student graph problems.

Guided Lesson Notes

Understanding Adjacency List

Adjacency List focuses on relationships, reachability, paths, cycles, connectivity, and optimization over edges. An adjacency list stores each vertex with its neighbors, usually saving space when most vertices connect to only a few others.

Model: draw vertices as dots and edges as connections; then decide whether direction, weight, or capacity matters.

Invariant: visited, distance, parent, component, or flow arrays must match what has actually been discovered so far.

Implementation: adjacency lists or matrices, queues, stacks, priority queues, union-find, and repeated edge relaxation.

Use Adjacency List when the problem mentions roads, networks, prerequisites, components, shortest paths, spanning costs, or dependency order.

Visual Model

A small picture for Adjacency List

ABCDE

Key Ideas

  • Neighbor lists
  • Sparse graph efficiency
  • Iteration over outgoing edges

Practice Prompts

  • Build adjacency lists using arrays of lists or maps.
  • Trace how BFS reads neighbors from an adjacency list.

Vocabulary

Terms students should be able to say clearly

Vertex

An object or state in the graph.

Edge

A relationship or move between vertices.

Path

A sequence of edges from one vertex to another.

Cycle

A path that returns to a previous vertex.

Component

A group of vertices connected by reachability.

Relaxation

Trying to improve a known distance or cost through an edge.

Worked Example

Worked example: tracing Adjacency List

Trace a small input, focusing on neighbor lists.

  1. 1Write a small input where Adjacency List is clearly useful.
  2. 2Label the part of the input related to Neighbor lists.
  3. 3Perform one operation and explain which invariant is still true afterward.
  4. 4Run a second operation that touches an edge case, such as an empty side, duplicate value, boundary index, disconnected vertex, or tie.
  5. 5Finish by saying which operation dominates the runtime and why.

Complexity Check

Costs students should be able to explain

OperationTypical CostReason
Adjacency-list traversalO(V + E)DFS and BFS visit vertices and inspect edges.
Adjacency matrix storageO(V^2)A matrix reserves space for every possible pair.
Weighted optimizationVariesMST and shortest-path algorithms depend on sorting, heaps, and graph density.

Common Mistakes

What to watch while practicing

  • Coding before drawing the structure or state changes.
  • Forgetting the invariant that makes the algorithm correct.
  • Testing only the sample input and skipping boundary cases.
  • Giving Big-O without explaining which operation dominates the work.

Interview-Style Coding Problem

Adjacency List interview problem: Number of Islands

Interview medium

Problem

Given a grid of 0s and 1s, count how many connected islands of 1s exist using four-direction adjacency.

Input

The first line contains n and m. The next n lines contain m values, each 0 or 1.

Output

Print the island count.

Sample Input

4 5
1 1 0 0 0
1 0 0 1 1
0 0 0 0 0
1 0 1 1 0

Sample Output

4

Why the sample works

There are four separate groups of land cells.

Approach

  1. 1Scan every cell in the grid.
  2. 2When an unseen land cell is found, start a DFS or BFS and mark its entire island.
  3. 3Increase the island count once per new traversal.
  4. 4Water cells and already visited land cells are skipped.

O(nm) time and O(nm) extra space in the worst case.

Java Solution

import java.util.*;

public class Main {
  static int n;
  static int m;
  static int[][] grid;
  static boolean[][] seen;
  static int[] dr = {1, -1, 0, 0};
  static int[] dc = {0, 0, 1, -1};

  static void dfs(int r, int c) {
    seen[r][c] = true;
    for (int d = 0; d < 4; d++) {
      int nr = r + dr[d];
      int nc = c + dc[d];
      if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;
      if (grid[nr][nc] == 1 && !seen[nr][nc]) {
        dfs(nr, nc);
      }
    }
  }

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    n = sc.nextInt();
    m = sc.nextInt();
    grid = new int[n][m];
    seen = new boolean[n][m];

    for (int r = 0; r < n; r++) {
      for (int c = 0; c < m; c++) {
        grid[r][c] = sc.nextInt();
      }
    }

    int islands = 0;
    for (int r = 0; r < n; r++) {
      for (int c = 0; c < m; c++) {
        if (grid[r][c] == 1 && !seen[r][c]) {
          islands++;
          dfs(r, c);
        }
      }
    }

    System.out.println(islands);
  }
}

Python Solution

n, m = map(int, input().split())
grid = [list(map(int, input().split())) for _ in range(n)]
seen = [[False] * m for _ in range(n)]

def dfs(r, c):
    seen[r][c] = True
    for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
        nr, nc = r + dr, c + dc
        if 0 <= nr < n and 0 <= nc < m:
            if grid[nr][nc] == 1 and not seen[nr][nc]:
                dfs(nr, nc)

islands = 0
for r in range(n):
    for c in range(m):
        if grid[r][c] == 1 and not seen[r][c]:
            islands += 1
            dfs(r, c)

print(islands)

Practice Challenge

Make the idea your own

Create a two-minute explanation of Adjacency List: define it, trace one example, name one edge case, and give the runtime of the main operation.

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