From e92eb158311e728b62d11115d261f328debc7927 Mon Sep 17 00:00:00 2001 From: Sainik Khaddar <73096052+playboySeductor@users.noreply.github.com> Date: Sun, 2 Jul 2023 17:02:28 +0530 Subject: [PATCH] Create README - LeetHub --- Number of Provinces - GFG/README.md | 29 +++++++++++++++++++++++++++++ 1 file changed, 29 insertions(+) create mode 100644 Number of Provinces - GFG/README.md diff --git a/Number of Provinces - GFG/README.md b/Number of Provinces - GFG/README.md new file mode 100644 index 0000000..ea1d4cd --- /dev/null +++ b/Number of Provinces - GFG/README.md @@ -0,0 +1,29 @@ +# Number of Provinces +## Medium +

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.

Note: A province is a group of directly or indirectly connected cities and no other cities outside of the group.

+

Example 1:

+
Input:
+[
+ [1, 0, 1],
+ [0, 1, 0],
+ [1, 0, 1]
+]
+
+Output:
+2
+Explanation:
+The graph clearly has 2 Provinces [1,3] and [2]. As city 1 and city 3 has a path between them they belong to a single province. City 2 has no path to city 1 or city 3 hence it belongs to another province.
+
+
Example 2:
+
Input:
+[
+ [1, 1],
+ [1, 1]
+]
+
+Output :
+1
+
+


Your Task:  
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 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.

+


Expected Time Complexity: O(V2)
Expected Auxiliary Space: O(V)

+


Constraints:
1 ≤ V ≤ 500

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