Disjoint Set Union (Union-Find)

6. Practice Problems

1. Number of Provinces — given an n×n adjacency matrix, count connected groups.
   n×n adjacency matrix থেকে connected group-এর সংখ্যা বের করুন।
   Show Answer

   provinces.cpp

   #include <bits/stdc++.h>
   using namespace std;
   int p[200];
   int f(int x) { return p[x] == x ? x : p[x] = f(p[x]); }
   int main() {
       vector<vector<int>> M = {{1,1,0},{1,1,0},{0,0,1}};
       int n = M.size();
       iota(p, p + n, 0);
       for (int i = 0; i < n; ++i)
           for (int j = i + 1; j < n; ++j)
               if (M[i][j]) p[f(i)] = f(j);
       int ans = 0;
       for (int i = 0; i < n; ++i) if (f(i) == i) ++ans;
       cout << ans << "\n";
   }

   Run Copy

2. Redundant Connection — given an undirected graph that is a tree plus one extra edge, find the extra edge.
   একটি tree-এর সাথে অতিরিক্ত একটি edge যোগ করা আছে — সেই অতিরিক্ত edge-টি বের করুন।
   Show Answer

   redundant.cpp

   #include <bits/stdc++.h>
   using namespace std;
   int p[1010];
   int f(int x) { return p[x] == x ? x : p[x] = f(p[x]); }
   int main() {
       vector<pair<int,int>> e = {{1,2},{2,3},{3,4},{1,4},{1,5}};
       iota(p, p + 1010, 0);
       for (auto& [u, v] : e) {
           if (f(u) == f(v)) { cout << u << " " << v << "\n"; return 0; }
           p[f(u)] = f(v);
       }
   }

   Run Copy

3. Accounts Merge — merge accounts that share at least one email; output unified email list per person.
   যেসব account-এ অন্তত একটি email common, সেগুলো merge করুন এবং প্রতি ব্যক্তির জন্য combined email list বের করুন।
   Show Answer

   accounts.cpp

   #include <bits/stdc++.h>
   using namespace std;
   int p[5000];
   int f(int x){return p[x]==x?x:p[x]=f(p[x]);}
   int main() {
       vector<vector<string>> A = {
           {"Arif","a@x","a@y"},
           {"Arif","a@y","a@z"},
           {"Mim","m@x"}
       };
       iota(p, p + 5000, 0);
       unordered_map<string,int> idx;
       unordered_map<string,string> owner;
       for (int i = 0; i < (int)A.size(); ++i) {
           for (int j = 1; j < (int)A[i].size(); ++j) {
               if (!idx.count(A[i][j])) { idx[A[i][j]] = idx.size(); owner[A[i][j]] = A[i][0]; }
               p[f(idx[A[i][1]])] = f(idx[A[i][j]]);
           }
       }
       map<int, set<string>> g;
       for (auto& [e, i] : idx) g[f(i)].insert(e);
       for (auto& [k, s] : g) {
           cout << owner[*s.begin()] << ": ";
           for (auto& e : s) cout << e << " ";
           cout << "\n";
       }
   }

   Run Copy

4. Minimum Cost to Connect Cities — Kruskal's MST using DSU.
   শহরগুলো connect করার সর্বনিম্ন খরচ — Kruskal MST।
   Show Answer

   kruskal.cpp

   #include <bits/stdc++.h>
   using namespace std;
   int p[100];
   int f(int x){return p[x]==x?x:p[x]=f(p[x]);}
   int main() {
       int n = 4;
       vector<tuple<int,int,int>> e = {{1,0,1},{4,0,2},{2,1,2},{3,1,3},{5,2,3}};
       sort(e.begin(), e.end());
       iota(p, p + n, 0);
       int cost = 0;
       for (auto& [w, u, v] : e) {
           if (f(u) != f(v)) { p[f(u)] = f(v); cost += w; }
       }
       cout << "MST cost = " << cost << "\n";
   }

   Run Copy

5. Smallest Equivalent String — given pairs of equivalent characters, output the lexicographically smallest version of a string.
   কিছু character-জোড়া equivalent দেওয়া আছে; একটি string-এর lexicographically smallest version বের করুন।
   Show Answer

   eqstr.cpp

   #include <bits/stdc++.h>
   using namespace std;
   int p[26];
   int f(int x){return p[x]==x?x:p[x]=f(p[x]);}
   void u(int a,int b){a=f(a);b=f(b);if(a==b)return;if(a<b)p[b]=a;else p[a]=b;}
   int main() {
       string s1 = "parker", s2 = "morris", base = "parser";
       iota(p, p + 26, 0);
       for (int i = 0; i < (int)s1.size(); ++i) u(s1[i] - 'a', s2[i] - 'a');
       for (char& c : base) c = f(c - 'a') + 'a';
       cout << base << "\n";
   }

   Run Copy

6. Satisfiability of Equality Equations — given equations like "a==b" and "b!=c", say whether all are satisfiable.
   "a==b" ও "b!=c" এর মতো সমীকরণ দেওয়া আছে — সবগুলো একসাথে সঙ্গতিপূর্ণ কিনা বলুন।
   Show Answer

   equations.cpp

   #include <bits/stdc++.h>
   using namespace std;
   int p[26];
   int f(int x){return p[x]==x?x:p[x]=f(p[x]);}
   int main() {
       vector<string> E = {"a==b","b==c","a!=c"};
       iota(p, p + 26, 0);
       for (auto& e : E) if (e[1] == '=') p[f(e[0] - 'a')] = f(e[3] - 'a');
       for (auto& e : E)
           if (e[1] == '!' && f(e[0] - 'a') == f(e[3] - 'a')) {
               cout << "unsatisfiable\n"; return 0;
           }
       cout << "satisfiable\n";
   }

   Run Copy

7. Number of Islands II — given a grid and a stream of "add land" operations, output the island count after each.
   গ্রিডে একটির পর একটি ভূমি যোগ হচ্ছে — প্রতিবার যোগের পর island-এর সংখ্যা প্রিন্ট করুন।
   Show Answer

   islands2.cpp

   #include <bits/stdc++.h>
   using namespace std;
   int p[10000];
   int f(int x){return p[x]==x?x:p[x]=f(p[x]);}
   int main() {
       int R = 3, C = 3;
       vector<int> land(R*C, 0);
       iota(p, p + R*C, 0);
       vector<pair<int,int>> ops = {{0,0},{0,1},{1,2},{2,1}};
       int cnt = 0, dr[] = {-1,1,0,0}, dc[] = {0,0,-1,1};
       for (auto& [r, c] : ops) {
           int id = r*C + c;
           if (land[id]) { cout << cnt << " "; continue; }
           land[id] = 1; ++cnt;
           for (int k = 0; k < 4; ++k) {
               int nr = r + dr[k], nc = c + dc[k];
               if (nr < 0 || nr >= R || nc < 0 || nc >= C) continue;
               int nid = nr*C + nc;
               if (land[nid] && f(id) != f(nid)) { p[f(id)] = f(nid); --cnt; }
           }
           cout << cnt << " ";
       }
       cout << "\n";
   }

   Run Copy