Graph Representations & Traversals (BFS/DFS)
Graph — উপস্থাপনা ও traversal
1. Graphs Are Everywhere
A graph G = (V, E) is a set of vertices and edges. Roads, social networks, web links, dependency graphs, the Internet, even chess positions — all are graphs. Almost every Computer Science problem reduces to a graph in disguise.
2. Three Common Representations
| Representation | Memory | Edge query | List neighbours | Best for |
|---|---|---|---|---|
Adjacency matrix m[u][v] | O(V²) | O(1) | O(V) | Dense graphs, small V |
Adjacency list vector<int> adj[V] | O(V + E) | O(deg(u)) | O(deg(u)) | Sparse graphs, almost everything |
| Edge list | O(E) | O(E) | O(E) | Kruskal MST, Bellman-Ford |
3. BFS — Layer by Layer
Breadth-first search uses a queue. Mark source as visited, push, then repeatedly pop u and push every unvisited neighbour. Each vertex enters the queue once → O(V + E). BFS computes the shortest path in unweighted graphs.
#include <bits/stdc++.h>
using namespace std;
int main() {
int V = 6;
vector<vector<int>> adj(V);
vector<pair<int,int>> edges = {{0,1},{0,2},{1,3},{2,3},{3,4},{4,5}};
for (auto [u, v] : edges) {
adj[u].push_back(v);
adj[v].push_back(u);
}
vector<int> dist(V, -1);
queue<int> q;
dist[0] = 0; q.push(0);
while (!q.empty()) {
int u = q.front(); q.pop();
for (int v : adj[u]) if (dist[v] == -1) {
dist[v] = dist[u] + 1;
q.push(v);
}
}
for (int i = 0; i < V; i++)
cout << "dist[0..." << i << "] = " << dist[i] << "\n";
}4. DFS — Go Deep First
Depth-first search uses a stack (or recursion). Useful for cycle detection, topological sort, articulation points, SCC, and traversal-order tasks. Like BFS, O(V + E).
#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> adj;
vector<bool> vis;
void dfs(int u) {
vis[u] = true;
cout << u << " ";
for (int v : adj[u]) if (!vis[v]) dfs(v);
}
int main() {
int V = 6;
adj.assign(V, {}); vis.assign(V, false);
vector<pair<int,int>> edges = {{0,1},{0,2},{1,3},{2,3},{3,4},{4,5}};
for (auto [u, v] : edges) { adj[u].push_back(v); adj[v].push_back(u); }
dfs(0);
}std::stack for very deep graphs.
5. Cycle Detection
- Undirected graph: during DFS, if you see a visited neighbour that is not your parent, there is a cycle.
- Directed graph: use 3 colours — WHITE (unvisited), GRAY (in current DFS path), BLACK (finished). Seeing GRAY → cycle.
- BFS in DAG: Kahn's algorithm — if you can't pop V vertices, a cycle exists.
6. Practice Problems
-
Number of islands in a 0/1 grid using DFS or BFS.0/1 grid-এ islands গুনুন।
✨ Show Answer (উত্তর দেখুন)
a1.cpp#include <bits/stdc++.h> using namespace std; vector<string> g; int R, C; void dfs(int i, int j) { if (i<0||j<0||i>=R||j>=C||g[i][j]=='0') return; g[i][j] = '0'; dfs(i+1,j); dfs(i-1,j); dfs(i,j+1); dfs(i,j-1); } int main() { g = {"11000","11000","00100","00011"}; R = g.size(); C = g[0].size(); int cnt = 0; for (int i=0; i<R; i++) for (int j=0; j<C; j++) if (g[i][j] == '1') { cnt++; dfs(i,j); } cout << cnt; } -
Detect cycle in an undirected graph.Undirected graph-এ cycle।
✨ Show Answer (উত্তর দেখুন)
Approach: DFS with parent tracking. If during DFS at u, you see neighbour v that is visited and v ≠ parent[u], cycle exists. Run from every component.
-
Detect cycle in a directed graph (3-colour DFS).Directed graph-এ cycle — 3-colour DFS।
✨ Show Answer (উত্তর দেখুন)
a3.cpp#include <bits/stdc++.h> using namespace std; vector<vector<int>> adj; vector<int> col; // 0 white, 1 gray, 2 black bool cyc(int u) { col[u] = 1; for (int v : adj[u]) { if (col[v] == 1) return true; if (col[v] == 0 && cyc(v)) return true; } col[u] = 2; return false; } int main() { adj = {{1},{2},{0},{2}}; col.assign(4, 0); bool hasCycle = false; for (int i = 0; i < 4; i++) if (col[i] == 0 && cyc(i)) hasCycle = true; cout << (hasCycle ? "YES" : "NO"); } -
Bipartite check via BFS 2-colouring.Bipartite যাচাই — BFS 2-colouring।
✨ Show Answer (উত্তর দেখুন)
Approach: BFS, alternating colours. If you ever reach a neighbour with the same colour as yourself, it is not bipartite. O(V + E).
-
Word Ladder — shortest transformation from begin to end where each step changes one letter and is in the dictionary.Word Ladder — BFS দিয়ে shortest transformation।
✨ Show Answer (উত্তর দেখুন)
Approach: BFS where neighbours of a word are all dictionary words at Hamming distance 1. Use intermediate "wildcard" buckets (like "h*t") for O(L · 26) neighbour generation per word.
-
Number of connected components in an undirected graph.Undirected graph-এ connected components।
✨ Show Answer (উত্তর দেখুন)
Approach: for each unvisited vertex, run BFS/DFS and count one. Or use DSU (Module 22).
-
Flood fill of a 2D image at coordinate (sr, sc) with color newColor.Flood fill — DFS।
✨ Show Answer (উত্তর দেখুন)
Approach: DFS from (sr, sc), only spreading to cells with the original colour. Replace with newColour as you visit.
-
Shortest path in an unweighted grid (BFS) from top-left to bottom-right.Unweighted grid-এ shortest path — BFS।
✨ Show Answer (উত্তর দেখুন)
Approach: standard BFS on a grid with 4-directional moves. dist[r][c] = dist[parent] + 1.
Summary — Module 26
Adjacency list is the default representation. BFS finds shortest paths in unweighted graphs in O(V+E). DFS powers cycle detection, topological sort, SCC, articulation points. Cycle detection differs for undirected (parent check) and directed (3-colour). Phase 6 begins.