Cycle Detection in Directed Graph in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the visited array to zero.

DSA C

int visited[V] = {0};
for (int i = 0; i < V; i++) {
    visited[i] = [1];
}

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A1

B0

C-1

DV

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing visited array with 1 instead of 0.

Using uninitialized values causing unpredictable behavior.

The visited array should be initialized to 0 to mark all nodes as unvisited at the start.

2fill in blank

medium

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

DSA C

void dfs(int node) {
    visited[node] = [1];
    // rest of DFS code
}

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A1

B2

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Marking visited[node] as 0 or 2 incorrectly.

Not marking the node at all.

Marking the node as 1 means it is currently being visited in the DFS path.

3fill in blank

hard

Fix the error in the condition to detect a cycle in DFS.

DSA C

for (int i = 0; i < adj[node].size(); i++) {
    int next = adj[node][i];
    if (visited[next] == [1]) {
        return 1; // cycle detected
    }
}

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A1

B2

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using wrong values like 2 or -1.

If visited[next] == 1, it means the node is in the current recursion stack, indicating a cycle.

4fill in blank

hard

Fill both blanks to correctly mark nodes after DFS call.

DSA C

visited[node] = [1];
for (int i = 0; i < adj[node].size(); i++) {
    int next = adj[node][i];
    if (visited[next] == 0 && dfs(next)) {
        return 1;
    }
}
visited[node] = [2];
return 0;

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A1

B0

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Not marking node as 2 after DFS.

Marking node as 0 instead of 1 initially.

Mark node as 1 when visiting, and 2 after all descendants are processed to indicate done.

5fill in blank

hard

Fill all three blanks to complete the cycle detection function.

DSA C

int isCyclic() {
    for (int i = 0; i < V; i++) {
        if (visited[i] == [1]) {
            if (dfs(i) == [2]) {
                return [3];
            }
        }
    }
    return 0;
}

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A0

B1

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting DFS from visited nodes.

Returning wrong values on cycle detection.

Start DFS from unvisited nodes (visited[i] == 0). If dfs returns 1 (cycle found), return 1.