Given an undirected and unweighted graph and two nodes as **source** and **destination**, the task is to print all the paths of the shortest length between the given source and destination.

**Examples:**

Input:source = 0, destination = 5

Output:

0 -> 1 -> 3 -> 5

0 -> 2 -> 3 -> 5

0 -> 1 -> 4 -> 5

Explanation:

All the above paths are of length 3, which is the shortest distance between 0 and 5.

**Approach:** The is to do a Breadth First Traversal (BFS) for a graph. Below are the steps:

- Start BFS traversal from source vertex.
- While doing BFS, store the shortest distance to each of the other nodes and also maintain a parent vector for each of the nodes.
- Make the parent of source node as
**“-1”**. For each node, it will store all the parents for which it has the shortest distance from the source node. - Recover all the paths using parent array. At any instant, we will push one vertex in the path array and then call for all its parents.
- If we encounter “-1” in the above steps, then it means a path has been found and can be stored in the paths array.

Below is the implementation of the above approach:

`// Cpp program for the above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to form edge between ` `// two vertices src and dest ` `void` `add_edge(vector<` `int` `> adj[], ` ` ` `int` `src, ` `int` `dest) ` `{ ` ` ` `adj[src].push_back(dest); ` ` ` `adj[dest].push_back(src); ` `} ` ` ` `// Function which finds all the paths ` `// and stores it in paths array ` `void` `find_paths(vector<vector<` `int` `> >& paths, ` ` ` `vector<` `int` `>& path, ` ` ` `vector<` `int` `> parent[], ` ` ` `int` `n, ` `int` `u) ` `{ ` ` ` `// Base Case ` ` ` `if` `(u == -1) { ` ` ` `paths.push_back(path); ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `// Loop for all the parents ` ` ` `// of the given vertex ` ` ` `for` `(` `int` `par : parent[u]) { ` ` ` ` ` `// Insert the current ` ` ` `// vertex in path ` ` ` `path.push_back(u); ` ` ` ` ` `// Recursive call for its parent ` ` ` `find_paths(paths, path, parent, ` ` ` `n, par); ` ` ` ` ` `// Remove the current vertex ` ` ` `path.pop_back(); ` ` ` `} ` `} ` ` ` `// Function which performs bfs ` `// from the given souce vertex ` `void` `bfs(vector<` `int` `> adj[], ` ` ` `vector<` `int` `> parent[], ` ` ` `int` `n, ` `int` `start) ` `{ ` ` ` `// dist will contain shortest distance ` ` ` `// from start to every other vertex ` ` ` `vector<` `int` `> dist(n, INT_MAX); ` ` ` ` ` `queue<` `int` `> q; ` ` ` ` ` `// Insert source vertex in queue and make ` ` ` `// its parent -1 and distance 0 ` ` ` `q.push(start); ` ` ` `parent[start] = { -1 }; ` ` ` `dist[start] = 0; ` ` ` ` ` `// Untill Queue is empty ` ` ` `while` `(!q.empty()) { ` ` ` `int` `u = q.front(); ` ` ` `q.pop(); ` ` ` `for` `(` `int` `v : adj[u]) { ` ` ` `if` `(dist[v] > dist[u] + 1) { ` ` ` ` ` `// A shorter distance is found ` ` ` `// So erase all the previous parents ` ` ` `// and insert new parent u in parent[v] ` ` ` `dist[v] = dist[u] + 1; ` ` ` `q.push(v); ` ` ` `parent[v].clear(); ` ` ` `parent[v].push_back(u); ` ` ` `} ` ` ` `else` `if` `(dist[v] == dist[u] + 1) { ` ` ` ` ` `// Another candidate parent for ` ` ` `// shortes path found ` ` ` `parent[v].push_back(u); ` ` ` `} ` ` ` `} ` ` ` `} ` `} ` ` ` `// Function which prints all the paths ` `// from start to end ` `void` `print_paths(vector<` `int` `> adj[], ` ` ` `int` `n, ` `int` `start, ` `int` `end) ` `{ ` ` ` `vector<vector<` `int` `> > paths; ` ` ` `vector<` `int` `> path; ` ` ` `vector<` `int` `> parent[n]; ` ` ` ` ` `// Function call to bfs ` ` ` `bfs(adj, parent, n, start); ` ` ` ` ` `// Function call to find_paths ` ` ` `find_paths(paths, path, parent, n, end); ` ` ` ` ` `for` `(` `auto` `v : paths) { ` ` ` ` ` `// Since paths contain each ` ` ` `// path in reverse order, ` ` ` `// so reverse it ` ` ` `reverse(v.begin(), v.end()); ` ` ` ` ` `// Print node for the current path ` ` ` `for` `(` `int` `u : v) ` ` ` `cout << u << ` `" "` `; ` ` ` `cout << endl; ` ` ` `} ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `// Number of vertices ` ` ` `int` `n = 6; ` ` ` ` ` `// array of vectors is used ` ` ` `// to store the graph ` ` ` `// in the form of an adjacency list ` ` ` `vector<` `int` `> adj[n]; ` ` ` ` ` `// Given Graph ` ` ` `add_edge(adj, 0, 1); ` ` ` `add_edge(adj, 0, 2); ` ` ` `add_edge(adj, 1, 3); ` ` ` `add_edge(adj, 1, 4); ` ` ` `add_edge(adj, 2, 3); ` ` ` `add_edge(adj, 3, 5); ` ` ` `add_edge(adj, 4, 5); ` ` ` ` ` `// Given source and destination ` ` ` `int` `src = 0; ` ` ` `int` `dest = n - 1; ` ` ` ` ` `// Function Call ` ` ` `print_paths(adj, n, src, dest); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

0 1 3 5 0 2 3 5 0 1 4 5

**Time Complexity:** *O(V + E)* where V is the number of vertices and E is the number of edges.

**Auxiliary Space:** *O(V)* where V is the number of vertices.

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