花花酱 LeetCode 547. Friend Circles

zxi — Fri, 22 Sep 2017 04:29:41 +0000

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 direct friend 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 i_thand j_th 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

Example 2:

Input: 
[[1,1,0],
 [1,1,1],
 [0,1,1]]
Output: 1

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

Idea:

Find all connected components using DFS

Solution: DFS

C++

// Author: Huahua
// Time complexity: O(n^2)
// Space complexity: O(n)
// Running Time: 25 ms
class Solution {
public:
    int findCircleNum(vector>& M) {
        if (M.empty()) return 0;
        int n = M.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (!M[i][i]) continue;
            ++ans;
            dfs(M, i, n);
        }
        return ans;
    }
private:
    void dfs(vector>& M, int curr, int n) {        
        // Visit all friends (neighbors)
        for (int i = 0; i < n; ++i) {
            if (!M[curr][i]) continue;
            M[curr][i] = M[i][curr] = 0;
            dfs(M, i, n);
        }
    }
};

Java

// Author: Huahua
// Time complexity: O(n^2)
// Space complexity: O(n)
// Running Time: 20 ms
class Solution {
    public int findCircleNum(int[][] M) {
        int n = M.length;
        if (n == 0) return 0;
        
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (M[i][i] == 0) continue;            
            ++ans;
            dfs(M, i, n);
        }
        return ans;
    }
    
    private void dfs(int[][] M, int curr, int n) {
        for (int i = 0; i < n; ++i) {
            if (M[curr][i] == 0) continue;
            M[curr][i] = M[i][curr] = 0;
            dfs(M, i, n);
        }
    }
}

Python

"""
Author: Huahua
Time complexity: O(n^2)
Space complexity: O(n)
Running Time: 162 ms
"""
class Solution(object):
    def findCircleNum(self, M):
        """
        :type M: List[List[int]]
        :rtype: int
        """
        
        def dfs(M, curr, n):
            for i in xrange(n):
                if M[curr][i] == 1:
                    M[curr][i] = M[i][curr] = 0
                    dfs(M, i, n)
        
        n = len(M)
        ans = 0
        for i in xrange(n):
            if M[i][i] == 1:
                ans += 1
                dfs(M, i, n)
        
        return ans

Solution 2: Union Find

C++

// Author: Huahua
// Runtime: 19 ms
class UnionFindSet {
public:
    UnionFindSet(int n) {
        parents_ = vector(n + 1, 0);
        ranks_ = vector(n + 1, 0);
        
        for (int i = 0; i < parents_.size(); ++i)
            parents_[i] = i;
    }
    
    bool Union(int u, int v) {
        int pu = Find(u);
        int pv = Find(v);
        if (pu == pv) return false;
        
        if (ranks_[pu] > ranks_[pv]) {
            parents_[pv] = pu;
        } else if (ranks_[pv] > ranks_[pu]) {
            parents_[pu] = pv;
        } else {
            parents_[pu] = pv;
            ++ranks_[pv];
        }
 
        return true;
    }
    
    int Find(int id) {        
        if (id != parents_[id])
            parents_[id] = Find(parents_[id]);        
        return parents_[id];
    }
    
private:
    vector parents_;
    vector ranks_;
};
 
class Solution {
public:
    int findCircleNum(vector>& M) {
        int n = M.size();
        UnionFindSet s(n);
        for (int i = 0; i < n; ++i)
            for (int j = i + 1; j < n; ++j)
                if (M[i][j] == 1) s.Union(i, j);
        
        unordered_set circles;
        for (int i = 0; i < n; ++i)
            circles.insert(s.Find(i));
        
        return circles.size();
    }
};

SCC Archives - Huahua's Tech Road

花花酱 LeetCode 547. Friend Circles

Solution: DFS

C++

Java

Python

Solution 2: Union Find

C++

Related Problems