Number of Unicolored Paths between two nodes

Difficulty Level : Hard
Last Updated : 29 Apr, 2019

Given an undirected colored graph(edges are colored), with a source vertex ‘s’ and a destination vertex ‘d’, print number of paths which from given ‘s’ to ‘d’ such that the path is UniColored(edges in path having same color).

The edges are colored, here Colors are represented with numbers. At maximum, number of different colors will be number of edges.
Capture

Input : Graph
       u, v, color
        1, 2, 1
        1, 3, 2
        2, 3, 3
        2, 4, 2
        2, 5, 4
        3, 5, 3
        4, 5, 2
source = 2   destination = 5             

Output : 3
Explanation : There are three paths from 2 to 5
2 -> 5 with color red
2 -> 3 - > 5 with color sky blue
2 -> 4 - > 5  with color green

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Algorithm :
1. Do dfs traversal on the neighbour nodes of source node.
2. The color between source node and neighbour nodes is known, if the DFS traversal also have same color, proceed, else stop going on that path.
3. After reaching destination node, increment count by 1.

NOTE : Number of Colors will always be less than number of edges.

C++

// C++ code to find unicolored paths
#include <bits/stdc++.h>
using namespace std;
  
const int MAX_V = 100;
  
int color[MAX_V];
bool vis[MAX_V];
  
// Graph class represents a udirected graph
// using adjacency list representation
class Graph
{
    // vertices, edges, adjancy list
    int V;
    int E;
    vector<pair<int, int> > adj[MAX_V];
  
    // function used by UniColorPaths
    // DFS traversal o from x to y
    void dfs(int x, int y, int z);
  
// Constructor
public:
    Graph(int V, int E);
  
    // function to add an edge to graph
    void addEdge(int v, int w, int z);
  
    // finds paths between a and b having
    // same color edges
    int UniColorPaths(int a, int b);
};
  
Graph::Graph(int V, int E)
{
    this -> V = V;
    this -> E = E;
}
  
void Graph::addEdge(int a, int b, int c)
{
    adj[a].push_back({b, c}); // Add b to a’s list.
    adj[b].push_back({a, c}); // Add c to b’s list.
}
  
void Graph::dfs(int x, int y, int col)
{
    if (vis[x])
        return;
    vis[x] = 1;
  
    // mark this as a possible color to reach s to d
    if (x == y)
    {
        color[col] = 1;
        return;
    }
  
    // if the next edge is also of same color
    for (int i = 0; i < int(adj[x].size()); i++)
        if (adj[x][i].second == col)
            dfs(adj[x][i].first, y, col);
}
  
// function that finds paths between a and b
// such that all edges are same colored
// It uses recursive dfs()
int Graph::UniColorPaths(int a, int b)
{
  
    // dfs on nodes directly connected to source
    for (int i = 0; i < int(adj[a].size()); i++)
    {
        dfs(a, b, adj[a][i].second);
  
        // to visit again visited nodes
        memset(vis, 0, sizeof(vis));
    }
  
    int cur = 0;
    for (int i = 0; i <= E; i++)
        cur += color[i];
  
    return (cur);
}
  
// driver code
int main()
{
    // Create a graph given in the above diagram
    Graph g(5, 7);
    g.addEdge(1, 2, 1);
    g.addEdge(1, 3, 2);
    g.addEdge(2, 3, 3);
    g.addEdge(2, 4, 2);
    g.addEdge(2, 5, 4);
    g.addEdge(3, 5, 3);
    g.addEdge(4, 5, 2);
  
    int s = 2; // source
    int d = 5; // destination
  
    cout << "Number of unicolored paths : ";
    cout << g.UniColorPaths(s, d) << endl;
    return 0;
}

Python3

# Python3 code to find unicolored paths 
  
MAX_V = 100 
color = [0] * MAX_V
vis = [0] * MAX_V
  
# Graph class represents a udirected graph 
# using adjacency list representation 
class Graph: 
   
    def __init__(self, V, E):
        self.V = V
        self.E = E
        self.adj = [[] for i in range(MAX_V)]
      
    # Function used by UniColorPaths 
    # DFS traversal o from x to y 
    def dfs(self, x, y, col):
          
        if vis[x]: 
            return 
        vis[x] = 1 
      
        # mark this as a possible color to reach s to d 
        if x == y: 
            color[col] = 1 
            return 
           
        # if the next edge is also of same color 
        for i in range(0, len(self.adj[x])): 
            if self.adj[x][i][1] == col: 
                self.dfs(self.adj[x][i][0], y, col)
  
    def addEdge(self, a, b, c): 
       
        self.adj[a].append((b, c)) # Add b to a’s list. 
        self.adj[b].append((a, c)) # Add c to b’s list. 
  
    # Function that finds paths between a 
    # and b such that all edges are same 
    # colored. It uses recursive dfs() 
    def UniColorPaths(self, a, b): 
      
        global vis
          
        # dfs on nodes directly connected to source 
        for i in range(0, len(self.adj[a])): 
           
            self.dfs(a, b, self.adj[a][i][1]) 
      
            # to visit again visited nodes 
            vis = [0] * len(vis) 
           
        cur = 0 
        for i in range(0, self.E + 1): 
            cur += color[i] 
      
        return cur
  
# Driver code 
if __name__ == "__main__": 
   
    # Create a graph given in the above diagram 
    g = Graph(5, 7) 
    g.addEdge(1, 2, 1) 
    g.addEdge(1, 3, 2) 
    g.addEdge(2, 3, 3) 
    g.addEdge(2, 4, 2) 
    g.addEdge(2, 5, 4) 
    g.addEdge(3, 5, 3) 
    g.addEdge(4, 5, 2) 
  
    s = 2 # source 
    d = 5 # destination 
  
    print("Number of unicolored paths : ", end = "") 
    print(g.UniColorPaths(s, d)) 
  
# This code is contributed by Rituraj Jain

Output:

Number of unicolored paths : 3

Time Complexity : O(E * (E + V))

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