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DSA Javascriptprogramming~10 mins

Why Graphs Exist and What Trees Cannot Model in DSA Javascript - Test Your Knowledge

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to create a graph node with a value and an empty list of neighbors.

DSA Javascript
class GraphNode {
  constructor(value) {
    this.value = value;
    this.neighbors = [1];
  }
}
Drag options to blanks, or click blank then click option'
A[]
Bnull
C{}
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Using an object {} instead of an array for neighbors.
2fill in blank
medium

Complete the code to add an edge from one node to another in an undirected graph.

DSA Javascript
function addEdge(node1, node2) {
  node1.neighbors.push([1]);
  node2.neighbors.push(node1);
}
Drag options to blanks, or click blank then click option'
Anode2
Bnode1
Cnull
Dundefined
Attempts:
3 left
💡 Hint
Common Mistakes
Adding node1 to its own neighbors list.
3fill in blank
hard

Fix the error in the code to check if a graph has a cycle using a visited set.

DSA Javascript
function hasCycle(node, visited = new Set()) {
  if (visited.has(node)) {
    return [1];
  }
  visited.add(node);
  for (const neighbor of node.neighbors) {
    if (hasCycle(neighbor, visited)) {
      return true;
    }
  }
  return false;
}
Drag options to blanks, or click blank then click option'
Anull
Bfalse
Ctrue
Dundefined
Attempts:
3 left
💡 Hint
Common Mistakes
Returning false when a cycle is detected.
4fill in blank
hard

Fill both blanks to create a function that counts the total edges in an undirected graph.

DSA Javascript
function countEdges(graph) {
  let count = 0;
  for (const node of graph) {
    count += node.[1].length;
  }
  return count [2] 2;
}
Drag options to blanks, or click blank then click option'
Aneighbors
B*
C/
Dlength
Attempts:
3 left
💡 Hint
Common Mistakes
Multiplying by 2 instead of dividing.
Using wrong property instead of neighbors.
5fill in blank
hard

Fill all three blanks to create a function that finds if a path exists between two nodes using DFS.

DSA Javascript
function pathExists(start, target, visited = new Set()) {
  if (start.value === [1]) return true;
  visited.add(start);
  for (const neighbor of start.[2]) {
    if (!visited.has(neighbor) && pathExists(neighbor, [3], visited)) {
      return true;
    }
  }
  return false;
}
Drag options to blanks, or click blank then click option'
Atarget.value
Bneighbors
Ctarget
Dstart.value
Attempts:
3 left
💡 Hint
Common Mistakes
Comparing start.value with start.value instead of target.value.
Using wrong variable for recursion.

Practice

(1/5)
1. Why do we need graphs when we already have trees in data structures?
easy
A. Because graphs can represent connections between nodes in any direction, including cycles.
B. Because trees are faster to search than graphs.
C. Because trees can only store numbers, not other data types.
D. Because graphs use less memory than trees.

Solution

  1. Step 1: Understand tree limitations

    Trees are hierarchical and cannot represent cycles or multiple parent connections.
  2. Step 2: Understand graph capabilities

    Graphs allow nodes to connect in any way, including cycles and multiple parents.
  3. Final Answer:

    Because graphs can represent connections between nodes in any direction, including cycles. -> Option A
  4. Quick Check:

    Graphs model cycles and complex connections = C [OK]
Hint: Graphs allow cycles; trees do not [OK]
Common Mistakes:
  • Thinking trees can have cycles
  • Confusing memory use between trees and graphs
  • Assuming trees only store numbers
2. Which of the following is the correct way to represent an undirected graph using adjacency lists in JavaScript?
easy
A. const graph = [ [1, 2], [2, 3], [3, 1] ];
B. const graph = { 1: [2, 3], 2: [1], 3: [1] };
C. const graph = { 1: 2, 2: 3, 3: 1 };
D. const graph = [[1, 2], [2, 3], [3, 1]];

