Yandex

Binary TreesBinary Trees36

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree

GraphsGraphs46

  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

TriesTries1

Number of Provinces
Using Graph Traversal



Problem Statement

Given an undirected graph with V vertices represented as an adjacency matrix isConnected, a province is a group of directly or indirectly connected cities (vertices). A city is connected to itself.

Your task is to determine the total number of such provinces (i.e., the number of connected components in the graph).

Examples

isConnected (Adjacency Matrix) Number of Provinces Description
[[1,1,0],[1,1,0],[0,0,1]] 2 Cities 0 and 1 form one province, city 2 forms another.
[[1,0,0],[0,1,0],[0,0,1]] 3 Each city is isolated, hence 3 provinces.
[[1,1,1],[1,1,1],[1,1,1]] 1 All cities are connected, so only one province.
[[1]] 1 Only one city, so one province.
[] 0 No cities, hence no province.

Solution

Understanding the Problem

The problem involves identifying how many disconnected groups of cities exist, where a direct or indirect connection means two cities belong to the same group. These groups are known as 'provinces' in this context, and they represent the number of connected components in an undirected graph.

Case 1: All Cities Are Isolated

If none of the cities are connected to any other (i.e., the adjacency matrix has only 1s on the diagonal and 0s elsewhere), then each city is its own province. In this case, the number of provinces is equal to the number of cities.

Case 2: All Cities Are Fully Connected

If every city is connected either directly or through a chain of connections to every other city, then the graph is a single connected component. So the answer is 1 province.

Case 3: Multiple Connected Groups

When the graph has multiple subsets of cities where each subset is internally connected but disconnected from other subsets, each subset forms its own province. The algorithm identifies each such group through DFS or BFS, marking visited cities to avoid recounting.

Step-by-Step Traversal

The traversal (either DFS or BFS) starts from each unvisited city. If a city has not been visited, a new traversal begins from that city and marks all reachable cities as visited. This entire traversal corresponds to one province. Once all cities are checked, the total count of such traversals equals the number of provinces.

Algorithm Steps

  1. Initialize a visited array of size V with all values set to false.
  2. Initialize provinceCount = 0.
  3. For each city i from 0 to V-1:
    1. If visited[i] is false:
      1. Perform DFS/BFS starting from i.
      2. Mark all connected cities as visited.
      3. Increment provinceCount.
  4. Return provinceCount.

Code

JavaScript
function findCircleNum(isConnected) {
  const n = isConnected.length;
  const visited = new Array(n).fill(false);

  function dfs(city) {
    for (let neighbor = 0; neighbor < n; neighbor++) {
      if (isConnected[city][neighbor] === 1 && !visited[neighbor]) {
        visited[neighbor] = true;
        dfs(neighbor);
      }
    }
  }

  let provinces = 0;

  for (let i = 0; i < n; i++) {
    if (!visited[i]) {
      visited[i] = true;
      dfs(i);
      provinces++;
    }
  }

  return provinces;
}

console.log("Provinces:", findCircleNum([[1,1,0],[1,1,0],[0,0,1]])); // Output: 2
console.log("Provinces:", findCircleNum([[1,0,0],[0,1,0],[0,0,1]])); // Output: 3

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(V + E)Even if all cities are in a single province, DFS/BFS will still visit all vertices and edges to ensure full connectivity.
Average CaseO(V + E)Each node and edge is visited once across all connected components using DFS or BFS.
Worst CaseO(V + E)In the worst case (e.g., each city is isolated), we still visit every node and no edges, resulting in a traversal of size V.

Space Complexity

O(V)

Explanation: A visited array of size V is required to track which cities have already been checked during traversal.



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