Number of Unicolored Paths between two nodes

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))

My Personal Notes

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

C++

Python3

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