Bipartite Graph - Google Top Interview Questions
Problem Statement :
Given an undirected graph represented as an adjacency list, return whether the graph is bipartite.
Constraints
n, m ≤ 250 where n and m are the number of rows and columns in graph
Example 1
Input
graph = [
[1],
[0]
]
Output
True
Explanation
This is bipartite since the node 1 can belong in set A and node 2 can belong in set B. Then the edges 0 -
> 1 and 1 -> 0 has one node in A and one node in B
Example 2
Input
Visualize
graph = [
[2, 3],
[2, 3],
[0, 1],
[0, 1]
]
Output
True
Explanation
0 and 1 can belong in set A and 2 and 3 can belong in set B.
Example 3
Input
graph = [
[1, 2, 3],
[0, 2],
[0, 1, 3],
[0, 2]
]
Output
False
Explanation
No matter how the nodes are partitioned, an edge will belong to the same set.
Solution :
Solution in C++ :
bool isbipartite(vector<vector<int>> &graph, int src, vector<int> &color) {
for (int i = 0; i < graph[src].size(); i++) {
if (color[graph[src][i]] == -1) {
color[graph[src][i]] = !color[src];
if (!isbipartite(graph, graph[src][i], color)) return false;
} else if (color[graph[src][i]] == color[src])
return false;
}
return true;
}
bool solve(vector<vector<int>> &graph) {
// DFS by coloring nodes.
// Color root by R and subsequent by Blue if we found neighbour with same color it is not BPG.
vector<int> colors(graph.size(), -1);
for (int i = 0; i < graph.size(); i++) {
if (colors[i] == -1) {
colors[i] = 1;
if (!isbipartite(graph, i, colors)) return false;
}
}
return true;
}
Solution in Java :
import java.util.*;
class Solution {
private Set<Integer> visitedSet = new HashSet();
private Set<Integer> seta = new HashSet();
private Set<Integer> setb = new HashSet();
private boolean isBipartite = true;
public boolean solve(int[][] graph) {
if (graph == null || graph.length == 0)
return false;
for (int i = 0; i < graph.length; i++) {
if (!visitedSet.contains(i)) {
if (!isBipartite)
break;
;
dfs(graph, i, -1, 0);
}
}
return isBipartite;
}
private void dfs(int[][] graph, int vertex, int parent, int setIdentifier) {
addToSet(vertex, setIdentifier);
int[] adj = graph[vertex];
if (!isBipartite)
return;
for (int adjVertex : adj) {
if (adjVertex != parent) {
if (setIdentifier == 0 && seta.contains(adjVertex)) {
isBipartite = false;
return;
}
if (setIdentifier == 1 && setb.contains(adjVertex)) {
isBipartite = false;
return;
}
if (!seta.contains(adjVertex) && !setb.contains(adjVertex))
dfs(graph, adjVertex, vertex, (setIdentifier == 0 ? 1 : 0));
}
}
}
private void addToSet(int vertex, int setIdentifier) {
visitedSet.add(vertex);
if (setIdentifier == 0)
seta.add(vertex);
else
setb.add(vertex);
}
}
Solution in Python :
class Solution:
def solve(self, arr):
n = len(arr)
color_start = 1
seen = {}
def can_split(node, color):
if node in seen:
return seen[node]
seen[node] = color
for nei in arr[node]:
if nei not in seen:
if not can_split(nei, -color):
return False
elif seen[nei] != -color:
return False
return True
for i in range(n):
if not can_split(i, color_start):
return False
return True
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