doing a dfs and checking if neighbours are opposite colour

Author: Arjiit

DFS/isBipartite.java

@@ -0,0 +1,35 @@
+class Solution {
+    public boolean isBipartite(int[][] graph) {
+        /*
+        we have to colour the nodes such that the next node is of different colour.
+        basically index position in graph array is parent and the array elements is
+        the neighbours. So, ith position we colour by 1 and its neighbour we try to
+        colour differently. We note the colour in colours array.
+        
+        O(n) time complexity
+        */
+        int nodes = graph.length; // number of nodes
+        int[] colours = new int[nodes]; // to keep track whether the node is coloured or not
+        
+        for (int i=0; i<graph.length; i++) {
+            if ((colours[i] == 0) && (!dfs(graph, colours, 1, i)) ) { // if it's not coloured and we can't colour it.
+                return false;
+            }
+        }
+        return true;
+    }
+    // utility function which will tell if we can colour the node.
+    public boolean dfs(int[][] graph, int[] colours, int colour, int node) {
+        if (colours[node] != 0) { // if the node is already coloured
+            return (colours[node] == colour); // check if it is of the colour we want.
+        } else {
+            colours[node] = colour; // colour the node with the colour
+            for (int neighbour: graph[node]) { // try to colour it neighbours
+             if (!dfs(graph, colours, -colour, neighbour)) { // if we can't colour the neighbour to opposite colour
+                 return false; // return false if we can't colour it opposite.
+             }   
+            }
+        }
+        return true;
+    }
+}