Given a directed graph with N nodes and E edges where the weight of each of the edge is > 1, also given a source S and a destination D. The task is to find the path with the minimum product of edges from S to D. If there is no path from S to D then print -1.
Examples:
Input: N = 3, E = 3, Edges = {{{1, 2}, 5}, {{1, 3}, 9}, {{2, 3}, 1}}, S = 1, and D = 3
Output: 5
The path with smallest product of edges will be 1->2->3
with product as 5*1 = 5.Input: N = 3, E = 3, Edges = {{{3, 2}, 5}, {{3, 3}, 9}, {{3, 3}, 1}}, S = 1, and D = 3
Output: -1
Approach: The idea is to use Dijkstra’s shortest path algorithm with a slight variation.
The following steps can be followed to compute the result:
- If the source is equal to the destination then return 0.
- Initialise a priority-queue pq with S and its weight as 1 and a visited array v[].
- While pq is not empty:
- Pop the top-most element from pq. Let’s call it as curr and its product of distance as dist.
- If curr is already visited then continue.
- If curr is equal to D then return dist.
- Iterate all the nodes adjacent to curr and push into pq (next and dist + gr[nxt].weight)
- return -1.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the smallest // product of edges double dijkstra( int s, int d,
vector<vector<pair< int , double > > > gr)
{ // If the source is equal
// to the destination
if (s == d)
return 0;
// Initialise the priority queue
set<pair< int , int > > pq;
pq.insert({ 1, s });
// Visited array
bool v[gr.size()] = { 0 };
// While the priority-queue
// is not empty
while (pq.size()) {
// Current node
int curr = pq.begin()->second;
// Current product of distance
int dist = pq.begin()->first;
// Popping the top-most element
pq.erase(pq.begin());
// If already visited continue
if (v[curr])
continue ;
// Marking the node as visited
v[curr] = 1;
// If it is a destination node
if (curr == d)
return dist;
// Traversing the current node
for ( auto it : gr[curr])
pq.insert({ dist * it.second, it.first });
}
// If no path exists
return -1;
} // Driver code int main()
{ int n = 3;
// Graph as adjacency matrix
vector<vector<pair< int , double > > > gr(n + 1);
// Input edges
gr[1].push_back({ 3, 9 });
gr[2].push_back({ 3, 1 });
gr[1].push_back({ 2, 5 });
// Source and destination
int s = 1, d = 3;
// Dijkstra
cout << dijkstra(s, d, gr);
return 0;
} |
// Java implementation of the approach import java.util.ArrayList;
import java.util.PriorityQueue;
class Pair implements Comparable<Pair>
{ int first, second;
public Pair( int first, int second)
{
this .first = first;
this .second = second;
}
public int compareTo(Pair o)
{
if ( this .first == o.first)
{
return this .second - o.second;
}
return this .first - o.first;
}
} class GFG{
// Function to return the smallest // xor sum of edges static int dijkstra( int s, int d,
ArrayList<ArrayList<Pair>> gr)
{ // If the source is equal
// to the destination
if (s == d)
return 0 ;
// Initialise the priority queue
PriorityQueue<Pair> pq = new PriorityQueue<>();
pq.add( new Pair( 1 , s));
// Visited array
boolean [] v = new boolean [gr.size()];
// While the priority-queue
// is not empty
while (!pq.isEmpty())
{
// Current node
Pair p = pq.poll();
int curr = p.second;
// Current xor sum of distance
int dist = p.first;
// If already visited continue
if (v[curr])
continue ;
// Marking the node as visited
v[curr] = true ;
// If it is a destination node
if (curr == d)
return dist;
// Traversing the current node
for (Pair it : gr.get(curr))
pq.add( new Pair(dist ^ it.second, it.first));
}
// If no path exists
return - 1 ;
} // Driver code public static void main(String[] args)
{ int n = 3 ;
// Graph as adjacency matrix
ArrayList<ArrayList<Pair>> gr = new ArrayList<>();
for ( int i = 0 ; i < n + 1 ; i++)
{
gr.add( new ArrayList<Pair>());
}
// Input edges
gr.get( 1 ).add( new Pair( 3 , 9 ));
gr.get( 2 ).add( new Pair( 3 , 1 ));
gr.get( 1 ).add( new Pair( 2 , 5 ));
// Source and destination
int s = 1 , d = 3 ;
System.out.println(dijkstra(s, d, gr));
