README.md

547. 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.

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.

Example 1:

Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
Output: 2

Example 2:

Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3

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]

Solutions (Ruby)

1. DFS

# @param {Integer[][]} is_connected
# @return {Integer}
def find_circle_num(is_connected)
  seen = [false] * is_connected.size
  stack = []
  ret = 0

  (0...is_connected.size).each do |i|
    next if seen[i]

    ret += 1
    stack.push(i)

    until stack.empty?
      c = stack.pop
      seen[c] = true
      (i + 1...is_connected.size).each do |j|
        stack.push(j) if !seen[j] && is_connected[c][j] == 1
      end
    end
  end

  ret
end

Solutions (Rust)

1. DFS

impl Solution {
    pub fn find_circle_num(m: Vec<Vec<i32>>) -> i32 {
        let mut seen = vec![false; m.len()];
        let mut stack = vec![];
        let mut ret = 0;

        for i in 0..m.len() {
            if !seen[i] {
                ret += 1;
                stack.push(i);

                while let Some(c) = stack.pop() {
                    seen[c] = true;
                    for j in i + 1..m.len() {
                        if !seen[j] && m[c][j] == 1 {
                            stack.push(j);
                        }
                    }
                }
            }
        }

        ret
    }
}