Number of minimum length paths between 1 to N including each node
Given an undirected and unweighted graph of N nodes and M edges, the task is to count the minimum length paths between node 1 to N through each of the nodes. If there is doesn’t exist any such path, then print “-1”.
Note: The path can pass through a node any number of times.
Examples:
Input: N = 4, M= 4, edges = {{1, 2}, {2, 3}, {1, 3}, {2, 4}}
Output: 1 1 1 1
Explanation:
Total paths of minimum length from 1 to 4, passing from 1 is 1.
Total paths of minimum length from 1 to 4, passing from 2 is 1.
Total paths of minimum length from 1 to 4, passing from 3 is 1.
Total paths of minimum length from 1 to 4, passing from 4 is 1.Input: N = 5, M = 5, edges = {{1, 2}, {1, 4}, {1 3}, {2, 5}, {2, 4}}
Output: 1 1 0 1 1
Approach: The given problem can be solved by performing two BFS, one from node 1 excluding node N and another from node N excluding node 1 to find the minimum distance of all the nodes from 1 and N, and the product of both the minimum distances will be the total count of minimum length paths from 1 to N including the node. Follow the steps below to solve the problem:
- Initialize a queue, say queue1 to perform BFS from node 1 and a queue queue2 to perform BFS from node N.
- Initialize arrays, say dist[] to store the shortest distance and ways[] to count the number of ways to reach that node.
- Perform two BFS and perform the following steps in each case:
- Pop from the queue and store node in x and its distance in dis.
- If dist[x] is smaller than dis then continue.
- Traverse the adjacency list of x and for each child y, if dist[y] is greater than dis + 1 then update dist[y] equals dis + 1 and ways[y] equals ways[x]. Otherwise, if dist[y] equals dis +1 then add ways[x] to ways[y].
- Finally, iterate over the range N, and for each node print the count of minimum length paths as ways1[i]*ways2[i].
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; #define ll long long int // Function to calculate the distances // from node 1 to N void countMinDistance( int n, int m, int edges[][2]) { // Stores the number of edges vector<ll> g[10005]; // Storing the edges in vector for ( int i = 0; i < m; i++) { int a = edges[i][0] - 1; int b = edges[i][1] - 1; g[a].push_back(b); g[b].push_back(a); } // Initialize queue queue<pair<ll, ll> > queue1; queue1.push({ 0, 0 }); vector< int > dist(n, 1e9); vector< int > ways1(n, 0); dist[0] = 0; ways1[0] = 1; // BFS from 1st node using queue while (!queue1.empty()) { auto up = queue1.front(); // Pop from queue queue1.pop(); int x = up.first; int dis = up.second; if (dis > dist[x]) continue ; if (x == n - 1) continue ; // Traversing the adjacency list for (ll y : g[x]) { if (dist[y] > dis + 1) { dist[y] = dis + 1; ways1[y] = ways1[x]; queue1.push({ y, dis + 1 }); } else if (dist[y] == dis + 1) { ways1[y] += ways1[x]; } } } // Initialize queue queue<pair<ll, ll> > queue2; queue2.push({ n - 1, 0 }); vector< int > dist1(n, 1e9); vector< int > ways2(n, 0); dist1[n - 1] = 0; ways2[n - 1] = 1; // BFS from last node while (!queue2.empty()) { auto up = queue2.front(); // Pop from queue queue2.pop(); int x = up.first; int dis = up.second; if (dis > dist1[x]) continue ; if (x == 0) continue ; // Traverse the adjacency list for (ll y : g[x]) { if (dist1[y] > dis + 1) { dist1[y] = dis + 1; ways2[y] = ways2[x]; queue2.push({ y, dis + 1 }); } else if (dist1[y] == 1 + dis) { ways2[y] += ways2[x]; } } } // Print the count of minimum // distance for ( int i = 0; i < n; i++) { cout << ways1[i] * ways2[i] << " " ; } } // Driver Code int main() { int N = 5, M = 5; int edges[M][2] = { { 1, 2 }, { 1, 4 }, { 1, 3 }, { 2, 5 }, { 2, 4 } }; countMinDistance(N, M, edges); return 0; } |
