Graph

Graph algorithms are quite common and usually fall under the following algorithms. You will usually be required to modify these algorithms to fit the problem but it's good to know the fundamentals

Time complexities

Algorithm	Time Complexity	Space Complexity	Remarks
BFS	$O(E + V)$	$O(V)$
DFS	$O(E + V)$	$O(V)$	Space accounting for stack frames used in recursion
Topological Sorting	$O(E + V)$	$O(V)$ for storing the edges and frontier	Note that when iterating over every node, we only decrease the edges at most $O(E)$ times
Dijkstra	$O((V + E) \log V)$	$O(E)$ for priority queue
Bellman-Ford	$O(EV)$		For $O(V)$ vertices, we iterate through all $O(E)$ edges and compute the SSP
Prim’s Algorithm	$O(E \log V)$ or $O(V^2)$		Time complexity achieved if using Fibonacci heap AND iterating over the entire priority queue
Kruskal’s Algorithm	$O(E \log V)$

Algorithm

Time Complexity

Space Complexity

Remarks

BFS

$O(E + V)$

$O(V)$

DFS

$O(E + V)$

$O(V)$

Space accounting for stack frames used in recursion

Topological Sorting

$O(E + V)$

$O(V)$ for storing the edges and frontier

Note that when iterating over every node, we only decrease the edges at most $O(E)$ times

Dijkstra

$O((V + E) \log V)$

$O(E)$ for priority queue

Bellman-Ford

$O(EV)$

For $O(V)$ vertices, we iterate through all $O(E)$ edges and compute the SSP

Prim’s Algorithm

$O(E \log V)$ or $O(V^2)$

Time complexity achieved if using Fibonacci heap AND iterating over the entire priority queue

Kruskal’s Algorithm

$O(E \log V)$

Frequency in interviews

Common: BFS, DFS
Uncommon: Topological sort, Dijkstra
Almost never: Bellman-Ford, Floyd Warshall, Prim’s, Kruskal’s

BFS

Trees do not require a visited set since there is only 1 path between nodes (property of trees)

from collections import deque

def bfs(matrix):
  # Check for an empty matrix/graph.
  if not matrix:
    return []

  rows, cols = len(matrix), len(matrix[0])
  visited = set()
  directions = ((0, 1), (0, -1), (1, 0), (-1, 0))

  def traverse(i, j):
    queue = deque([(i, j)])
    while queue:
      curr_i, curr_j = queue.popleft()
      if (curr_i, curr_j) not in visited:
        visited.add((curr_i, curr_j))
        # Traverse neighbors.
        for direction in directions:
          next_i, next_j = curr_i + direction[0], curr_j + direction[1]
          if 0 <= next_i < rows and 0 <= next_j < cols:
            # Add in question-specific checks, where relevant.
            queue.append((next_i, next_j))

  for i in range(rows):
    for j in range(cols):
      traverse(i, j)

DFS

def dfs(matrix):
  # Check for an empty matrix/graph.
  if not matrix:
    return []

  rows, cols = len(matrix), len(matrix[0])
  visited = set()
  directions = ((0, 1), (0, -1), (1, 0), (-1, 0))

  def traverse(i, j):
    if (i, j) in visited:
      return

    visited.add((i, j))
    # Traverse neighbors.
    for direction in directions:
      next_i, next_j = i + direction[0], j + direction[1]
      if 0 <= next_i < rows and 0 <= next_j < cols:
        # Add in question-specific checks, where relevant.
        traverse(next_i, next_j)

  for i in range(rows):
    for j in range(cols):
      traverse(i, j)

Topological sorting

Used for job scheduling a sequence of jobs or tasks that have dependencies on other jobs/tasks
- Jobs represent vertices and edges from X to Y (directed) if X depends on Y

def graph_topo_sort(num_nodes, edges):
    from collections import deque
    nodes, order, queue = {}, [], deque()
    
    # O(V)
    for node_id in range(num_nodes):
        nodes[node_id] = { 'in': 0, 'out': set() }
		
    # O(E)
    for node_id, pre_id in edges:
        nodes[node_id]['in'] += 1
        nodes[pre_id]['out'].add(node_id)

    # O(V)
    for node_id in nodes.keys():
        if nodes[node_id]['in'] == 0:
            queue.append(node_id)

    # O(E), total number of decreases happen O(E) times at most
    while len(queue):  # At most O(V) elements
        node_id = queue.pop()
        for outgoing_id in nodes[node_id]['out']:  # At most O(V - 1) edges
            nodes[outgoing_id]['in'] -= 1
            if nodes[outgoing_id]['in'] == 0:
                queue.append(outgoing_id)
        order.append(node_id)
    return order if len(order) == num_nodes else None

print(graph_topo_sort(4, [[0, 1], [0, 2], [2, 1], [3, 0]]))
# [1, 2, 0, 3]

Dijkstra

Only works with non-negative edge weights. If the edges have a negative weight, use Bellman-Ford instead

Optimizations

If target node known, once we process target node (i.e. pop is target node), we can early return

import heapq
import math

class Solution:
    def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
        graph = {}
        # O(E)
        # Maintain the edge costs + edges
        for [source, target, time] in times:
            if (source - 1) not in graph:
                graph[source - 1] = []
            graph[source - 1].append((target - 1, time))
	       
	# All nodes start with inf cost to reach
        costs = [math.inf] * n
	# Start node has cost of 0 (naturally)
        costs[k - 1] = 0
        q = [(0, k - 1)]
        visited = set()
				
