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| 1 | +class Solution { |
| 2 | +public: |
| 3 | + bool isBipertite(vector < vector <int> > &graph, int src, vector <int> &visited){ |
| 4 | + // Create a color array to store colors |
| 5 | + // assigned to all veritces. Vertex |
| 6 | + // number is used as index in this array. |
| 7 | + // The value '-1' of colorArr[i] |
| 8 | + // is used to indicate that no color |
| 9 | + // is assigned to vertex 'i'. The value 1 |
| 10 | + // is used to indicate first color |
| 11 | + // is assigned and value 0 indicates |
| 12 | + // second color is assigned. |
| 13 | + |
| 14 | + vector <int> color(graph.size()); |
| 15 | + |
| 16 | + for(int i = 0; i<color.size(); i++){ |
| 17 | + color[i] = -1; |
| 18 | + } |
| 19 | + // Assign first color to source |
| 20 | + color[src] = 1; |
| 21 | + |
| 22 | + queue <int> q; |
| 23 | + q.push(src); |
| 24 | + |
| 25 | + while(!q.empty()){ |
| 26 | + |
| 27 | + int temp = q.front(); |
| 28 | + visited[temp] = 1; |
| 29 | + q.pop(); |
| 30 | + // Return false if there is a self-loop , not possible in this problem |
| 31 | + if(graph[temp][temp] == 1){ |
| 32 | + return false; |
| 33 | + } |
| 34 | + // Find all non-colored adjacent vertices |
| 35 | + for(int i = 0; i<graph.size(); i++){ |
| 36 | + // An edge from u to v exists and destination v is not colored |
| 37 | + if(graph[temp][i] && color[i] == -1){ |
| 38 | + // Assign alternate color to this adjacent v of u |
| 39 | + color[i] = 1 - color[temp]; |
| 40 | + q.push(i); |
| 41 | + } |
| 42 | + // An edge from u to v exists and destination v is colored with same color as u |
| 43 | + else if(graph[temp][i] && color[i] == color[temp]){ |
| 44 | + return false; |
| 45 | + } |
| 46 | + } |
| 47 | + } |
| 48 | + |
| 49 | + return true; |
| 50 | + } |
| 51 | + |
| 52 | + |
| 53 | + bool possibleBipartition(int N, vector<vector<int>>& dislikes) { |
| 54 | + vector < vector <int> > graph(N+1, vector <int> (N+1, 0)); |
| 55 | + for(int i = 0; i < dislikes.size(); i++){ |
| 56 | + int u = dislikes[i][0]; |
| 57 | + int v = dislikes[i][1]; |
| 58 | + graph[u][v] = 1; |
| 59 | + graph[v][u] = 1; |
| 60 | + } |
| 61 | + int res = 0; |
| 62 | + int src = 0; |
| 63 | + vector <int> visited(N+1,0); |
| 64 | + visited[0] = 1; |
| 65 | + res = isBipertite(graph, 1, visited); |
| 66 | + if(res == 1){ |
| 67 | + for(int i = 0; i<=N; i++){ |
| 68 | + if(visited[i] == 0){ |
| 69 | + res = isBipertite(graph, i, visited); |
| 70 | + } |
| 71 | + if(res == 0){ |
| 72 | + return 0; |
| 73 | + } |
| 74 | + } |
| 75 | + } |
| 76 | + return res; |
| 77 | + } |
| 78 | +}; |
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