Number of sink nodes in a graph

29 Nov, 2021
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++
Java
Python3
C#
Javascript
// 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 Programint main(){    int n = 4, m = 2;    int edgeFrom[] = { 2, 4 };    int edgeTo[] = { 3, 3 };      cout << countSink(n, m, edgeFrom, edgeTo) << endl;      return 0;}
// Java program to count number if sink nodesimport java.util.*;  class GFG{  // Return the number of Sink NOdes.static int countSink(int n, int m,                      int edgeFrom[], int edgeTo[]){    // Array for marking the non-sink node.    int []mark = new int[n + 1];      // 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] == 0)            count++;      return count;}  // Driver Codepublic static void main(String[] args){    int n = 4, m = 2;    int edgeFrom[] = { 2, 4 };    int edgeTo[] = { 3, 3 };      System.out.println(countSink(n, m,                        edgeFrom, edgeTo));}}  // This code is contributed by 29AjayKumar
# 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 Codeif __name__ == '__main__':           n = 4    m = 2    edgeFrom = [2, 4]     edgeTo = [3, 3]      print(countSink(n, m, edgeFrom, edgeTo))  # This code is contributed by PranchalK
// C# program to count number if sink nodesusing System;      class GFG{  // Return the number of Sink NOdes.static int countSink(int n, int m,                      int []edgeFrom,                     int []edgeTo){    // Array for marking the non-sink node.    int []mark = new int[n + 1];      // 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] == 0)            count++;      return count;}  // Driver Codepublic static void Main(String[] args){    int n = 4, m = 2;    int []edgeFrom = { 2, 4 };    int []edgeTo = { 3, 3 };      Console.WriteLine(countSink(n, m,                       edgeFrom, edgeTo));}}  // This code is contributed by PrinciRaj1992
<script>  // Javascript program to count number if sink nodes  // Return the number of Sink NOdes.function countSink(n, m, edgeFrom, edgeTo){          // Array for marking the non-sink node.    let mark = new Array(n + 1);    for(let i = 0; i < n + 1; i++)    {        mark[i] = 0;    }          // Marking the non-sink node.    for(let i = 0; i < m; i++)        mark[edgeFrom[i]] = 1;        // Counting the sink nodes.    let count = 0;    for(let i = 1; i <= n; i++)        if (mark[i] == 0)            count++;        return count;}  // Driver Codelet n = 4, m = 2;let edgeFrom = [ 2, 4 ];let edgeTo = [ 3, 3 ];  document.write(countSink(n, m,                          edgeFrom, edgeTo));  // This code is contributed by rag2127  </script>
Output: 
2
Time Complexity: O(m + n) where n is number of nodes and m is number of edges.
Related Article: The Celebrity Problem
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PranchalKatiyar
29AjayKumar
princiraj1992
javasick
rag2127
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