# 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.

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

**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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