Why Shortest Path Is a Graph Problem Not a Tree Problem in DSA Go - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize a graph adjacency list in Go.

DSA Go

graph := make(map[int][]int)
graph[1] = []int{2, 3}
graph[2] = []int{4}
graph[3] = []int{4}
graph[4] = [1]

Drag options to blanks, or click blank then click option'

A[]int{}

B[]int{5}

Cnil

D[]int{1}

Attempts:

3 left

💡 Hint

Common Mistakes

Using nil instead of an empty slice can cause runtime errors when iterating.

Adding edges back to node 1 creates cycles, which is not needed here.

Node 4 has no outgoing edges, so its adjacency list is an empty slice.

2fill in blank

medium

Complete the code to add an edge from node 2 to node 5 in the graph.

DSA Go

graph[2] = append(graph[2], [1])

Drag options to blanks, or click blank then click option'

A3

B5

C4

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Appending the wrong node number creates incorrect edges.

Appending to the wrong node's adjacency list.

We want to add an edge from node 2 to node 5, so append 5 to graph[2].

3fill in blank

hard

Fix the error in the BFS queue initialization for shortest path search.

DSA Go

queue := []int[1]

Drag options to blanks, or click blank then click option'

A0

Bnil

C[]int{}

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting BFS from node 0 when nodes start at 1.

Initializing queue as nil or empty slice causes no traversal.

The BFS starts from node 1, so the queue should start with 1.

4fill in blank

hard

Fill both blanks to check if a neighbor node has been visited before adding to the queue.

DSA Go

if !visited[[1]] {
    visited[neighbor] = true
    queue = append(queue, [2])
}

Drag options to blanks, or click blank then click option'

Aneighbor

Bnode

Cvisited

Dqueue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'node' instead of 'neighbor' causes logic errors.

Using 'visited' or 'queue' as variables in condition or append.

We check if 'neighbor' is visited and then add 'neighbor' to the queue.

5fill in blank

hard

Fill all three blanks to update the distance map during BFS shortest path search.

DSA Go

distance[[1]] = distance[[2]] [3] 1

Drag options to blanks, or click blank then click option'

Aneighbor

Bnode

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Subtracting 1 instead of adding causes wrong distances.

Swapping neighbor and node in distance update.

Distance to neighbor is distance to current node plus 1 edge.

Practice

1. Why is finding the shortest path usually considered a graph problem rather than a tree problem?

easy

A. Because graphs can have cycles and multiple paths between nodes, unlike trees

B. Because trees do not have nodes

C. Because shortest path algorithms only work on trees

D. Because trees always have weighted edges

Solution

1. Step 1: Understand the structure of trees and graphs

   Trees are a special type of graph with no cycles and exactly one path between any two nodes. Graphs can have cycles and multiple paths between nodes.

2. Step 2: Relate shortest path to multiple paths

   Shortest path problems require checking multiple possible paths to find the minimum distance. This is only possible in graphs with cycles or multiple paths, not in trees where only one path exists.

3. Final Answer:

   Because graphs can have cycles and multiple paths between nodes, unlike trees -> Option A

4. Quick Check:

   Graphs allow multiple paths = shortest path problem [OK]

Hint: Shortest path needs multiple paths, trees have only one [OK]

Common Mistakes:

• Thinking trees have cycles
• Assuming shortest path only applies to trees
• Confusing nodes with edges

2. Which of the following Go code snippets correctly declares a graph using adjacency lists for shortest path calculation?

easy

A. var graph [][]int = make([][]int, 0)

B. var graph []int = make([]int, 10)

C. var graph map[int][]int = make(map[int][]int)

D. var graph map[int]int = make(map[int]int)

Solution

1. Step 1: Identify correct graph representation

   Graphs for shortest path are often represented as adjacency lists: a map from node to list of connected nodes.

2. Step 2: Check each option

   var graph map[int][]int = make(map[int][]int) declares a map from int to slice of ints, perfect for adjacency lists. var graph []int = make([]int, 10) is a slice of ints, not adjacency list. var graph map[int]int = make(map[int]int) maps int to int, not adjacency list. var graph [][]int = make([][]int, 0) is a slice of slices but initialized empty, less common and incomplete.

