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Number of sink nodes in a graph

Given a Directed Acyclic Graph of n nodes (numbered from 1 to n) and m edges. The task is to find the number of sink nodes. A sink node is a node such that no edge emerges out of it.

Examples:

Input : n = 4, m = 2
        Edges[] = {{2, 3}, {4, 3}} 
Output : 2

Only node 1 and node 3 are sink nodes.

Input : n = 4, m = 2
        Edges[] = {{3, 2}, {3, 4}} 
Output : 3



The idea is to iterate through all the edges. And for each edge, mark the source node from which the edge emerged out. Now, for each node check if it is marked or not. And count the unmarked nodes.
Algorithm:

 
1. Make any array A[] of size equal to the
    number of nodes and initialize to 1.
2. Traverse all the edges one by one, say, 
   u -> v.
     (i) Mark A[u] as 1.
3. Now traverse whole array A[] and count 
   number of unmarked nodes.

Below is implementation of this approach:

C++

// C++ program to count number if sink nodes
#include<bits/stdc++.h>
using namespace std;
  
// Return the number of Sink NOdes.
int countSink(int n, int m, int edgeFrom[],
                        int edgeTo[])
{
    // Array for marking the non-sink node.
    int mark[n];
    memset(mark, 0, sizeof mark);
  
    // Marking the non-sink node.
    for (int i = 0; i < m; i++)
        mark[edgeFrom[i]] = 1;
  
    // Counting the sink nodes.
    int count = 0;
    for (int i = 1; i <= n ; i++)
        if (!mark[i])
            count++;
  
    return count;
}
  
// Driven Program
int main()
{
    int n = 4, m = 2;
    int edgeFrom[] = { 2, 4 };
    int edgeTo[] = { 3, 3 };
  
    cout << countSink(n, m, edgeFrom, edgeTo) << endl;
  
    return 0;
}

Python3

# Python3 program to count number if sink nodes
  
# Return the number of Sink NOdes. 
def countSink(n, m, edgeFrom, edgeTo):
      
    # Array for marking the non-sink node. 
    mark = [0] * (n + 1)
  
    # Marking the non-sink node.
    for i in range(m):
        mark[edgeFrom[i]] = 1
  
    # Counting the sink nodes. 
    count = 0
    for i in range(1, n + 1):
        if (not mark[i]): 
            count += 1
  
    return count
  
# Driver Code
if __name__ == '__main__'
      
    n = 4
    m = 2
    edgeFrom = [2, 4
    edgeTo = [3, 3]
  
    print(countSink(n, m, edgeFrom, edgeTo))
  
# This code is contributed by PranchalK


Output:

2

Time Complexity: O(m + n) where n is number of nodes and m is number of edges.

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Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

tags:

Graph Graph

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