DFS Depth First Search on Graph in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the DFS function parameter for the graph adjacency list.

DSA C

void dfs(int node, int visited[], vector<vector<int>> [1], int n) {

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Alist

Bgraph

Cadj

Dedges

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that is not defined or inconsistent with the rest of the code.

The adjacency list is commonly named 'adj' to represent graph connections.

2fill in blank

medium

Complete the code to mark the current node as visited inside DFS.

DSA C

visited[node] = [1];

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A0

B1

C-1

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Marking visited[node] as 0 which means unvisited.

Marking visited[node] as 1 means the node is visited.

3fill in blank

hard

Fix the error in the DFS recursive call to visit adjacent nodes.

DSA C

for(int i = 0; i < adj[node].size(); i++) {
    int adjNode = adj[node][i];
    if(!visited[[1]]) {
        dfs(adjNode, visited, adj, n);
    }
}

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AadjNode

Bnode

Ci

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Checking visited[node] instead of visited[adjNode].

Using the loop index 'i' as the node index.

We check if the adjacent node 'adjNode' is visited before calling DFS on it.

4fill in blank

hard

Fill both blanks to correctly initialize and call DFS for all nodes in the graph.

DSA C

int visited[n];
for(int i = 0; i < n; i++) {
    visited[i] = [1];
}
for(int i = 0; i < n; i++) {
    if(visited[i] == [2]) {
        dfs(i, visited, adj, n);
    }
}

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A0

B1

C-1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing visited array with 1 instead of 0.

Checking for visited[i] == 1 instead of 0.

Initialize all nodes as unvisited (0) and call DFS on nodes that are unvisited (0).

5fill in blank

hard

Fill all three blanks to complete the DFS function that prints nodes in DFS order.

DSA C

void dfs(int node, int visited[], vector<vector<int>> [1], int n) {
    visited[node] = [2];
    printf("%d ", [3]);
    for(int i = 0; i < adj[node].size(); i++) {
        int adjNode = adj[node][i];
        if(!visited[adjNode]) {
            dfs(adjNode, visited, adj, n);
        }
    }
}

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Aadj

B1

Cnode

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Printing visited[node] instead of node.

Using wrong variable name for adjacency list.

The adjacency list is 'adj', visited[node] is set to 1, and we print the current node.