Python3
# Python 3 program for the above approach # Function to calculate the distances # from node 1 to N def countMinDistance(n, m, edges): # Stores the number of edges g = [[] for i in range ( 10005 )] # Storing the edges in vector for i in range (m): a = edges[i][ 0 ] - 1 b = edges[i][ 1 ] - 1 g[a].append(b) g[b].append(a) # Initialize queue queue1 = [] queue1.append([ 0 , 0 ]) dist = [ 1e9 for i in range (n)] ways1 = [ 0 for i in range (n)] dist[ 0 ] = 0 ways1[ 0 ] = 1 # BFS from 1st node using queue while ( len (queue1)> 0 ): up = queue1[ 0 ] # Pop from queue queue1 = queue1[: - 1 ] x = up[ 0 ] dis = up[ 1 ] if (dis > dist[x]): continue if (x = = n - 1 ): continue # Traversing the adjacency list for y in g[x]: if (dist[y] > dis + 1 ): dist[y] = dis + 1 ways1[y] = ways1[x] queue1.append([y, dis + 1 ]) elif (dist[y] = = dis + 1 ): ways1[y] + = ways1[x] # Initialize queue queue2 = [] queue2.append([n - 1 , 0 ]) dist1 = [ 1e9 for i in range (n)] ways2 = [ 0 for i in range (n)] dist1[n - 1 ] = 0 ways2[n - 1 ] = 1 # BFS from last node while ( len (queue2)> 0 ): up = queue2[ 0 ] # Pop from queue queue2 = queue2[: - 1 ] x = up[ 0 ] dis = up[ 1 ] if (dis > dist1[x]): continue if (x = = 0 ): continue # Traverse the adjacency list for y in g[x]: if (dist1[y] > dis + 1 ): dist1[y] = dis + 1 ways2[y] = ways2[x] queue2.append([y, dis + 1 ]) elif (dist1[y] = = 1 + dis): ways2[y] + = ways2[x] # Print the count of minimum # distance ways1[n - 1 ] = 1 ways2[n - 1 ] = 1 for i in range (n): print (ways1[i] * ways2[i],end = " " ) # Driver Code if __name__ = = '__main__' : N = 5 M = 5 edges = [[ 1 , 2 ],[ 1 , 4 ],[ 1 , 3 ],[ 2 , 5 ],[ 2 , 4 ]] countMinDistance(N, M, edges) # This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script> // Javascript program for the above approach // Function to calculate the distances // from node 1 to N function countMinDistance(n, m, edges) { // Stores the number of edges let g = new Array(10005).fill(0).map(() => []); // Storing the edges in vector for (let i = 0; i < m; i++) { let a = edges[i][0] - 1; let b = edges[i][1] - 1; g[a].push(b); g[b].push(a); } // Initialize queue let queue1 = []; queue1.push([0, 0]); let dist = new Array(n).fill(1e9); let ways1 = new Array(n).fill(0); dist[0] = 0; ways1[0] = 1; // BFS from 1st node using queue while (queue1.length > 0) { let up = queue1[0]; // Pop from queue queue1.pop(); let x = up[0]; let dis = up[1]; if (dis > dist[x]) continue ; if (x == n - 1) continue ; // Traversing the adjacency list for (let y of g[x]) { if (dist[y] > dis + 1) { dist[y] = dis + 1; ways1[y] = ways1[x]; queue1.push([y, dis + 1]); } else if (dist[y] == dis + 1) ways1[y] += ways1[x]; } } // Initialize queue let queue2 = []; queue2.push([n - 1, 0]); let dist1 = new Array(n).fill(1e9); let ways2 = new Array(n).fill(0); dist1[n - 1] = 0; ways2[n - 1] = 1; // BFS from last node while (queue2.length > 0) { let up = queue2[0]; // Pop from queue queue2.pop(); let x = up[0]; let dis = up[1]; if (dis > dist1[x]) continue ; if (x == 0) continue ; // Traverse the adjacency list for (let y of g[x]) { if (dist1[y] > dis + 1) { dist1[y] = dis + 1; ways2[y] = ways2[x]; queue2.push([y, dis + 1]); } else if (dist1[y] == 1 + dis) ways2[y] += ways2[x]; } } // Print the count of minimum // distance ways1[n - 1] = 1; ways2[n - 1] = 1; for (let i = 0; i < n; i++) document.write(ways1[i] * ways2[i] + " " ); } // Driver Code let N = 5; let M = 5; let edges = [ [1, 2], [1, 4], [1, 3], [2, 5], [2, 4], ]; countMinDistance(N, M, edges); // This code is contributed by gfgking </script> |
1 1 0 1 1
Time Complexity: O(N + M)
Auxiliary Space: O(N)
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