Problem 547

README.md

@@ -158,6 +158,7 @@ In order to practice with similar data structures I'll be placing each problem i
 ### Graph
 
 - [Clone Graph #133](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/clone-graph-133.md)
+- [Number of Provinces #547](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/number-of-provinces-547.md)
 
 ### Heap
 
@@ -193,6 +194,7 @@ Within the problems above there are several patterns that often occur. I plan to
 - [01 Matrix #542](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/01-matrix-542.md)
 - [Binary Tree Level Order Traversal #102](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/binary-tree-level-102.md)
 - [Clone Graph #133](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/clone-graph-133.md)
+- [Number of Provinces #547](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/number-of-provinces-547.md)
 
 ### Depth First Search
 
@@ -203,6 +205,7 @@ Within the problems above there are several patterns that often occur. I plan to
 - [Diameter of Binary Tree #543](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/easy/diameter-binary-tree-543.md)
 - [Maximum Depth of Binary Tree #104](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/easy/depth-binary-tree-104.md)
 - [Clone Graph #133](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/clone-graph-133.md)
+- [Number of Provinces #547](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/number-of-provinces-547.md)
 
 ### Divide & Conquer
 

medium/number-of-provinces-547.md

@@ -0,0 +1,70 @@
+# Number of Provinces
+
+Page on leetcode: https://leetcode.com/problems/number-of-provinces/
+
+## Problem Statement
+
+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.
+
+A province is a group of directly or indirectly connected cities and no other cities outside of the group.
+
+You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.
+
+Return the total number of provinces.
+
+### Constraints
+
+- 1 <= n <= 200
+- n == isConnected.length
+- n == isConnected[i].length
+- isConnected[i][j] is 1 or 0.
+- isConnected[i][i] == 1
+- isConnected[i][j] == isConnected[j][i]
+
+### Example
+
+```
+Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
+Output: 2
+```
+
+## Solution
+
+### Initial Thoughts
+
+- iterate thru arrays
+- Have a counter, when you find a break add to counter
+- minimum province is 1
+- Can I check from i = 0 to i = n/2
+
+### Optimized Solution
+
+Time and space complexity for this solution would be O(n). This is a recursive DFS solution however it can also be solved with BFS and using a Union Find.
+You can see a good solution to this problem here: https://www.youtube.com/watch?v=S5UUvCTM0V4
+
+```javascript
+const findCircleNum = function (isConnected) {
+  const visited = new Set();
+  let numOfProv = 0;
+
+  // Iterate thru matrix
+  for (let i = 0; i < isConnected.length; i++) {
+    if (!visited.has(i)) {
+      dfs(i);
+      numOfProv++;
+    }
+  }
+
+  return numOfProv;
+
+  // Helper function for DFS search
+  function dfs(i) {
+    for (let j = 0; j < isConnected.length; j++) {
+      if (isConnected[i][j] === 1 && !visited.has(j)) {
+        visited.add(j);
+        dfs(j);
+      }
+    }
+  }
+};
+```