Problem of the day
Given a graph, check whether the graph is bipartite or not. Your function should return true if the given graph's vertices can be divided into two independent sets, ‘U’ and ‘V’ such that every edge (‘u’, ‘v’) either connects a vertex from ‘U’ to ‘V’ or a vertex from ‘V’ to ‘U’.
You are given a 2D array ‘edges’ which contains 0 and 1, where ‘edges[i][j]’ = 1 denotes a bi-directional edge from ‘i’ to ‘j’.
Note:If edges[i][j] = 1, that implies there is a bi-directional edge between ‘i’ and ‘j’, that means there exists both edges from ‘i’ to ‘j’ and to ‘j’ to ‘i’.
For example
Given:
‘N’ = 3
‘edges’ = [[0, 1, 1], [0, 0, 1], [0,0,0]].
The first line of input contains an integer ‘T’ denoting the number of test cases.
The first line of each test case contains two space-separated integers, ‘N,’ where ‘N’ is the number of rows in ‘edges’ and the number of columns in ‘edges’.
The next ‘N’ line of each test case contains ‘N’ space-separated integers which tell if there is an edge between ‘i’ and ‘j’.
Output Format :
For each test case, You are supposed to return a bool value determining whether the graph is bipartite or not.
Note:
You are not required to print the expected output; it has already been taken care of. Just implement the function.
1 <= ‘T’ <= 10
2 <= ‘N’ <= 300
0 <= ‘edges[i][j]’ <= 1.
Time Limit: 1sec.
2
4
0 1 0 0
0 0 0 1
0 0 0 0
0 0 0 0
3
0 1 1
0 0 1
0 0 0
1
0
In the first test case, the graph is visualized as below,
The graph can be divided into 2 disjointed sections, i.e. S1 = {0,2} and S2 = {1,3}. Therefore the answer is true.
In the second test case, the graph is visualized as below:
The answer is 0 since there is no way this graph can be divided into 2 disjoint sets of points.
2
4
0 0 1 0
0 0 0 1
0 0 0 0
0 0 0 0
3
0 1 1
0 0 0
0 0 0
1
1
Can we think of 2 sets as 2 different colors?
The idea is that a bipartite can be colored in such a way that only 2 colors in total would be used, where the neighbor nodes have different colors. Therefore we can do a breadth-first search and assign colors to each vertex and if at any point neighboring nodes have the same color we return false.
The steps are as follows:
O(N ^ 2), where ‘N’ is the number of nodes.
As we have to color all the nodes present, and for each node, we have to traverse its neighbors. In the worst cases, there can be at most ‘N’ neighbors. Hence, the overall time complexity is O(N ^ 2).
O(N), where ‘N’ is the number of nodes.
Since we are using an array to store the color of each node and a queue to do BFS, the overall space complexity will be O(N).