3. Final Answer:

   var graph map[int][]int = make(map[int][]int) -> Option C

4. Quick Check:

   Adjacency list = map[int][]int [OK]

Hint: Adjacency list uses map[int][]int in Go [OK]

Common Mistakes:

• Using map[int]int instead of map[int][]int
• Confusing slice of ints with adjacency list
• Not initializing the map

3. Consider the following Go code snippet for a graph with edges and weights. What is the output of the shortest path from node 0 to node 3?

package main
import "fmt"
func main() {
    graph := map[int]map[int]int{
        0: {1: 4, 2: 1},
        1: {3: 1},
        2: {1: 2, 3: 5},
        3: {},
    }
    dist := map[int]int{0: 0, 1: 999, 2: 999, 3: 999}
    visited := map[int]bool{}
    for len(visited) < 4 {
        u := -1
        minDist := 999
        for node, d := range dist {
            if !visited[node] && d < minDist {
                minDist = d
                u = node
            }
        }
        if u == -1 {
            break
        }
        visited[u] = true
        for v, w := range graph[u] {
            if dist[u]+w < dist[v] {
                dist[v] = dist[u] + w
            }
        }
    }
    fmt.Println(dist[3])
}

medium

A. 4

B. 7

C. 5

D. 6

Solution

1. Step 1: Trace Dijkstra's algorithm steps

   Start dist: 0=0,1=999,2=999,3=999. Pick node 0 (dist 0). Update neighbors: dist[1]=4, dist[2]=1.

2. Step 2: Pick next node with smallest dist

   Pick node 2 (dist 1). Update neighbors: dist[1]=min(4,1+2=3)=3, dist[3]=min(999,1+5=6)=6.

3. Step 3: Pick next node

   Pick node 1 (dist 3). Update neighbor: dist[3]=min(6,3+1=4)=4.

4. Step 4: Pick node 3 (dist 4)

   No neighbors to update. Algorithm ends.

5. Final Answer:

   4 -> Option A

6. Quick Check:

   Shortest path dist to 3 = 4 [OK]

Hint: Follow Dijkstra stepwise updating distances [OK]

Common Mistakes:

• Not updating distances correctly
• Choosing wrong next node
• Ignoring multiple paths

4. The following Go code attempts to find the shortest path in a graph but has a bug. What is the bug?

package main
import "fmt"
func main() {
    graph := map[int][]int{
        0: {1, 2},
        1: {3},
        2: {1, 3},
        3: {},
    }
    dist := map[int]int{0: 0, 1: 999, 2: 999, 3: 999}
    visited := map[int]bool{}
    for len(visited) < 4 {
        u := -1
        minDist := 999
        for node, d := range dist {
            if !visited[node] && d < minDist {
                minDist = d
                u = node
            }
        }
        if u == -1 {
            break
        }
        visited[u] = true
        for _, v := range graph[u] {
            if dist[u]+1 < dist[v] {
                dist[v] = dist[u] + 1
            }
        }
    }
    fmt.Println(dist[3])
}

medium

A. The graph edges are unweighted but code assumes weight 1, which is correct

B. The code does not handle cycles, causing infinite loop

C. The initial distances are set incorrectly

D. The code uses a map for visited nodes but does not check if all nodes are reachable

Solution

1. Step 1: Analyze visited and loop condition

   The loop runs until all nodes are visited. But if some nodes are unreachable, the loop may break early with u == -1.

2. Step 2: Check if code handles unreachable nodes

   Since unreachable nodes remain with dist=999 and unvisited, the loop breaks early, so dist for unreachable nodes stays 999, which is expected. But the code does not explicitly handle unreachable nodes or report them.

3. Final Answer:

   The code uses a map for visited nodes but does not check if all nodes are reachable -> Option D

4. Quick Check:

   Unreachable nodes not handled = bug [OK]

Hint: Check if unreachable nodes cause early loop break [OK]

Common Mistakes:

• Thinking unweighted edges cause error
• Assuming infinite loop occurs
• Ignoring unreachable nodes

5. You have a graph with nodes representing cities and edges representing roads with distances. Some roads form cycles. Why can't you use a tree-based shortest path algorithm here, and which graph algorithm is suitable instead?

hard

A. Because trees are faster; use Depth-First Search (DFS) instead

B. Because cycles can create multiple paths; use Dijkstra's algorithm for shortest path

C. Because trees have weighted edges; use Breadth-First Search (BFS) for shortest path

D. Because graphs have no cycles; use Prim's algorithm for shortest path

Solution

1. Step 1: Understand why trees don't fit

   Trees have no cycles and only one path between nodes. Graphs with cycles can have multiple paths, so tree algorithms can't handle all cases.

2. Step 2: Identify suitable algorithm for graphs with cycles

   Dijkstra's algorithm works on graphs with weighted edges and cycles to find shortest paths efficiently.

3. Final Answer:

   Because cycles can create multiple paths; use Dijkstra's algorithm for shortest path -> Option B

4. Quick Check:

   Cycles need graph algorithm = Dijkstra [OK]

Hint: Cycles need graph algorithms like Dijkstra, not tree methods [OK]

Common Mistakes:

• Using DFS or BFS for weighted shortest path
• Confusing Prim's algorithm (for MST) with shortest path
• Assuming trees can represent cyclic graphs