547. Number of Provinces #

Problem #

There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a direct friend of B, and B is a directfriend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.

Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are direct friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.

Example 1:

Input: 
[[1,1,0],
 [1,1,0],
 [0,0,1]]
Output: 2
Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. 
The 2nd student himself is in a friend circle. So return 2.

Example 2:

Input: 
[[1,1,0],
 [1,1,1],
 [0,1,1]]
Output: 1
Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, 
so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.

Note:

1. N is in range [1,200].
2. M[i][i] = 1 for all students.
3. If M[i][j] = 1, then M[j][i] = 1.

Problem Summary #

There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive. If it is known that A is a friend of B, and B is a friend of C, then we can consider A to also be a friend of C. A so-called friend circle refers to the set of all friends.

Given an N * N matrix M representing the friend relationships between students in the class. If M[i][j] = 1, it means the ith and jth students are known to be friends with each other; otherwise, it is unknown. You must output the total number of known friend circles among all the students.

Note:

• N is in the range [1,200].
• For all students, M[i][i] = 1.
• If M[i][j] = 1, then M[j][i] = 1.

Solution Ideas #

• Given a two-dimensional matrix, the rows and columns of the matrix indicate whether two people are friends. If the value is 1, it means the two people are friends. Since a person is certainly friends with themselves, the diagonal entries are all 1, and the matrix is also symmetric about the diagonal from the top-left to the bottom-right.
• There are 2 solutions to this problem. The first solution is Union Find. Scan the matrix in order; if two people know each other and their roots are not equal, perform a union operation. After scanning the entire matrix, count how many different roots remain, and that is the final answer. The second solution is DFS or BFS. Use the idea of FloodFill to color nodes; each time one coloring is completed, increment the counter by one. After scanning the entire matrix, the counter’s value is the final result.

Code #

package leetcode

import (
	"github.com/halfrost/leetcode-go/template"
)

// Solution 1: Union Find

func findCircleNum(M [][]int) int {
	n := len(M)
	if n == 0 {
		return 0
	}
	uf := template.UnionFind{}
	uf.Init(n)
	for i := 0; i < n; i++ {
		for j := 0; j <= i; j++ {
			if M[i][j] == 1 {
				uf.Union(i, j)
			}
		}
	}
	return uf.TotalCount()
}

// Solution 2: FloodFill DFS brute-force solution
func findCircleNum1(M [][]int) int {
	if len(M) == 0 {
		return 0
	}
	visited := make([]bool, len(M))
	res := 0
	for i := range M {
		if !visited[i] {
			dfs547(M, i, visited)
			res++
		}
	}
	return res
}

func dfs547(M [][]int, cur int, visited []bool) {
	visited[cur] = true
	for j := 0; j < len(M[cur]); j++ {
		if !visited[j] && M[cur][j] == 1 {
			dfs547(M, j, visited)
		}
	}
}