# HelloKoding

Practical coding guides

# Topological Sort Algorithms

In this article, you will learn to implement a Topological sort algorithm by using Depth-First Search and In-degree algorithms

Topological sort is an algorithm which takes a directed acyclic graph and returns a list of vertices in the linear ordering where each vertex has to precede all vertices it directs to

There are no topological orderings exist in a directed cyclic graph and more than one of them can exist in one directed acyclic graph

Let’s get started!

## Problem

• Given a directed graph G = (V, E)
• Write an algorithm to return a sorted vertex list in topological ordering
• Return an empty list if no topological orderings exist

## Example 1

• Input: Given a directed graph G = (V, E) with V = 4 and E = [[0, 1], [2, 0], [2, 1], [3, 2]] • Output: [3, 2, 0, 1]
• Explanation: [3, 2, 1, 0] is also a valid topological ordering

## Example 2

• Input: Given a directed graph G = (V, E) with V = 5 and E = [[0, 1], [1, 2], [2, 0], [1, 3], [2, 4]] • Output: []
• Explanation: No topological orderings exist as [0, 1], [1, 2], and [2, 0] form a cycle. Vertex 0 can not both precede and follow vertex 2

## Approach 1: DFS Iterative with Color

• Run DFS Iterative with Color to find a vertex with no outgoing edges (Black vertex) and add it to a stack S. Stop finding when found a cycle. Repeat for all other vertices in the graph
• Convert S to a list to get the final result

TopologicalSortByDFSIterativeColor.java

``````package com.hellokoding.datastructure.graph;

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;

public class TopologicalSortByDFSIterativeColor {
static final int WHITE = 0, GRAY = 1, BLACK = 2;

int[] color = new int[g.getV()];
Deque<Integer> stack = new ArrayDeque<>(g.getV());
Deque<Integer> sorted = new ArrayDeque<>(g.getV());

for (int i = 0; i < g.getV(); i++) {
if (color[i] != WHITE) continue;
stack.push(i);

while (!stack.isEmpty()) {
int v = stack.peek();

if (color[v] == WHITE) {
color[v] = GRAY;
} else {
color[v] = BLACK;
sorted.push(stack.pop());
}

for (int w : g.getAdjacencyList().get(v)) {
if (color[w] == GRAY) {
// Found a cycle
return new ArrayList<>();
} else if (color[w] == WHITE) {
stack.push(w);
}
}
}
}

return new ArrayList<>(sorted);
}

static void printResult(List<Integer> sorted) {
if (sorted.isEmpty()) {
System.out.println("There are no topological orderings as the input graph is cyclic");
} else {
sorted.forEach((x) -> System.out.printf("%d ", x));
System.out.println();
}
}

public static void main(String[] args) {
printResult(topologicalSort(g1));

printResult(topologicalSort(g2));
}
}
``````

`GraphDirectedByAdjacencyList` is defined in Graph Data Structure

• Output
``````There are no topological orderings as the input graph is cyclic
3 2 0 1``````
• Time complexity: O(V+E)
• Space complexity: O(V)

## Approach 2: DFS Recursive with Color

• Run DFS Recursive with Color to find a vertex with no outgoing edges (Black vertex) and add it to a stack S. Stop finding when found a cycle. Repeat for all other vertices in the graph
• Convert S to a list to get the final result

TopologicalSortByDFSRecursiveColor.java

``````package com.hellokoding.datastructure.graph;

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;

public class TopologicalSortByDFSRecursiveColor {
static final int WHITE = 0, GRAY = 1, BLACK = 2;
static boolean hasCycle = false;

static void topologicalSortByDFSRecursive(GraphDirectedByAdjacencyList g, int v, int[] color, Deque<Integer> stack) {
if (hasCycle) {
return;
}

color[v] = GRAY;

for (Integer w : g.getAdjacencyList().get(v)) {
if (color[w] == GRAY) {
hasCycle = true;
}

if (color[w] == WHITE) {
topologicalSortByDFSRecursive(g, w, color, stack);
}
}

color[v] = BLACK;
stack.push(v);
}

int[] color = new int[g.getV()];
Deque<Integer> stack = new ArrayDeque<>();
hasCycle = false;

for (int i = 0; i < g.getV(); i++) {
if (color[i] == WHITE) {
topologicalSortByDFSRecursive(g, i, color, stack);
}
}

return hasCycle ? new ArrayList<>() : new ArrayList<>(stack);
}

static void printResult(List<Integer> sorted) {
if (sorted.isEmpty()) {
System.out.println("There are no topological orderings as the input graph is cyclic");
} else {
sorted.forEach((x) -> System.out.printf("%d ", x));
System.out.println();
}
}

public static void main(String[] args) {
printResult(topologicalSort(g1));

printResult(topologicalSort(g2));
}
}
``````

`GraphDirectedByAdjacencyList` is defined in Graph Data Structure

• Output
``````There are no topological orderings as the input graph is cyclic
3 2 0 1``````
• Time complexity: O(V+E)
• Space complexity: O(V)

## Approach 3: In-degree

• Find a vertex with no incoming edges (in-degree = 0) and add it to a queue Q. Repeat for all other vertices in the graph
• Convert Q to a list to get the final result

TopologicalSortByInDegree.java

``````package com.hellokoding.datastructure.graph;

import java.util.*;

public class TopologicalSortByInDegree {
List<Integer> sorted = new ArrayList<>();

int[] inDegree = new int[g.getV()];
for (int v = 0; v < g.getV(); v++) {
inDegree[w] += 1;
}
}

Queue<Integer> queue = new ArrayDeque<>();
for (int v = 0; v < g.getV(); v++) {
if (inDegree[v] == 0) {
queue.offer(v);
}
}

while (!queue.isEmpty()) {
int v = queue.poll();

for (int w : g.getAdjacencyList().get(v)) {
inDegree[w]--;

if (inDegree[w] == 0) {
queue.offer(w);
}
}
}

return sorted.size() == g.getV() ? sorted : new ArrayList<>();
}

static void printResult(List<Integer> sorted) {
if (sorted.isEmpty()) {
System.out.println("There are no topological orderings as the input graph is cyclic");
} else {
sorted.forEach((x) -> System.out.printf("%d ", x));
System.out.println();
}
}

public static void main(String[] args) {
printResult(topologicalSortByInDegree(g1));

`GraphDirectedByAdjacencyList` is defined in Graph Data Structure
``````There are no topological orderings as the input graph is cyclic