Search results
Number of Provinces - There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.
You don't need to read input or print anything. Your task is to complete the function numProvinces () which takes an integer V and an adjacency matrix adj (as a 2d vector) as input and returns the number of provinces. adj [i] [j] = 1, if nodes i and j are connected and adj [i] [j] = 0, if not connected.
In-depth solution and explanation for LeetCode 547. Number of Provinces in Python, Java, C++ and more. Intuitions, example walk through, and complexity analysis.
Number of Provinces Leetcode Solution – We are given an adjacency matrix representation of a graph and need to find the number of provinces. Here province is a group of directly or indirectly connected cities and no other cities outside of the group. Example. Example 1: Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]] Output: 2. Example 2:
Jun 4, 2023 · Today, we’ll tackle problem 547 from LeetCode, titled “Number of Provinces.” This problem revolves around determining the total number of provinces in a given network of cities.
LeetCode Solutions in C++20, Java, Python, MySQL, and TypeScript.
Can you solve this real interview question? Number of Provinces - Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.
LeetCode: Number Of Provinces Leetcode Solution. Difficulty: Medium. Topics: breadth-first-search union-find depth-first-search graph. Problem Statement: There are n cities. Some of them are connected, while some are not.
Apr 10, 2024 · Given an undirected graph with V vertices. We say two vertices u and v belong to a single province if there is a path from u to v or v to u. Your task is to find the number of provinces....
Jul 4, 2024 · Number of Provinces | LeetCode. # java # beginners # algorithms # leetcode. class Solution { public int findCircleNum(int[][] isConnected) { int n = isConnected.length; int visited[] = new int[n]; int count = 0; for(int i=0; i<n; i++){ if(visited[i]==0){ dfs(isConnected, i, visited, n); count++; } } return count; } void dfs(int grid[][], int u ...
