isGraphBipartite.swift

/*
 Given an undirected graph, return true if and only if it is bipartite.

 Recall that a graph is bipartite if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

 The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists.  Each node is an integer between 0 and graph.length - 1.  There are no self edges or parallel edges: graph[i] does not contain i, and it doesn't contain any element twice.

 Example 1:
 Input: [[1,3], [0,2], [1,3], [0,2]]
 Output: true
 Explanation: 
 The graph looks like this:
 0----1
 |    |
 |    |
 3----2
 We can divide the vertices into two groups: {0, 2} and {1, 3}.

 Example 2:
 Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
 Output: false
 Explanation: 
 The graph looks like this:
 0----1
 | \  |
 |  \ |
 3----2
 We cannot find a way to divide the set of nodes into two independent subsets.

 Note:

1. graph will have length in range [1, 100].
2. graph[i] will contain integers in range [0, graph.length - 1].
3. graph[i] will not contain i or duplicate values.
4. The graph is undirected: if any element j is in graph[i], then i will be in graph[j].
*/

class Solution {
        enum Color {
        case blue
        case red
        case uncolor
    }

    func isBipartite(_ graph: [[Int]]) -> Bool {

        var color = [Int:Color]()
        var current:Color = .blue

        for i in 0..<graph.count {
            color[i] = .uncolor
        }

        for var i in 0..<graph.count {
            if color[i] == .uncolor {
                var stack = [i]
                current = current == .blue ? .red : .blue
                color[i] = current

                while !stack.isEmpty {
                    i = stack.removeLast()
                    for n in graph[i] {
                        if color[n] == .uncolor {
                            stack.append(n)
                            current = color[i] == .blue ? .red : .blue
                            color[n] = current
                        } else if color[n] == color[i] { return false }
                    }
                }
            }
        }
        return true
    }
}