Solution

  1. Step 1: Understand adjacency list format

    An adjacency list uses an object with keys as nodes and values as arrays of connected nodes.
  2. Step 2: Check each option

    const graph = { 1: [2, 3], 2: [1], 3: [1] }; correctly uses an object with arrays for neighbors; others use incorrect formats.
  3. Final Answer:

    const graph = { 1: [2, 3], 2: [1], 3: [1] }; -> Option B
  4. Quick Check:

    Adjacency list = object with arrays [OK]
Hint: Adjacency lists use objects with arrays of neighbors [OK]
Common Mistakes:
  • Using arrays of pairs instead of adjacency lists
  • Assigning single values instead of arrays
  • Confusing syntax for objects and arrays
3. What will be the output of the following JavaScript code that models a graph with cycles?
const graph = {
  A: ['B'],
  B: ['C', 'D'],
  C: ['A'],
  D: []
};

function hasCycle(node, visited, stack) {
  if (!visited[node]) {
    visited[node] = true;
    stack[node] = true;
    for (const neighbor of graph[node]) {
      if (!visited[neighbor] && hasCycle(neighbor, visited, stack)) {
        return true;
      } else if (stack[neighbor]) {
        return true;
      }
    }
  }
  stack[node] = false;
  return false;
}

console.log(hasCycle('A', {}, {}));
medium
A. TypeError
B. false
C. undefined
D. true

Solution

  1. Step 1: Analyze graph structure

    Graph has nodes A->B, B->C,D, C->A forming a cycle A->B->C->A.
  2. Step 2: Trace hasCycle function

    Function detects cycles using recursion and stack tracking; it returns true when cycle found.
  3. Final Answer:

    true -> Option D
  4. Quick Check:

    Cycle detected in graph = true [OK]
Hint: Cycle detection returns true if back edge found [OK]
Common Mistakes:
  • Ignoring the cycle from C back to A
  • Assuming no cycles in directed graphs
  • Confusing visited and stack usage
4. The following code attempts to check if a graph has cycles but has a bug. What is the bug?
const graph = {
  1: [2],
  2: [3],
  3: [1]
};

function detectCycle(node, visited) {
  if (visited[node]) return true;
  visited[node] = true;
  for (const neighbor of graph[node]) {
    if (detectCycle(neighbor, visited)) return true;
  }
  return false;
}

console.log(detectCycle(1, {}));
medium
A. The graph object keys should be strings, not numbers.
B. The visited object is not initialized properly.
C. The function does not track recursion stack to detect back edges.
D. The function should return false immediately after visiting a node.

Solution

  1. Step 1: Understand cycle detection logic

    Detecting cycles in directed graphs requires tracking recursion stack to find back edges.
  2. Step 2: Identify missing recursion stack tracking

    Code only uses visited set, so it cannot distinguish between visited nodes in current path and others.
  3. Final Answer:

    The function does not track recursion stack to detect back edges. -> Option C
  4. Quick Check:

    Missing recursion stack tracking = bug [OK]
Hint: Track recursion stack to detect cycles in directed graphs [OK]
Common Mistakes:
  • Assuming visited alone detects cycles
  • Confusing graph keys types
  • Returning false too early
5. You want to model a social network where users can follow each other, including mutual follows and cycles. Which data structure best models this, and why can't a tree be used here?
hard
A. Use a graph because it supports cycles and multiple connections; trees cannot model cycles or multiple parents.
B. Use a tree because it is simpler and supports mutual connections.
C. Use an array because it can store all users linearly.
D. Use a linked list because it supports cycles naturally.

Solution

  1. Step 1: Understand social network connections

    Users can follow each other mutually, creating cycles and multiple parent connections.
  2. Step 2: Compare data structures

    Trees are hierarchical and cannot represent cycles or multiple parents; graphs can model these complex relationships.
  3. Final Answer:

    Use a graph because it supports cycles and multiple connections; trees cannot model cycles or multiple parents. -> Option A
  4. Quick Check:

    Social network needs graph for cycles = A [OK]
Hint: Social networks need graphs for cycles and mutual links [OK]
Common Mistakes:
  • Choosing trees for mutual connections
  • Using arrays or lists for complex relations
  • Ignoring cycles in social networks
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