} } // This code is contributed by sanjeev2552 |
# Python3 implementation of the approach # Function to return the smallest # product of edges def dijkstra(s, d, gr) :
# If the source is equal
# to the destination
if (s = = d) :
return 0 ;
# Initialise the priority queue
pq = [];
pq.append(( 1 , s ));
# Visited array
v = [ 0 ] * ( len (gr) + 1 );
# While the priority-queue
# is not empty
while ( len (pq) ! = 0 ) :
# Current node
curr = pq[ 0 ][ 1 ];
# Current product of distance
dist = pq[ 0 ][ 0 ];
# Popping the top-most element
pq.pop();
# If already visited continue
if (v[curr]) :
continue ;
# Marking the node as visited
v[curr] = 1 ;
# If it is a destination node
if (curr = = d) :
return dist;
# Traversing the current node
for it in gr[curr] :
if it not in pq :
pq.insert( 0 ,( dist * it[ 1 ], it[ 0 ] ));
# If no path exists
return - 1 ;
# Driver code if __name__ = = "__main__" :
n = 3 ;
# Graph as adjacency matrix
gr = {};
# Input edges
gr[ 1 ] = [( 3 , 9 ) ];
gr[ 2 ] = [ ( 3 , 1 ) ];
gr[ 1 ].append(( 2 , 5 ));
gr[ 3 ] = [];
#print(gr);
# Source and destination
s = 1 ; d = 3 ;
# Dijkstra
print (dijkstra(s, d, gr));
# This code is contributed by AnkitRai01 |
//C# code for the above approach using System;
using System.Collections.Generic;
class Pair : IComparable<Pair>
{ public int first, second;
public Pair( int first, int second)
{
this .first = first;
this .second = second;
}
public int CompareTo(Pair o)
{
if ( this .first == o.first)
{
return this .second - o.second;
}
return this .first - o.first;
}
} class GFG
{ // Function to return the smallest
// xor sum of edges
static int Dijkstra( int s, int d,
List<List<Pair>> gr)
{
// If the source is equal
// to the destination
if (s == d)
return 0;
// Initialise the priority queue
SortedSet<Pair> pq = new SortedSet<Pair>();
pq.Add( new Pair(1, s));
// Visited array
bool [] v = new bool [gr.Count];
// While the priority-queue
// is not empty
while (pq.Count != 0)
{
// Current node
Pair p = pq.Min;
pq.Remove(p);
int curr = p.second;
// Current xor sum of distance
int dist = p.first;
// If already visited continue
if (v[curr])
continue ;
// Marking the node as visited
v[curr] = true ;
// If it is a destination node
if (curr == d)
return dist;
// Traversing the current node
foreach (Pair it in gr[curr])
pq.Add( new Pair(dist ^ it.second, it.first));
}
// If no path exists
return -1;
}
// Driver code
public static void Main( string [] args)
{
int n = 3;
// Graph as adjacency matrix
List<List<Pair>> gr = new List<List<Pair>>();
for ( int i = 0; i < n + 1; i++)
{
gr.Add( new List<Pair>());
}
// Input edges
gr[1].Add( new Pair(3, 9));
gr[2].Add( new Pair(3, 1));
gr[1].Add( new Pair(2, 5));
// Source and destination
int s = 1, d = 3;
Console.WriteLine(Dijkstra(s, d, gr));
}
} |
<script> // Javascript implementation of the approach // Function to return the smallest // product of edges function dijkstra(s, d, gr)
{ // If the source is equal
// to the destination
if (s == d)
return 0;
// Initialise the priority queue
var pq = [];
pq.push([1, s]);
// Visited array
var v = Array(gr.length).fill(0);
// While the priority-queue
// is not empty
while (pq.length!=0) {
// Current node
var curr = pq[0][1];
// Current product of distance
var dist = pq[0][0];
// Popping the top-most element
pq.shift();
// If already visited continue
if (v[curr])
continue ;
// Marking the node as visited
v[curr] = 1;
// If it is a destination node
if (curr == d)
return dist;
// Traversing the current node
for ( var it of gr[curr])
pq.push([dist * it[1], it[0] ]);
pq.sort();
}
// If no path exists
return -1;
} // Driver code var n = 3;
// Graph as adjacency matrix var gr = Array.from(Array(n + 1), ()=> Array());
// Input edges gr[1].push([3, 9]); gr[2].push([3, 1]); gr[1].push([2, 5]); // Source and destination var s = 1, d = 3;
// Dijkstra document.write(dijkstra(s, d, gr)); // This code is contributed by rrrtnx. </script> |
5
Time complexity: O((E + V) logV)
Auxiliary Space: O(V).