Create Day 27 Possible Bipartition.cpp

Author: tusharjaiswal123

Day 27 Possible Bipartition.cpp

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+PROBLEM:
+
+
+
+
+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. 1 <= N <= 2000
+2. 0 <= dislikes.length <= 10000
+3. 1 <= dislikes[i][j] <= N
+4. dislikes[i][0] < dislikes[i][1]
+5. There does not exist i != j for which dislikes[i] == dislikes[j].
+
+
+
+
+
+SOLUTION:
+
+
+
+class Solution {
+public:
+    
+    bool dfs(vector<int> adj[],vector<bool>& vis,int color[],int s)
+    {
+        vis[s]=true;
+        
+        for(auto k:adj[s])
+        {
+            if(!vis[k])
+            {
+                color[k]=1-color[s];
+                
+                if(dfs(adj,vis,color,k)==false)
+                return false;
+            
+            }
+            else if(color[k]==color[s])
+                return false;
+        }
+        
+        return true;
+    }
+    
+    bool possibleBipartition(int N, vector<vector<int>>& dislikes) {
+        
+        int n,m,i,j,a,b;
+        n=dislikes.size();
+        
+        vector<int> adj[N+1];
+        vector<bool> vis(N+1,false);
+        int color[N+1];
+        
+        for(i=0;i<n;i++)
+        {
+            a=dislikes[i][0];
+            b=dislikes[i][1];
+            
+            adj[a].push_back(b);
+            adj[b].push_back(a);
+        }
+        
+        for(i=1;i<=N;i++)
+        {
+            if(!vis[i])
+            {   
+                color[i]=0;
+                if(dfs(adj,vis,color,i)==false)
+                    return false;
+            }
+        }
+        
+        
+        return true;
+        
+    }
+};