	# O(V log V) + O(E log V)
	# O((E + V) log V)
	# O(E log V)
        while q:  # At most O(V)
	    # O(log V)
            node_cost, node = heapq.heappop(q)
            if node in visited:
                continue
            visited.add(node)
            if node not in graph:
                continue
	    # O(E log V)
            for neighbor, time in graph[node]: # Visit at most O(E) nodes
		# Only update cost and re-queue if the cost to reach neighbor decreases
                if node_cost + time < costs[neighbor]:
                    costs[neighbor] = node_cost + time
                    heapq.heappush(q, (node_cost + time, neighbor))
        
        if len(visited) != n:
            # Disjoint component
	    # Alternative way to check is to check if any cost is still inf
            return -1

        return max(costs)

Bellman-Ford

Works with negative weight edges but does not work if there are negative weight cycles. For those cases, we can detect that a cycle exists but cannot do anything about it

Loop for v - 1 times and for each loop, relax all edges

To detect negative weight edges, check if the cost of the same node decreases twice

Optimizations

Track if any costs decreased, if none did, then we can early terminate since that means we found the lowest possible cost for all edges

import heapq
import math

class Solution:
    def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
        costs = [math.inf] * n
        costs[k - 1] = 0
        # O(V)
        for _ in range(n - 1):
            has_change = False

	# O(E)
            for source, target, weight in times:
                if costs[target - 1] > costs[source - 1] + weight:
                    costs[target - 1] = costs[source - 1] + weight
                    has_change = True

            if not has_change:
                break
        
        if any([cost == math.inf for cost in costs]):
            return -1
        
        return max(costs)

Prim’s algorithm

Finds the Minimum Spanning Tree of a graph and is easier to implement than Kruskal’s algorithm

The general intuition for Prim's algorithm is as such: Given a node i, go through all connected points and add the edge weights to a min heap (weight, other point). For every element in the min heap, if visited before, discard, otherwise, use the topmost point’s edge. Iterate at most N - 1 times as that is the maximum number of edges of a tree.

For this specific implementation, the time complexity is $O(EV \log V)$

import heapq

# edges are: [from, to, weight]
# n points in total
def prim(n: int, edges: List[List[int]]):
	graph = {}
	# O(E)
	for f, t, w in edges:
		if f not in graph: graph[f] = []
		if t not in graph: graph[t] = []
		graph[f].append((t, w))
		graph[t].append((f, w))

	visited = [False] * n
	h = []
	node = 0
	min_weight = 0
	
	# O(VE log V)
	for i in range(N - 1):
		visited[node] = True
		# O(E log V)
		for neighbor, weight in graph[node]: # Total O(E)
			if not visited[neighbor]:
				heapq.heappush(h, (weight, neighbor))

		# O(log V)
		while visited[h[0][1]]:
			heapq.heappop(h)

		# O(log V)
		weight, node = heapq.heappop(h)
		min_weight += weight

	return min_weight

Fully-connected graphs

For fully connected graphs, instead of using a heap, use a min_d array, tracking the minimum weight to reach each point in the graph.

The general intuition for this variation is: For each node node, we update the edge weight to all connected points. Then the point we choose to use would be the one that has not yet been visited and has the shortest edge weight.

Once a node is visited, set the weight to reach to be inf so no extra visited is needed

import math

def prim(n: int, edges: List[List[int]]):
	graph = {}
	for f, t, w in edges:
		if f not in graph: graph[f] = []
		if t not in graph: graph[t] = []
		graph[f].append((t, w))
		graph[t].append((f, w))

	min_d = [10*8] * n
	node = 0
	min_weight = 0
	for i in range(N - 1):
		min_d[node] = math.inf
		min_j = node
		for neighbor, weight in graph[node]:
			if min_d[neighbor] != math.inf:
				min_d[neighbor] = min(min_d[neighbor], weight)
				min_j = neighbor if min_d[neighbor] < min_d[min_j] else min_j
		min_weight += min_d[min_j]
		node = min_j
	
	return min_weight

Kruskal’s algorithm

Use Union-Find Disjoint Set (UFDS) to determine which edges are redundant and use a min heap to store the weights of the edges. Redundant edges are those whose points already exist in the same set (meaning that there exists another path between these two points in the MST so far).

This implementation has a time complexity of $O(E \log E + E \log V)$ because we’re not sorting the entire graph at once. If we sorted, the time complexity would be similar but $O(\log E) = O(\log V^2) = O(2 \log V) = O(\log V)$ dominates the term so we take that instead

import heapq

def find(ds, i):
	# O(log V)
	if ds[i] == i:
		return i
	
	ds[i] = find(ds, ds[i])
	return ds[i]

def union(ds, p, q):
	# O(log V)
	root_p = find(p)
	root_q = find(q)
	ds[root_p] = root_q

def kruskal(n: int, edges: List[List[int]]):
	ds = list(range(n))
	h = []
	min_weight = 0
	# O(E)
	for f, t, w in edges:
		h.append((w, f, t))
	# O(E)
	heapq.heapify(h)
	used = 0	

	# O(E log E + E log V)
	# O(E)
	while h:
		# O(log E)
		weight, i, j = heapq.heappop(h)
		# O(log V)
		i = find(ds, i)
		j = find(ds, j)
		if i != j:
			min_weight += weight
			union(ds, i, j)
			used += 1
			if used == n - 1:
				break

	return min_weight

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Last updated 11 months ago