# Daily Coding Challenge #18

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

``````/*
LeetCode - Possible Bipartition

Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size.

Each person may dislike some other people, and they should not go into the same group.

Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.

Return true if and only if it is possible to split everyone into two groups in this way.

Example 1:

Input: N = 4, dislikes = [[1,2],[1,3],[2,4]]
Output: true
Explanation: group1 [1,4], group2 [2,3]
Example 2:

Input: N = 3, dislikes = [[1,2],[1,3],[2,3]]
Output: false
Example 3:

Input: N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
Output: false

Note:

1 <= N <= 2000
0 <= dislikes.length <= 10000
1 <= dislikes[i][j] <= N
dislikes[i] < dislikes[i]
There does not exist i != j for which dislikes[i] == dislikes[j].
*/

class Solution {
public:
bool possibleBipartition(int N, vector<vector<int>>& dislikes) {
vector<vector<int>> graph(N,vector<int>(N,0));
vector<int> group(N,0);
for(vector<int> d:dislikes){
// build a bi-directional graph
graph[d-1][d-1]=graph[d-1][d-1]=1;
}

// traverse each person
for(int i=0;i<N;i++){
// check if this person does not belong to any group and its dfs result
// group[i]==0  : no group
// group[i]==1  : group 1
// group[i]==-1 : group 2
if(!group[i]&&!dfs(graph,group,i,1)){
return false;
}
}
return true;
}
private:
bool dfs(vector<vector<int>>& graph, vector<int>& group, int index, int grp){
// assign this person to group grp
group[index]=grp;
// traverse each graph
for(int i=0;i<graph.size();i++){
// if person index dislikes person i
if(graph[index][i]){
// and if person i belongs to the same group
if(group[i]==grp){
// then return false
return false;
}
// check if person i does not belong to any group and its dfs result
if(!group[i]&&!dfs(graph,group,i,-grp)){
return false;
}
}
}
return true;
}
};
``````

The source code is available in corresponding repo below. Star and watch for timely updates!

## wingkwong / atcoder

### Discussion   